3.6 Entrance Corner: Trigonometric Ratios and Identities

\(
y=f(x)=\cot x
\)
Domain \(\rightarrow R-n \pi, n \in I ;\) Range \(\rightarrow(-\infty, \infty)\); Period \(\rightarrow \pi\); 
Discontinuous at \(x=n \pi, n \in I\)
\(\cot ^2 x,|\cot x| \in[0, \infty)\)
\(
\cot x=0 \Rightarrow x=(2 n+1) \pi / 2, n \in I
\)

Case -5: \(f(x)=\sec x\)
\(
\begin{aligned}
& y=f(x)=\sec x \\
& \text { Domain } \rightarrow R \sim(2 n+1) \pi / 2, n \in I ; \text { Range } \rightarrow(-\infty,-1] \cup[1, \infty) \\
& \text { Period } \rightarrow 2 \pi, \sec ^2 x,|\sec x| \in[1, \infty)
\end{aligned}
\)

Case -6: \(f(x)=\operatorname{cosec} x\)
\(
\begin{aligned}
& y=f(x)=\operatorname{cosec} x \\
& \text { Domain } \rightarrow R \sim n \pi, n \in I ; \\
& \text { Range } \rightarrow(-\infty,-1] \cup[1, \infty) \\
& \text { Period } \rightarrow 2 \pi, \operatorname{cosec}^2 x,|\operatorname{cosec} x| \in[1, \infty)
\end{aligned}
\)

Signs of the Trigonometric Ratios or Functions

The signs of trigonometric functions depend on the quadrant in which the terminal side of the angle lies. We always take the length \(O P=r\) to be positive. Thus, \(\sin \theta=y / r\) has the \(\operatorname{sign}\) of \(y\) and \(\cos \theta=x / r\) has the sign of \(x\). The sign of \(\tan \theta\) depends on the signs of \(x\) and \(y\) and similarly the signs of other trigonometric ratios are determined by the signs of \(x\) and/or \(y\). Sign can also be determined by the graphs. Thus, we have the following:
\(
\begin{array}{|c|c|c|c|c|}
\hline \text { Function } & \begin{array}{c}
\text { 1st } \\
\text { quadrant }
\end{array} & \begin{array}{c}
\text { 2nd } \\
\text { quadrant }
\end{array} & \begin{array}{c}
\text { 3rd } \\
\text { quadrant }
\end{array} & \begin{array}{c}
\text { 4th } \\
\text { quadrant }
\end{array} \\
\hline \begin{array}{c}
\sin \theta \\
\operatorname{cosec} \theta
\end{array} & +\mathrm{ve} & +\mathrm{ve} & -\mathrm{ve} & -\mathrm{ve} \\
\hline \begin{array}{c}
\cos \theta \\
\sec \theta
\end{array} & +\mathrm{ve} & -\mathrm{ve} & -\mathrm{ve} & +\mathrm{ve} \\
\hline \begin{array}{c}
\tan \theta \\
\cot \theta
\end{array} & +\mathrm{ve} & -\mathrm{ve} & +\mathrm{ve} & -\mathrm{ve} \\
\hline
\end{array}
\)

Variations in the Values of Trigonometric Functions in Different Quadrants

\(
\begin{array}{|l|l|l|l|l|}
\hline & 1^{\text {st }} \text { quadrant } & 2^{\text {nd }} \text { quadrant } & 3^{\text {rt }} \text { quadrant } & \text { 4 }^{\text {th }} \text { quadrant } \\
\hline \sin \theta & \uparrow \text { from } 0 \text { to } 1 & \downarrow \text { from } 1 \text { to } 0 & \downarrow \text { from } 0 \text { to }-1 & \uparrow \text { from }-1 \text { to } 0 \\
\hline \cos \theta & \downarrow \text { from } 1 \text { to } 0 & \downarrow \text { from } 0 \text { to-1 } & \uparrow \text { from }-1 \text { to } 0 & \uparrow \text { from } 0 \text { to } 1 \\
\hline \tan \theta & \uparrow \text { from } 0 \text { to } \infty & \uparrow \text { from }-\infty \text { to } 0 & \uparrow \text { from } 0 \text { to } \infty & \uparrow \text { from }-\infty \text { to } 0 \\
\hline \cot \theta & \downarrow \text { from } \infty \text { to } 0 & \downarrow \text { from } 0 \text { to }-\infty & \downarrow \text { from } \infty \text { to } 0 & \downarrow \text { from } 0 \text { to }-\infty \\
\hline \sec \theta & \uparrow \text { from } 1 \text { to } \infty & \uparrow \text { from }-\infty \text { to-1 } & \downarrow-1 \text { to }-\infty & \downarrow \text { from } \infty \text { to } 1 \\
\hline \operatorname{cosec} \theta & \downarrow \text { from } \infty \text { to } 1 & \uparrow \text { from } 1 \text { to } \infty & \uparrow \text { from }-\infty \text { to -1 } & \downarrow \text { from }-1 \text { to }-\infty \\
\hline
\end{array}
\)

Note: \(+\infty\) and \(-\infty\) are two symbols. These are not real numbers. When we say’ that tan \(\theta\) increases from 0 to \(\infty\) as \(\theta\) varies from 0 to \(\pi / 2\), it means that tan \(\theta\) increases in the interval \((0, \pi / 2)\) and it attains arbitrarily large positive values as \(\theta\) tends to \(\pi / 2\). Similarly, this happens for other trigonometrical functions as well.

\(
\begin{aligned}
&\text { Trigonometric Ratios of Standard Angles }\\
&\begin{array}{|c|c|c|c|}
\hline \begin{array}{c}
\text { Angle }(\boldsymbol{\theta}) \rightarrow \\
\text { T-Ratio } \downarrow
\end{array} & \mathbf{3 0 ^ { \circ }} & \mathbf{4 5 ^ { \circ }} & \mathbf{6 0 ^ { \circ }} \\
\hline \sin \theta & 1 / 2 & 1 / \sqrt{2} & \sqrt{3} / 2 \\
\hline \cos \theta & \sqrt{3} / 2 & 1 / \sqrt{2} & 1 / 2 \\
\hline \tan \theta & 1 / \sqrt{3} & 1 & \sqrt{3} \\
\hline \operatorname{cosec} \theta & 2 & \sqrt{2} & 2 / \sqrt{3} \\
\hline \sec \theta & 2 / \sqrt{3} & \sqrt{2} & 2 \\
\hline \cot \theta & \sqrt{3} & 1 & 1 / \sqrt{3} \\
\hline
\end{array}
\end{aligned}
\)

Trigonometric Ratios of any Angle

\(
\begin{array}{ll}
\sin (-\theta)=-\sin \theta & \cos (-\theta)=\cos \theta \\
\sin \left(\frac{\pi}{2}-\theta\right)=\cos \theta & \cos \left(\frac{\pi}{2}-\theta\right)=\sin \theta \\
\tan (-\theta)=-\tan \theta & \cot (-\theta)=-\cot \theta \\
\tan \left(\frac{\pi}{2}-\theta\right)=\cot \theta & \cot \left(\frac{\pi}{2}-\theta\right)=\tan \theta
\end{array}
\)
\(
\begin{array}{ll}
\operatorname{cosec}(-\theta)=-\operatorname{cosec} \theta & \sec (-\theta)=\sec \theta \\
\sec \left(\frac{\pi}{2}-\theta\right)=\operatorname{cosec} \theta & \operatorname{cosec}\left(\frac{\pi}{2}-\theta\right)=\sec \theta \\
\sin \left(\frac{\pi}{2}+\theta\right)=\cos \theta & \cos \left(\frac{\pi}{2}+\theta\right)=-\sin \theta \\
\sin (\pi-\theta)=\sin \theta & \cos (\pi-\theta)=-\cos \theta
\end{array}
\)
\(
\begin{array}{ll}
\tan \left(\frac{\pi}{2}+\theta\right)=-\cot \theta & \cot \left(\frac{\pi}{2}+\theta\right)=-\tan \theta \\
\tan (\pi-\theta)=-\tan \theta & \cot (\pi-\theta)=-\cot \theta \\
\sec \left(\frac{\pi}{2}+\theta\right)=-\operatorname{cosec} \theta & \operatorname{cosec}\left(\frac{\pi}{2}+\theta\right)=\sec \theta \\
\sec (\pi-\theta)=-\sec \theta & \operatorname{cosec}(\pi-\theta)=\operatorname{cosec} \theta \\
\sin (\pi+\theta)=-\sin \theta & \cos (\pi+\theta)=-\cos \theta \\
\sin \left(\frac{3 \pi}{2}-\theta\right)=-\cos \theta & \cos \left(\frac{3 \pi}{2}-\theta\right)=-\sin \theta
\end{array}
\)
\(
\begin{array}{ll}
\tan (\pi+\theta)=\tan \theta & \cot (\pi+\theta)=\cot \theta \\
\tan \left(\frac{3 \pi}{2}-\theta\right)=\cot \theta & \cot \left(\frac{3 \pi}{2}-\theta\right)=\tan \theta \\
\sec (\pi+\theta)=-\sec \theta & \operatorname{cosec}(\pi+\theta)=-\operatorname{cosec} \theta \\
\sec \left(\frac{3 \pi}{2}-\theta\right)=-\operatorname{cosec} \theta & \operatorname{cosec}\left(\frac{3 \pi}{2}-\theta\right)=-\sec \theta \\
\sin \left(\frac{3 \pi}{2}+\theta\right)=-\cos \theta & \cos \left(\frac{3 \pi}{2}+\theta\right)=\sin \theta \\
\sin (2 \pi-\theta)=-\sin \theta & \cos (2 \pi-\theta)=\cos \theta \\
\tan \left(\frac{3 \pi}{2}+\theta\right)=-\cot \theta & \cot \left(\frac{3 \pi}{2}+\theta\right)=-\tan \theta
\end{array}
\)
\(
\begin{array}{ll}
\tan (2 \pi-\theta)=-\tan \theta & \cot (2 \pi-\theta)=-\cot \theta \\
\sec \left(\frac{3 \pi}{2}+\theta\right)=\operatorname{cosec} \theta & \operatorname{cosec}\left(\frac{3 \pi}{2}+\theta\right)=-\sec \theta \\
\sec (2 \pi-\theta)=\sec \theta & \operatorname{cosec}(2 \pi-\theta)=-\operatorname{cosec} \theta \\
\sin (2 \pi+\theta)=\sin \theta & \cos (2 \pi+\theta)=\cos \theta \\
\tan (2 \pi+\theta)=\tan \theta & \cot (2 \pi+\theta)=\cot \theta \\
\sec (2 \pi+\theta)=\sec \theta & \operatorname{cosec}(2 \pi+\theta)=\operatorname{cosec} \theta
\end{array}
\)

Transformation of the Graphs of Trigonometric Functions

• To draw the graph of \(y=f(x+a) ;(a>0)\) from the graph of \(y=f(x)\), shift the graph of \(y=f(x), a\) units left along the \(x\)-axis. 

• To draw the graph of \(y=f(x-a) ;(a>0)\) from the graph of \(y=f(x)\), shift the graph of \(y=f(x), a\) units right along the \(x\)-axis.

• To draw the graph of \(y=f(x)+a ;(a>0)\) from the graph of \(y=f(x)\), shift the graph of \(y=f(x), a\) units upward along the \(y\)-axis..
  To draw the graph of \(y=f(x)-a ;(a>0)\) from the graph of \(y=f(x)\), shift the graph of \(y=f(x), a\) units downward along the \(y\)-axis.

• If \(y=f(x)\) has period \(T\), then period of \(y=f(a x)\) is \(T /|a|\).
  Period of \(y=\sin (2 x)\) is \(2 \pi / 2=\pi\)

• Period of \(y=\sin (x / 2)\) is \(2 \pi /(1 / 2)=4 \pi\)

• Since \(y=|f(x)| \geq 0\), to draw the graph of \(y=|f(x)|\), take the mirror of the graph of \(y=f(x)\) in \(x\)-axis for \(f(x)<0\), retaining the graph for \(f(x)>0\). Here period of \(f(x)=|\sin x|\) is \(\pi\).

• Here period of \(f(x)=|\cos x|\) is \(\pi\)

• Graph of \(y=a f(x)\) from the graph of \(y f(x)\)

Example 2.11: If \(f(\theta)=4 \sin \theta+\cos ^2 \theta\), then what is maximum and minimum value of \(f(\theta)\)?

Solution:

\(
\begin{aligned}
f(\theta) & =4 \sin \theta+\cos ^2 \theta=4 \sin \theta+1-\sin ^2 \theta \\
& =5-\left(4-4 \sin \theta+\sin ^2 \theta\right)=5-(\sin \theta-2)^2
\end{aligned}
\)
Now maximum value of \(f(\theta)\) occurs when \((\sin \theta-2)^2\) is minimum.
Minimum value of \((\sin \theta-2)^2\) occurs when \(\sin \theta=1\), then maximum value of \(f(\theta)\) is \(5-(1-2)^2=4\).
Also minimum value of \(f(\theta)\) occurs when \((\sin \theta-2)^2\) is maximum.
Maximum value of \((\sin \theta-2)^2\) occurs when \(\sin \theta=-1\), then minimum value of \(f(\theta)\) is \(5-(-1-2)^2=-4\).

Example 2.12: Is the equation \(\sec ^2 \theta=\frac{4 x y}{(x+y)^2}\) possible for real values of \(x\) and \(y\)?

Solution: Given, \(\sec ^2 \theta=\frac{4 x y}{(x+y)^2}\)
Since \(\sec ^2 \theta \geq 1\), we get \(\frac{4 x y}{(x+y)^2} \geq 1\)
\(
\begin{aligned}
& \Rightarrow \quad(x+y)^2 \leq 4 x y \\
& \Rightarrow \quad(x+y)^2-4 x y \leq 0 \text { or }(x-y)^2 \leq 0
\end{aligned}
\)
But for real values of \(x\) and \(y,(x-y)^2 \geq 0\)
Since \((x-y)^2=0, x=y\). Also \(x+y \neq 0 \Rightarrow x \neq 0, y \neq 0\)
Therefore, the given equation \(\sec ^2 \theta=\frac{4 x y}{(x+y)^2}\) is possible for real values of \(x\) and \(y\) only when \(x=y(x \neq 0)\).

Example 2.13: Show that the equation \(\sin \theta=x+\frac{1}{x}\) is impossible if \(x\) is real.

Solution: Given, \(\sin \theta=x+\frac{1}{x}\)
\(
\therefore \quad \sin ^2 \theta=x^2+\frac{1}{x^2}+2 x \frac{1}{x}=x^2+\frac{1}{x^2}+2=\left(x-\frac{1}{x}\right)^2+4 \geq 4
\)
which is not possible since \(\sin \theta^2 \leq 1\).

Example 2.14: If \(\sin ^2 \theta_1+\sin ^2 \theta_2+\sin ^2 \theta_3=0\), then which of the following is not the possible value of \(\cos \theta_1\) \(+\cos \theta_2+\cos \theta_3\)
(a) 3                (b) -3             (c) -1              (d) -2

Solution:
\(
\begin{aligned}
& \sin ^2 \theta_1+\sin ^2 \theta_2+\sin ^2 \theta_3=0 \\
& \Rightarrow \sin ^2 \theta_1=\sin ^2 \theta_2=\sin ^2 \theta_3=0 \quad \Rightarrow \quad \cos ^2 \theta_1, \cos ^2 \theta_2, \cos ^2 \theta_3=1 \quad \Rightarrow \quad \cos \theta_1, \cos \theta_2, \cos \theta_3= \pm 1 \\
& \cos \theta_1+\cos \theta_2+\cos \theta_3 \text { can be }-3 \text { (when all are }-1 \text { ) } \\
& \text { or } 3 \quad \text { (when all are }+1 \text { ) } \\
& \text { or }-1 \quad \text { (when any two are }-1 \text { and one }+1 \text { ) } \\
& \text { or } 1 \quad \text { (when any two are }+1 \text { and one }-1 \text { ) } \\
& \text { but }-2 \text { is not a possible value. }
\end{aligned}
\)

Example 2.15: Find the range of \(f(x)=\frac{1}{4 \cos x-3}\).

Solution:
\(
\begin{aligned}
& -1 \leq \cos x \leq 1 \\
& \Rightarrow-4 \leq 4 \cos x \leq 4 \\
& \Rightarrow-7 \leq 4 \cos x-3 \leq 1
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow-7 \leq 4 \cos x-3<0 \text { or } 0<4 \cos x-3 \leq 1 \quad(\because 4 \cos x-3 \neq 0) \\
& \Rightarrow-\frac{1}{7} \geq \frac{1}{4 \cos x-3}>-\infty \text { or } \infty>\frac{1}{4 \cos x-3} \geq 1 \\
& \Rightarrow \frac{1}{4 \cos x-3} \in\left(-\infty,-\frac{1}{7}\right] \cup[1, \infty)
\end{aligned}
\)

Example 2.16: Find the range of \(f(x)=\frac{1}{5 \sin x-6}\).

Solution:
\(
\begin{aligned}
& -1 \leq \sin x \leq 1 \\
& \Rightarrow-5 \leq 5 \sin x \leq 5 \\
& \Rightarrow-11 \leq 5 \sin x-6 \leq-1 \\
& \Rightarrow-1 \leq \frac{1}{5 \sin x-6} \leq-1 / 11 \\
& \Rightarrow \frac{1}{5 \sin x-6} \in[-1,-1 / 11]
\end{aligned}
\)

Example 2.17: Find the range of \(f(x)=\cos ^2 x+\sec ^2 x\)

Solution: We have
\(
\begin{aligned}
f(x) & =\cos ^2 x+\sec ^2 x \\
& =(\cos x-\sec x)^2+2 \cos x \sec x \\
& =2+(\cos x-\sec x)^2 \geq 2
\end{aligned}
\)

Example 2.18: Find the range of \(f(x)=\sin ^2 x-3 \sin x+2\)

Solution:
\(
\begin{aligned}
f(x)= & \sin ^2 x-3 \sin x+2 \\
= & (\sin x-3 / 2)^2+2-9 / 4 \\
= & (\sin x-3 / 2)^2-1 / 4 \\
& -1 \leq \sin x \leq 1 \\
\Rightarrow & -5 / 2 \leq \sin x-3 / 2 \leq-1 / 2 \\
\Rightarrow & 1 / 4 \leq(\sin x-3 / 2)^2 \leq 25 / 4 \\
\Rightarrow & 0 \leq(\sin x-3 / 2)^2-1 / 4 \leq 6
\end{aligned}
\)

Example 2.19: Find the range of \(f(x)=\sqrt{\sin ^2 x-6 \sin x+9}+3\).

Solution:
\(
\begin{aligned}
& \begin{aligned}
f(x) & =\sqrt{\sin ^2 x-6 \sin x+9}+3 \\
& =\sqrt{(\sin x-3)^2}+3 \\
& =|\sin x-3|+3
\end{aligned} \\
& \text { Now }-1 \leq \sin x \leq 1 \\
& \Rightarrow-4 \leq \sin x-3 \leq-2 \\
& \Rightarrow 2 \leq|\sin x-3| \leq 4 \\
& \Rightarrow 5 \leq|\sin x-3|+3 \leq 7
\end{aligned}
\)

Example 2.20: Find the range of \(f(x)=\operatorname{cosec}^2 x+25 \sec ^2 x\).

Solution:
\(
\begin{aligned}
f(x) & =\left(1+\cot ^2 x\right)+25\left(1+\tan ^2 x\right) \\
& =26+\cot ^2 x+25 \tan ^2 x \\
& =36+10+\left(\cot ^2 x+25 \tan ^2 x-2 \cot x 5 \tan x\right) \\
& =36+(\cot x-5 \tan x)^2 \geq 36
\end{aligned}
\)

Example 2.21: Find the value of \(x\) for which \(f(x)=\sqrt{\sin x-\cos x}\) is defined, \(x \in[0,2 \pi]\).

Solution: \(f(x)=\sqrt{\sin x-\cos x}\) is defined if \(\sin x \geq \cos x\)

Problems Based On Trigonometric Identities

Example 2.22: Show that \(2\left(\sin ^6 x+\cos ^6 x\right)-3\left(\sin ^4 x+\cos ^4 x\right)+1=0\).

Solution:
\(
\begin{aligned}
& 2\left(\sin ^6 x+\cos ^6 x\right)-3\left(\sin ^4 x+\cos ^4 x\right)+1 \\
= & 2\left[\left(\sin ^2 x\right)^3+\left(\cos ^2 x\right)^3\right]-3\left(\sin ^4 x+\cos ^4 x\right)+1=2\left[\left(\sin ^2 x+\cos ^2 x\right)^3-3 \sin ^2 x \cos ^2 x\left(\sin ^2 x+\cos ^2 x\right)\right] \\
& -3\left[\left(\sin ^2 x+\cos ^2 x\right)^2-2 \sin ^2 x \cos ^2 x\right]+1=2\left[1-3 \sin ^2 x \cos ^2 x\right]-3\left[1-2 \sin ^2 x \cos ^2 x\right]+1=0
\end{aligned}
\)

Example 2.23: If \(3 \sin \theta+5 \cos \theta=5\), then show that \(5 \sin \theta-3 \cos \theta= \pm 3\).

Solution: Given, \(3 \sin \theta+5 \cos \theta=5\)
Let \(5 \sin \theta-3 \cos \theta=x\)
By squaring and adding, we get
\(
\begin{aligned}
& \left(9 \sin ^2 \theta+25 \cos ^2 \theta+30 \sin \theta \cos \theta\right)+\left(25 \sin ^2 \theta+9 \cos ^2 \theta-30 \sin \theta \cos \theta\right)=25+x^2 \\
& \Rightarrow \quad 9\left(\sin ^2 \theta+\cos ^2 \theta\right)+25\left(\sin ^2 \theta+\cos ^2 \theta\right)=25+x^2 \\
& \Rightarrow \quad 34=25+x^2 \text { or } x^2=9
\end{aligned}
\)
\(
\Rightarrow \quad x= \pm 3
\)

Example 2.24: If \((\sec A+\tan A)(\sec B+\tan B)(\sec C+\tan C)=(\sec A-\tan A)(\sec B-\tan B)(\sec C-\tan C)\), prove that the value of each side is \(\pm 1\).

Solution:
\(
\begin{aligned}
& \text { Let }(\sec A+\tan A)(\sec B+\tan B)(\sec C+\tan C)=x \dots(i) \\
& (\sec A-\tan A)(\sec B-\tan B)(\sec C-\tan C)=x \dots(ii)
\end{aligned}
\)
Multiplying Eqs. (i) and (ii), we get.
\(
\begin{aligned}
& \left(\sec ^2 A-\tan ^2 A\right)\left(\sec ^2 B-\tan ^2 B\right)\left(\sec ^2 C-\tan ^2 C\right)=x^2 \\
& \text { or } x^2=1 \\
& \therefore x= \pm 1
\end{aligned}
\)
Hence, each side is equal to \(\pm 1\).

Example 2.25: If \(\tan \theta+\sec \theta=1.5\), find \(\sin \theta, \tan \theta\), and \(\sec \theta\).

Solution: Given, \(\sec \theta+\tan \theta=\frac{3}{2}\)

Now, \(\sec \theta-\tan \theta=\frac{1}{\sec \theta+\tan \theta}=\frac{2}{3}\)
Adding Eqs. (i) and (ii), we get \(2 \sec \theta=\frac{3}{2}+\frac{2}{3}=\frac{13}{6}\)
\(\therefore \sec \theta=\frac{13}{12}\)
\(\therefore \quad \tan \theta=\frac{5}{12}\)
and \(\sin \theta=\frac{5}{13}\)

Example 2.26: If \(x=\sec \theta-\tan \theta\) and \(y=\operatorname{cosec} \theta+\cot \theta\), then prove that \(x y+1=y-x\).

Solution:
\(
\begin{aligned}
x y+1 & =\frac{1-\sin \theta}{\cos \theta} \frac{1+\cos \theta}{\sin \theta}+1=\frac{1-\sin \theta+\cos \theta}{\sin \theta \cos \theta} \\
& =\frac{\left(\sin ^2 \theta+\cos ^2 \theta\right)}{\sin \theta \cos \theta}-\frac{(\sin \theta-\cos \theta)}{\sin \theta \cos \theta} \\
& =(\tan \theta+\cot \theta)-(\sec \theta-\operatorname{cosec} \theta) \\
& =(\operatorname{cosec} \theta+\cot \theta)-(\sec \theta-\tan \theta)=y-x
\end{aligned}
\)

Trigonometric Ratios Of Complementary and Supplementary Angles

\(
\begin{aligned}
& \sin (-\theta)=-\sin \theta \\
& \cos (-\theta)=\cos \theta
\end{aligned}
\)
\(
\begin{aligned}
&\sin \left(90^{\circ}-\theta\right)=\cos \theta\\
&\cos \left(90^{\circ}-\theta\right)=\sin \theta
\end{aligned}
\)
\(
\begin{aligned}
& \sin \left(90^{\circ}+\theta\right)=\cos \theta \\
& \cos \left(90^{\circ}+\theta\right)=-\sin \theta
\end{aligned}
\)
\(
\begin{aligned}
& \sin \left(180^{\circ}-\theta\right)=\sin \theta \\
& \cos \left(180^{\circ}-\theta\right)=-\cos \theta
\end{aligned}
\)
\(
\begin{aligned}
& \sin \left(180^{\circ}+\theta\right)=-\sin \theta \\
& \cos \left(180^{\circ}+\theta\right)=-\cos \theta
\end{aligned}
\)

Since the terminal sides of co-terminal angles coincide, hence their trigonometrical ratios are the same. Clearly, \(360^{\circ}-\theta\) and \(-\theta\) are coterminal angles.
Therefore, \(\sin \left(360^{\circ}-\theta\right)=\sin (-\theta)=-\sin \theta, \cos \left(360^{\circ}-\theta\right)=\cos (-\theta)=\cos \theta\), and \(\tan \left(360^{\circ}-\theta\right)=\tan (-\theta)\) \(=-\tan \theta\).
Similarly, \(\operatorname{cosec}\left(360^{\circ}-\theta\right)=-\operatorname{cosec} \theta, \sec \left(360^{\circ}-\theta\right)=\sec \theta\) and \(\cot \left(360^{\circ}-\theta\right)=-\cot \theta\).

Also \(\theta\) and \(360^{\circ}+\theta\) are co-terminal angles. Therefore, \(\sin \left(360^{\circ}+\theta\right)=\sin \theta, \cos \left(360^{\circ}+\theta\right)=\cos \theta, \tan \left(360^{\circ}\right.\) \(+\theta)=\tan \theta, \sec \left(360^{\circ}+\theta\right)=\sec \theta, \operatorname{cosec}\left(360^{\circ}+\theta\right)=\operatorname{cosec} \theta\) and \(\cot \left(360^{\circ}+\theta\right)=\cot \theta\).

In fact, for any positive integer \(n,\left(360^{\circ} \times n+\theta\right)\) is co-terminal to \(\theta\). Therefore, for any positive integer \(n\), we have \(\sin \left(360^{\circ} \times n+\theta\right)=\sin \theta, \cos \left(360^{\circ} \times n+\theta\right)=\cos \theta, \tan \left(360^{\circ} \times n+\theta\right)=\tan \theta, \operatorname{cosec}\left(360^{\circ} \times n+\theta\right)=\operatorname{cosec} \theta\), \(\sec \left(360^{\circ} \times n+\theta\right)=\sec \theta\) and \(\cot \left(360^{\circ} \times n+\theta\right)=\cot \theta\).

Example 2.27: Prove that \(\sin \left(-420^{\circ}\right)\left(\cos 390^{\circ}\right)+\cos \left(-660^{\circ}\right)\left(\sin 330^{\circ}\right)=-1\).

Solution:
\(
\begin{aligned}
\text { L.H.S. } & =\sin \left(-420^{\circ}\right)\left(\cos 390^{\circ}\right)+\cos \left(-660^{\circ}\right)\left(\sin 330^{\circ}\right) \\
& =-\sin 420^{\circ} \cos 390^{\circ}+\cos 660^{\circ} \sin 330^{\circ}
\end{aligned}
\)
\(
=-\sin 420^{\circ} \cos 390^{\circ}+\cos 660^{\circ} \sin 330^{\circ} \quad[\because \sin (-\theta)=-\sin \theta, \cos (-\theta)=\cos \theta]
\)
\(
\begin{aligned}
& =-\sin \left(90^{\circ} \times 4+60^{\circ}\right) \cos \left(90^{\circ} \times 4+30^{\circ}\right)+\cos \left(90^{\circ} \times 7+30^{\circ}\right) \sin \left(90^{\circ} \times 3+60^{\circ}\right) \\
& =-\left(\sin 60^{\circ}\right)\left(\cos 30^{\circ}\right)+\left(\sin 30^{\circ}\right)\left(-\cos 60^{\circ}\right) \\
& =-\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}+\frac{1}{2}\left(-\frac{1}{2}\right)=-1=\text { R.H.S. }
\end{aligned}
\)

Example 2.28: Prove that \(\frac{\cos \left(90^{\circ}+\theta\right) \sec (-\theta) \tan \left(180^{\circ}-\theta\right)}{\sec \left(360^{\circ}-\theta\right) \sin \left(180^{\circ}+\theta\right) \cot \left(90^{\circ}-\theta\right)}=-1\).

Solution:
L.H.S \(=\frac{\cos \left(90^{\circ}+\theta\right) \sec (-\theta) \tan \left(180^{\circ}-\theta\right)}{\sec \left(360^{\circ}-\theta\right) \sin \left(180^{\circ}+\theta\right) \cot \left(90^{\circ}-\theta\right)}=\frac{(-\sin \theta)(\sec \theta)(-\tan \theta)}{(\sec \theta)(-\sin \theta)(\tan \theta)}=-1=\) R.H.S.

Example 2.29: If \(A, B, C, D\) are angles of a cyclic quadrilateral, then prove that
\(
\cos A+\cos B+\cos C+\cos D=0
\)

Solution: We know that the opposite angles of a cyclic quadrilateral are supplementary, i.e., \(A+C=\pi\) and \(B+D=\pi\).
\(
\begin{aligned}
& \therefore \quad A=\pi-C \text { and } B=\pi-D \\
& \Rightarrow \quad \cos A=\cos (\pi-C)=-\cos C \\
& \text { and } \cos B=\cos (\pi-D)=-\cos D \\
& \therefore \quad \cos A+\cos B+\cos C+\cos D=-\cos C-\cos D+\cos C+\cos D=0
\end{aligned}
\)

Example 2.30: Show that \(\tan 1^{\circ} \tan 2^{\circ} \ldots \tan 89^{\circ}=1\).

Solution:
\(
\begin{aligned}
\text { L.H.S. } & =\left(\tan 1^{\circ} \tan 89^{\circ}\right)\left(\tan 2^{\circ} \tan 88^{\circ}\right) \ldots\left(\tan 44^{\circ} \tan 46^{\circ}\right) \tan 45^{\circ} \\
& =\left[\tan 1^{\circ} \tan \left(90^{\circ}-1^{\circ}\right)\right]\left[\tan 2^{\circ} \tan \left(90^{\circ}-2^{\circ}\right)\right] \ldots\left[\tan 44^{\circ} \tan \left(90^{\circ}-44^{\circ}\right)\right] \tan 45^{\circ} \\
& =\left(\tan 1^{\circ} \cot 1^{\circ}\right)\left(\tan 2^{\circ} \cot 2^{\circ}\right) \ldots\left(\tan 44^{\circ} \cot 44^{\circ}\right) \tan 45^{\circ} \\
& =1 \quad\left[\because \tan \theta \cot \theta=1 \text { and } \tan 45^{\circ}=1\right]
\end{aligned}
\)

Example 2.31: Show that \(\sin ^2 5^{\circ}+\sin ^2 10^{\circ}+\sin ^2 15^{\circ}+\cdots+\sin ^2 90^{\circ}=9 \frac{1}{2}\).

Solution:
\(
\begin{aligned}
\text { L.H.S. } & =\left(\sin ^2 5^{\circ}+\sin ^2 85^{\circ}\right)+\left(\sin ^2 10^{\circ}+\sin ^2 80^{\circ}\right)+\cdots+\left(\sin ^2 40^{\circ}+\sin ^2 50^{\circ}\right)+\sin ^2 45^{\circ}+\sin ^2 90^{\circ} \\
& =\left(\sin ^2 5^{\circ}+\cos ^2 5^{\circ}\right)+\left(\sin ^2 10^{\circ}+\cos ^2 10^{\circ}\right)+\cdots+\left(\sin ^2 40^{\circ}+\cos ^2 40^{\circ}\right)+\sin ^2 45^{\circ}+\sin ^2 90^{\circ} \\
& =(1+1+1+1+1+1+1+1)+\left(\frac{1}{\sqrt{2}}\right)^2+1=9 \frac{1}{2}
\end{aligned}
\)

Example 2.32: Find the value of \(\cos ^2 \frac{\pi}{16}+\cos ^2 \frac{3 \pi}{16}+\cos ^2 \frac{5 \pi}{16}+\cos ^2 \frac{7 \pi}{16}\).

Solution:
\(
\begin{aligned}
\text { L.H.S. } & =\cos ^2 \frac{\pi}{16}+\cos ^2 \frac{3 \pi}{16}+\cos ^2\left(\frac{\pi}{2}-\frac{3 \pi}{16}\right)+\cos ^2\left(\frac{\pi}{2}-\frac{\pi}{16}\right) \\
& =\cos ^2 \frac{\pi}{16}+\cos ^2 \frac{3 \pi}{16}+\sin ^2 \frac{3 \pi}{16}+\sin ^2 \frac{\pi}{16} \\
& =\left(\cos ^2 \frac{\pi}{16}+\sin ^2 \frac{\pi}{16}\right)+\left(\cos ^2 \frac{3 \pi}{16}+\sin ^2 \frac{3 \pi}{16}\right) \\
& =1+1=2
\end{aligned}
\)

Example 2.33: If \(\sin \left(120^{\circ}-\alpha\right)=\sin \left(120^{\circ}-\beta\right), 0<\alpha, \beta<\pi\), then find the relation between \(\alpha\) and \(\beta\).

Solution:
\(
\begin{aligned}
& \text { If } \sin A=\sin B \text {, where } A=120^{\circ}-\alpha \text { and } B=120^{\circ}-\beta \\
& \Rightarrow A=B \text { or } A=\pi-B \text {, i.e., } A+B=\pi \\
& \Rightarrow 120^{\circ}-\alpha=120^{\circ}-\beta \text {, or } 120^{\circ}-\alpha+120^{\circ}-\beta=180^{\circ} \\
& \Rightarrow \alpha=\beta \text { or } \alpha+\beta=60^{\circ}
\end{aligned}
\)

Trigonometric Rations Of Compound Angles

Cosine of the Difference and Sum of Two Angles
1. \(\cos (A-B)=\cos A \cos B+\sin A \sin B\)
2. \(\cos (A+B)=\cos A \cos B-\sin A \sin B\)
for all angles \(A\) and \(B\).

Sine of the Difference and Sum of Two Angles
1. \(\sin (A-B)=\sin A \cos B-\cos A \sin B\)
2. \(\sin (A+B)=\sin A \cos B+\cos A \sin B\)

Tangent of the Difference and Sum of Two Angles
1. \(\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}\)
2. \(\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}\)

Similarly, it can be proved that
\(
\begin{aligned}
\cot (A+B) & =\frac{\cot A \cot B-1}{\cot B+\cot A} \\
\text { and } \cot (A-B) & =\frac{\cot A \cot B+1}{\cot B-\cot A}
\end{aligned}
\)

Some More Results
1. \(\sin (A+B) \sin (A-B)=\sin ^2 A-\sin ^2 B=\cos ^2 B-\cos ^2 A\)
2. \(\cos (A+B) \cos (A-B)=\cos ^2 A-\sin ^2 B=\cos ^2 B-\sin ^2 A\)
3. \(\sin (A+B+C)=\sin A \cos B \cos C+\cos A \sin B \cos C+\cos A \cos B \sin C-\sin A \sin B \sin C\)
4. \(\cos (A+B+C)=\cos A \cos B \cos C-\cos A \sin B \sin C-\sin A \cos B \sin C-\sin A \sin B \cos C\)
5. \(\tan (A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A \tan B \tan C}{1-\tan A \tan B-\tan B \tan C-\tan C \tan A}\)

Example 2.34: Prove that \(\frac{\sin (B-C)}{\cos B \cos C}+\frac{\sin (C-A)}{\cos C \cos A}+\frac{\sin (A-B)}{\cos A \cos B}=0\).

Solution: First term of L.H.S. is
\(
\frac{\sin (B-C)}{\cos B \cos C}=\frac{\sin B \cos C-\cos B \sin C}{\cos B \cos C}=\frac{\sin B \cos C}{\cos B \cos C}-\frac{\cos B \sin C}{\cos B \cos C}=\tan B-\tan C
\)
Similarly, second term of L.H.S. \(=\tan C-\tan A\)
and, third term of L.H.S. \(=\tan A-\tan B\)
Now L.H.S. \(=(\tan B-\tan C)+(\tan C-\tan A)+(\tan A-\tan B)=0\).

Example 2.35: If \(\sin \alpha \sin \beta-\cos \alpha \cos \beta+1=0\), then prove that \(1+\cot \alpha \tan \beta=0\).

Solution: Given, \(3 \tan \theta \tan \varphi=1\) or \(\cot \theta \cot \varphi=3\)
or, \(\frac{\cos \theta \cos \varphi}{\sin \theta \sin \varphi}=\frac{3}{1}\)
By componendo and dividendo, we get
\(
\begin{aligned}
& \frac{\cos \theta \cos \varphi+\sin \theta \sin \varphi}{\cos \theta \cos \varphi-\sin \theta \sin \varphi}=\frac{3+1}{3-1} \\
& \Rightarrow \frac{\cos (\theta-\varphi)}{\cos (\theta+\varphi)}=2 \\
& \Rightarrow 2 \cos (\theta+\varphi)=\cos (\theta-\varphi)
\end{aligned}
\)

Example 2.36: If \(\sin (A-B)=\frac{1}{\sqrt{10}}, \cos (A+B)=\frac{2}{\sqrt{29}}\), find the value of \(\tan 2 A\) where \(A\) and \(B\) lie between 0 and \(\pi / 4\).

Solution: \(\tan 2 A=\tan [(A+B)+(A-B)]=\frac{\tan (A+B)+\tan (A-B)}{1-\tan (A+B) \tan (A-B)} \dots(i)\)
Given that, \(0<A<\frac{\pi}{4}\) and \(0<B<\frac{\pi}{4}\)
\(
\begin{aligned}
& \therefore \quad 0<A+B<\frac{\pi}{2} \\
& \text { Also, }-\frac{\pi}{4}<A-B<\frac{\pi}{4} \text { and } \sin (A-B)=\frac{1}{\sqrt{10}}=(+) \text { ve } \\
& \therefore \quad 0<A-B<\frac{\pi}{4} \\
& \text { Now, } \sin (A-B)=\frac{1}{\sqrt{10}} \\
& \Rightarrow \quad \tan (A-B)=\frac{1}{3} \dots(ii)
\end{aligned}
\)
\(
\begin{aligned}
&\begin{aligned}
\cos (A+B) & =\frac{2}{\sqrt{29}} \\
\Rightarrow \quad \tan (A+B) & =\frac{5}{2} \dots(iii)
\end{aligned}\\
&\text { From Eqs. (i), (ii) and (iii), we get }\\
&\tan 2 A=\frac{\frac{5}{2}+\frac{1}{3}}{1-\frac{5}{2} \times \frac{1}{3}}=\frac{17}{6} \times \frac{6}{1}=17
\end{aligned}
\)

Example 2.37: Prove that \(\frac{\cos 10^{\circ}+\sin 10^{\circ}}{\cos 10^{\circ}-\sin 10^{\circ}}=\tan 55^{\circ}\).

Solution: 

\(
\frac{\cos 10^{\circ}+\sin 10^{\circ}}{\cos 10^{\circ}-\sin 10^{\circ}}=\frac{1+\tan 10^{\circ}}{1-\tan 10^{\circ}}=\frac{\tan 45^{\circ}+\tan 10^{\circ}}{1-\tan 45^{\circ} \tan 10^{\circ}}=\tan \left(45^{\circ}+10^{\circ}\right)=\tan 55^{\circ} \quad\left(\text { dividing by } \cos 10^{\circ}\right)
\)

Example 2.38: \(\text { Prove that } \tan 70^{\circ}=2 \tan 50^{\circ}+\tan 20^{\circ} \text {. }\)

Solution: \(\tan 70^{\circ}=\tan \left(50^{\circ}+20^{\circ}\right)=\frac{\tan 50^{\circ}+\tan 20^{\circ}}{1-\tan 50^{\circ} \tan20^{\circ}}\)
\(
\begin{aligned}
& \tan 70^{\circ}\left(1-\tan 50^{\circ} \tan 20^{\circ}\right)=\tan 50^{\circ}+\tan 20^{\circ} \\
& \tan 70^{\circ}-\tan 50^{\circ} \tan 20^{\circ} \tan 70^{\circ}=\tan 50^{\circ}+\tan 20^{\circ} \\
& \begin{aligned}
\tan 70^{\circ} & =\tan 70^{\circ} \tan 50^{\circ} \tan 20^{\circ}+\tan 50^{\circ}+\tan 20^{\circ} \\
& =\tan \left(90^{\circ}-20^{\circ}\right) \tan 50^{\circ} \tan 20^{\circ}+\tan 50^{\circ}+\tan 20^{\circ} \\
& =\cot 20^{\circ} \tan 50^{\circ} \tan 20^{\circ}+\tan 50^{\circ}+\tan 20^{\circ} \\
& =\tan 50^{\circ}+\tan 50^{\circ}+\tan 20^{\circ}=2 \tan 50^{\circ}+\tan 20^{\circ}
\end{aligned}
\end{aligned}
\)

Example 2.39: Let \(A, B, C\) be three angles such that \(A+B+C=\pi\). If \(\tan A \cdot \tan B=2\). Then find the value of \(\frac{\cos A \cos B}{\cos C}\).

Solution: Given \(\tan A \cdot \tan B=2\)
\(
\begin{aligned}
\text { Let } y & =\frac{\cos A \cos B}{\cos C}=-\frac{\cos A \cdot \cos B}{\cos (A+B)}=\frac{\cos A \cdot \cos B}{\sin A \sin B-\cos A \cos B} \\
& =\frac{1}{\tan A \tan B-1}=\frac{1}{2-1}=1
\end{aligned}
\)

Range of \(f(\theta)=a \cos \theta+b \sin \theta\)

\(
\begin{aligned}
& \text { Now } f(\theta)=a \cos \theta+b \sin \theta=r \sin \alpha \cos \theta+r \cos \alpha \sin \theta=r \sin (\theta+\alpha)=\sqrt{a^2+b^2} \sin \left(\theta+\tan ^{-1} \frac{a}{b}\right) \\
& \text { Now }-1 \leq \sin \left(\theta+\tan ^{-1} \frac{a}{b}\right) \leq 1 \\
& \Rightarrow-\sqrt{a^2+b^2} \leq \sqrt{a^2+b^2} \sin \left(\theta+\tan ^{-1} \frac{a}{b}\right) \leq \sqrt{a^2+b^2}
\end{aligned}
\)
Hence, range is \(\left[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}\right]\).

Example 2.40: Find the maximum value of \(\sqrt{3} \sin x+\cos x\) and \(x\) for which a maximum value occurs.

Solution:
\(
\begin{aligned}
&\sqrt{3} \sin x+\cos x=2\left(\frac{\sqrt{3}}{2} \sin x+\frac{1}{2} \cos x\right)=2 \sin (x+\pi / 6)\\
&\text { which is maximum when } x+\pi / 6=\pi / 2 \text { or } x=60^{\circ} \text { and has a maximum value } 2 \text {. }
\end{aligned}
\)

Example 2.41: Find the maximum and minimum values of \(\cos ^2 \theta-6 \sin \theta \cos \theta+3 \sin ^2 \theta+2\).

Solution: \(\cos ^2 \theta-6 \sin \theta \cos \theta+3 \sin ^2 \theta+2\)
\(
\begin{aligned}
& =\frac{1+\cos 2 \theta}{2}-3 \sin 2 \theta+3 \frac{(1-\cos 2 \theta)}{2}+2 \\
& =4-\cos 2 \theta-3 \sin 2 \theta \\
& \text { Now, }-\cos 2 \theta-3 \sin 2 \theta \in[-\sqrt{10}, \sqrt{10}] \\
& \Rightarrow 4-\cos 2 \theta-3 \sin 2 \theta \in[4-\sqrt{10}, 4+\sqrt{10}]
\end{aligned}
\)

TRANSFORMATION FORMULAE

Formulae to Transform the Product into Sum or Difference

We know that
\(
\begin{aligned}
& \sin A \cos B+\cos A \sin B=\sin (A+B) \\
& \sin A \cos B-\cos A \sin B=\sin (A-B) \\
& \cos A \cos B-\sin A \sin B=\cos (A+B) \\
& \cos A \cos B+\sin A \sin B=\cos (A-B)
\end{aligned}
\)
\(
2 \sin A \cos B=\sin (A+B)+\sin (A-B) .
\)
\(
2 \cos A \sin B=\sin (A+B)-\sin (A-B)
\)
\(
2 \cos A \cos B=\cos (A+B)+\cos (A-B)
\)
\(
2 \sin A \sin B=\cos (A-B)-\cos (A+B)
\)

Formulae to Transform the Sum or Difference into Product

Let \(A+B=C\) and \(A-B=D\). Then, \(A=\frac{C+D}{2}\) and \(B=\frac{C-D}{2}\)
\(
\begin{aligned}
& \sin C+\sin D=2 \sin \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right) \\
& \sin C-\sin D=2 \sin \left(\frac{C-D}{2}\right) \cos \left(\frac{C+D}{2}\right) \\
& \cos C+\cos D=2 \cos \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)
\end{aligned}
\)
\(
\left.\begin{array}{l}
\cos D-\cos C=2 \sin \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right) \\
\cos C-\cos D=-2 \sin \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right) \\
\cos C-\cos D=2 \sin \left(\frac{C+D}{2}\right) \sin \left(\frac{D-C}{2}\right)
\end{array}\right\}
\)
These above formulae are used to convert the sum or difference of two sines or two cosines into the product of sines and cosines.

Example 2.42: Prove that \(\cos 55^{\circ}+\cos 65^{\circ}+\cos 175^{\circ}=0\).

Solution: \(\text { L.H.S. }=\cos 55^{\circ}+\cos 65^{\circ}+\cos 175^{\circ}\)
\(
=2 \cos \frac{55^{\circ}+65^{\circ}}{2} \cos \frac{55^{\circ}-65^{\circ}}{2}+\cos 175^{\circ} \left(\cos C+\cos D=2 \cos \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)\right)
\)
\(
=2 \cos 60^{\circ} \cos \left(-5^{\circ}\right)+\cos 175^{\circ}=2 \times \frac{1}{2} \cos 5^{\circ}+\cos \left(180^{\circ}-5^{\circ}\right)=\cos 5^{\circ}-\cos 5^{\circ}=0
\)

Example 2.43: Prove that \(\cos 18^{\circ}-\sin 18^{\circ}=\sqrt{2} \sin 27^{\circ}\).

Solution: L.H.S. \(=\cos 18^{\circ}-\sin 18^{\circ}=\cos 18^{\circ}-\sin \left(90^{\circ}-72^{\circ}\right)=\cos 18^{\circ}-\cos 72^{\circ}\)
\(\begin{aligned} & =2 \sin \frac{18^{\circ}+72^{\circ}}{2} \sin \frac{72^{\circ}-18^{\circ}}{2} \\ & =2 \sin 45^{\circ} \sin 27^{\circ}=2 \frac{1}{\sqrt{2}} \sin 27^{\circ}=\sqrt{2} \sin 27^{\circ}\end{aligned}\)

Example 2.44: Prove that \(\frac{\sin 5 A-\sin 3 A}{\cos 5 A+\cos 3 A}=\tan A\)

Solution:
\(
\text { L.H.S. }=\frac{\sin 5 A-\sin 3 A}{\cos 5 A+\cos 3 A}=\frac{2 \sin \left(\frac{5 A-3 A}{2}\right) \cos \left(\frac{5 A+3 A}{2}\right)}{2 \cos \left(\frac{5 A+3 A}{2}\right) \cos \left(\frac{5 A-3 A}{2}\right)}=\frac{2 \sin A \cos 4 A}{2 \cos 4 A \cos A}=\tan A
\)
\(\left(\sin C-\sin D=2 \sin \left(\frac{C-D}{2}\right) \cos \left(\frac{C+D}{2}\right)\right)\)

Example 2.45: Prove that \(\frac{\sin A+\sin 3 A}{\cos A+\cos 3 A}=\tan 2 A\)

Solution:
L.H.S. \(=\frac{\sin 3 A+\sin A}{\cos 3 A+\cos A}=\frac{2 \sin \left(\frac{3 A+A}{2}\right) \cos \left(\frac{3 A-A}{2}\right)}{2 \cos \left(\frac{3 A+A}{2}\right) \cos \left(\frac{3 A-A}{2}\right)}=\frac{\sin 2 A \cos A}{\cos 2 A \cos A}=\tan 2 A\)
\(\left(\sin C+\sin D=2 \sin \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)\right)\)

Example 2.46: Prove that \(\cos \alpha+\cos \beta+\cos \gamma+\cos (\alpha+\beta+\gamma)=4 \cos \frac{\alpha+\beta}{2} \cos \frac{\beta+\gamma}{2} \cos \frac{\gamma+\alpha}{2}\).

Solution: L.H.S. \(=\cos \alpha+\cos \beta+\cos \gamma+\cos (\alpha+\beta+\gamma)\)
\(
\begin{aligned}
& =(\cos \alpha+\cos \beta)+[\cos \gamma+\cos (\alpha+\beta+\gamma)] \\
& =2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)+2 \cos \left(\frac{\alpha+\beta+\gamma+\gamma}{2}\right) \cos \left(\frac{\alpha+\beta+\gamma-\gamma}{2}\right) \\
& =2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)+2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha+\beta+2 \gamma}{2}\right) \\
& =2 \cos \left(\frac{\alpha+\beta}{2}\right)\left\{\cos \left(\frac{\alpha-\beta}{2}\right)+\cos \left(\frac{\alpha+\beta+2 \gamma}{2}\right)\right\}
\end{aligned}
\)
\(
\begin{aligned}
& =2 \cos \left(\frac{\alpha+\beta}{2}\right)\left\{2 \cos \left(\frac{\frac{\alpha-\beta}{2}+\frac{\alpha+\beta+2 \gamma}{2}}{2}\right) \cos \left(\frac{\frac{\alpha+\beta+2 \gamma}{2}-\frac{\alpha-\beta}{2}}{2}\right)\right\} \\
& =2 \cos \left(\frac{\alpha+\beta}{2}\right)\left\{2 \cos \left(\frac{\alpha+\gamma}{2}\right) \cos \left(\frac{\beta+\gamma}{2}\right)\right\}=4 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\beta+\gamma}{2}\right) \cos \left(\frac{\gamma+\alpha}{2}\right)=\text { R.H.S. }
\end{aligned}
\)

Example 2.47: Prove that \(\frac{\sin A+\sin 2 A+\sin 4 A+\sin 5 A}{\cos A+\cos 2 A+\cos 4 A+\cos 5 A}=\tan 3 A\).

Solution:
\(
\begin{aligned}
& \frac{\sin A+\sin 2 A+\sin 4 A+\sin 5 A}{\cos A+\cos 2 A+\cos 4 A+\cos 5 A} \\
& =\frac{(\sin 5 A+\sin A)+(\sin 4 A+\sin 2 A)}{(\cos 5 A+\cos A)+(\cos 4 A+\cos 2 A)} \\
& =\frac{2 \sin 3 A \cos 2 A+2 \sin 3 A \cos A}{2 \cos 3 A \cos 2 A+2 \cos 3 A \cos A} \\
& =\frac{2 \sin 3 A(\cos 2 A+\cos A)}{2 \cos 3 A(\cos 2 A+\cos A)}=\tan 3 A
\end{aligned}
\)

Example 2.48: Prove that \(\left(\frac{\cos A+\cos B}{\sin A-\sin B}\right)^n+\left(\frac{\sin A+\sin B}{\cos A-\cos B}\right)^n=2 \cot ^n \frac{A-B}{2}\) or 0, accordingly as \(n\) is even or odd.

Solution:
\(
\text { L.H.S. }=\left(\frac{2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}}{2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}}\right)^n+\left(\frac{2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}}{2 \sin \frac{A+B}{2} \sin \frac{B-A}{2}}\right)^n
\)
\(
=\left(\cot \frac{A-B}{2}\right)^n+\left(-\cot \frac{A-B}{2}\right)^n [\because \sin (-\theta)=-\sin \theta]
\)
\(
=\cot ^n \frac{A-B}{2}+(-1)^n \cot ^n \frac{A-B}{2}=\cot ^n \frac{A-B}{2}\left[1+(-1)^n\right]
\)
\(
=\left\{\begin{array}{l}
0, \text { if } n \text { is odd } \\
2 \cot ^n \frac{A-B}{2}, \text { if } n \text { is even. }
\end{array}\right.
\)

Example 2.49: Prove that \((\cos \alpha+\cos \beta)^2+(\sin \alpha+\sin \beta)^2=4 \cos ^2\left(\frac{\alpha-\beta}{2}\right)\).

Solution:
\(
\begin{aligned}
\text { L.H.S. } & =(\cos \alpha+\cos \beta)^2+(\sin \alpha+\sin \beta)^2 \\
& =\left\{2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)\right\}^2+\left\{2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)\right\}^2 \\
& =4 \cos ^2\left(\frac{\alpha-\beta}{2}\right)\left\{\cos ^2 \frac{\alpha+\beta}{2}+\sin ^2 \frac{\alpha+\beta}{-2}\right\}
\end{aligned}
\)
\(
\begin{aligned}
& =4 \cos ^2\left(\frac{\alpha-\beta}{2}\right) \quad\left[\because \cos ^2 \frac{\alpha+\beta}{2}+\sin ^2 \frac{\alpha+\beta}{2}=1\right] \\
& =\text { R.H.S. }
\end{aligned}
\)

Example 2.50: If \(\sec (\theta+\alpha)+\sec (\theta-\alpha)=2 \sec \theta\), then show that \(\cos ^2 \theta=1+\cos \alpha\).

Solution: \(\sec (\theta+\alpha)+\sec (\theta-\alpha)=2 \sec \theta\)
\(
\begin{aligned}
& \Rightarrow \frac{1}{\cos (\theta+\alpha)}+\frac{1}{\cos (\theta-\alpha)}=\frac{2}{\cos \theta} \\
& \Rightarrow \frac{\cos (\theta-\alpha)+\cos (\theta+\alpha)}{\cos (\theta+\alpha) \cos (\theta-\alpha)}=\frac{2}{\cos \theta} \\
& \Rightarrow \frac{2 \cos \theta \cos \alpha}{\cos ^2 \theta-\sin ^2 \alpha}=\frac{2}{\cos \theta} \\
& \Rightarrow \cos ^2 \theta \cos \alpha=\cos ^2 \theta-\sin ^2 \alpha \\
& \Rightarrow \sin ^2 \alpha=\cos ^2 \theta(1-\cos \alpha) \\
& \Rightarrow 1-\cos ^2 \alpha=\cos ^2 \theta(1-\cos \alpha) \quad \Rightarrow 1+\cos \alpha=\cos ^2 \theta
\end{aligned}
\)

Example 2.51: In quadrilateral \(A B C D\) if \(\sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)+\sin \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)=2\), then find the value of \(\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \sin \frac{D}{2}\).

Solution:
\(
\begin{aligned}
& \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)+\sin \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)=2 \\
& \Rightarrow \frac{1}{2}[\sin A+\sin B+\sin C+\sin D]=2 \\
& \Rightarrow \sin A+\sin B+\sin C+\sin D=4 \\
& \Rightarrow A=B=C=D=90^{\circ} \\
& \Rightarrow \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \sin \frac{D}{2}=1 / 4
\end{aligned}
\)

TRIGONOMETRIC RATIOS OF MULTIPLES AND SUB-MULTIPLE ANGLES

Formulae for Multiple Angles

• \(\cos 2 A=\cos (A+A)=\cos ^2 A-\sin ^2 A=1-2 \sin ^2 A=2 \cos ^2 A-1\)
• \(\text { Also } \sin ^2 A=\frac{1}{2}(1-\cos 2 A), \cos ^2 A=\frac{1}{2}(1+\cos 2 A)\)
• \(\sin 2 A=\sin (A+A)=\sin A \cos A+\sin A \cos A=2 \sin A \cos A\)
• \(\tan 2 A=\tan (A+A)=\frac{\tan A+\tan A}{1-\tan A \tan A}=\frac{2 \tan A}{1-\tan ^2 A}\)
• \(\sin 3 A=\sin (2 A+A)\)
  \(
  \begin{aligned}
  & =\sin 2 A \cos A+\cos 2 A \sin A=2 \sin A \cos A \cos A+\left(1-2 \sin ^2 A\right) \sin A \\
  & =2 \sin A \cos ^2 A+\sin A-2 \sin ^3 A \\
  & =2 \sin A\left(1-\sin ^2 A\right)+\sin A-2 \sin ^3 A \\
  & =2 \sin A-2 \sin ^3 A+\sin A-2 \sin ^3 A \\
  & =3 \sin A-4 \sin ^3 A
  \end{aligned}
  \)
• \(\cos 3 A=\cos (2 A+A)\)
  \(
  \begin{aligned}
  & =\cos 2 A \cos A-\sin 2 A \sin A=\left(2 \cos ^2 A-1\right) \cos A-2 \sin A \cos A \sin A \\
  & =2 \cos ^3 A-\cos A-2 \cos A\left(1-\cos ^2 A\right) \\
  & =2 \cos ^3 A-\cos A-2 \cos A+2 \cos ^3 A \\
  & =4 \cos ^3 A-3 \cos A
  \end{aligned}
  \)
• \(\sin 2 A \text { and } \cos 2 A \text { in terms of } \tan A\)
  \(
  \sin 2 A=2 \sin A \cos A=\frac{2 \sin A \cos A}{\cos ^2 A+\sin ^2 A}=\frac{2 \tan A}{1+\tan ^2 A} \text { [dividing numerator and denominator by } \cos ^2 A \text { ] }
  \)
  \(
  \cos 2 A=\cos ^2 A-\sin ^2 A=\frac{\cos ^2 A-\sin ^2 A}{\cos ^2 A+\sin ^2 A}=\frac{1-\tan ^2 A}{1+\tan ^2 A} \text { [dividing numerator and denominator by } \cos ^2 A \text { ] }
  \)
  \(
  \text { Also } \tan ^2 A=\frac{1-\cos 2 A}{1+\cos 2 A}
  \)
• In the formula of \(\tan (A+B+C)\), putting \(B=A\) and \(C=A\), we get
  \(
  \tan 3 A=\frac{3 \tan A-\tan ^3 A}{1-3 \tan ^2 A}
  \)
  Similarly, we can prove that \(\cot 3 A=\frac{\cot ^3 A-3 \cot A}{3 \cot ^2 A-1}\)
• \(\tan \left(A_1+A_2+\cdots A_n\right)=\frac{S_1-S_3+S_5-S_7+\cdots}{1-S_2+S_4-S_6+\cdots}\),
  where
  \(S_1=\tan A_1+\tan A_2+\cdots+\tan A_n=\) Sum of the tangents of the separate angles, \(S_2=\tan A_1 \tan A_2+\tan A_1 \tan A_3+\ldots=\) Sum of the product of tangents taken two at a time, \(S_3=\tan A_1 \tan A_2 \tan A_3+\tan A_2 \tan A_3 \tan A_4+\ldots=\) Sum of the product of tangents taken three at a time, and so on.
  If \(A_1=A_2=\ldots=A_n=A\), then we have
  \(
  S_1=n \tan A, S_2={ }^n C_2 \tan ^2 A, S_3={ }^n C_3 \tan ^3 A, \cdots
  \)

Example 2.52: Prove that
a. \(\frac{\sin 2 \theta}{1+\cos 2 \theta}=\tan \theta\)
b. \(\frac{\sin 2 \theta}{1-\cos 2 \theta}=\cot \theta\)
c. \(\frac{1+\sin 2 \theta+\cos 2 \theta}{1+\sin 2 \theta-\cos 2 \theta}=\cot \theta\)
d \(\frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}=\tan \theta / 2\)
e. \(\frac{\cos 2 \theta}{1+\sin 2 \theta}=\tan (\pi / 4-\theta)\)
f. \(\frac{\cos \theta}{1+\sin \theta}=\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)\)

Solution:
a L.H.S. \(=\frac{\sin 2 \theta}{1+\cos 2 \theta}=\frac{2 \sin \theta \cos \theta}{2 \cos ^2 \theta}=\tan \theta=\) R.H.S.
b. L.H.S. \(=\frac{\sin 2 \theta}{1-\cos 2 \theta}=\frac{2 \sin \theta \cos \theta}{2 \sin ^2 \theta}=\cot \theta=\) R.H.S.
c. L.H.S. \(=\frac{1+\sin 2 \theta+\cos 2 \theta}{1+\sin 2 \theta-\cos 2 \theta}=\frac{(1+\cos 2 \theta)+\sin 2 \theta}{(1-\cos 2 \theta)+\sin 2 \theta}\)
\(
\begin{aligned}
& =\frac{2 \cos ^2 \theta+2 \sin \theta \cos \theta}{2 \sin ^2 \theta+2 \sin \theta \cos \theta} \\
& =\frac{2 \cos \theta(\cos \theta+\sin \theta)}{2 \sin \theta(\cos \theta+\sin \theta)}=\frac{\cos \theta}{\sin \theta}=\cot \theta=\text { R.H.S. }
\end{aligned}
\)
d L.H.S. \(=\frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}=\frac{(1-\cos \theta)+\sin \theta}{(1+\cos \theta)+\sin \theta}\)
\(
\begin{aligned}
& =\frac{2 \sin ^2 \frac{\theta}{2}+2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}+2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}=\frac{2 \sin \frac{\theta}{2}\left(\sin \frac{\theta}{2}+\cos \frac{\theta}{2}\right)}{2 \cos \frac{\theta}{2}\left(\sin \frac{\theta}{2}+\cos \frac{\theta}{2}\right)} \\
& =\tan \frac{\theta}{2}=\text { R.H.S. }
\end{aligned}
\)
e.
\(
\begin{aligned}
\text { L.H.S. } & =\frac{\cos 2 \theta}{1+\sin 2 \theta} \\
& =\frac{\sin \left(\frac{\pi}{2}-2 \theta\right)}{1+\cos \left(\frac{\pi}{2}-2 \theta\right)} \\
& =\frac{2 \sin \left(\frac{\pi}{4}-\theta\right) \cos \left(\frac{\pi}{4}-\theta\right)}{2 \cos ^2\left(\frac{\pi}{4}-\theta\right)} \\
& =\tan \left(\frac{\pi}{4}-\theta\right)=\text { R.H.S. }
\end{aligned}
\)
f. L.H.S. \(=\frac{\cos \theta}{1+\sin \theta}=\frac{\sin \left(\frac{\pi}{2}-\theta\right)}{1+\cos \left(\frac{\pi}{2}-\theta\right)}=\frac{2 \sin \left(\frac{\pi}{4}-\frac{\theta}{2}\right) \cos \left(\frac{\pi}{4}-\frac{\theta}{2}\right)}{2 \cos ^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}=\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)=\) R.H.S.

Example 2.53: Prove that \(\frac{1+\sin 2 \theta}{1-\sin 2 \theta}=\left(\frac{1+\tan \theta}{1-\tan \theta}\right)^2\).

Solution:
\(\begin{aligned} \text { L.H.S. } & =\frac{1+\sin 2 \theta}{1-\sin 2 \theta} \\ & =\frac{\sin ^2 \theta+\cos ^2 \theta+2 \sin \theta \cos \theta}{\sin ^2 \theta+\cos ^2 \theta-2 \sin \theta \cos \theta} \\ & =\left(\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}\right)^2=\left(\frac{1+\tan \theta}{1-\tan \theta}\right)^2\end{aligned}\)
\(
\text { (dividing numerator and denominator by } \cos \theta \text { ) }
\)

Example 2.54: If \(\alpha+\beta=90^{\circ}\), find the maximum value of \(\sin \alpha \sin \beta\).

Solution:
\(
\begin{aligned}
&\text { Let } y=\sin \alpha \sin \beta=\sin \alpha \sin \left(90^{\circ}-\alpha\right)=\sin \alpha \cos \alpha=\frac{1}{2} \sin 2 \alpha\\
&\text { which has the maximum value } 1 / 2 \text { when } \sin 2 \alpha=1 \text {. }
\end{aligned}
\)

Example 2.55: Prove that \(\frac{1-\tan ^2\left(\frac{\pi}{4}-A\right)}{1+\tan ^2\left(\frac{\pi}{4}-A\right)}=\sin 2 A\).

Solution:
\(
\frac{1-\tan ^2\left(\frac{\pi}{4}-A\right)}{1+\tan ^2\left(\frac{\pi}{4}-A\right)}=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\left(\text { where } \frac{\pi}{4}-A=\theta\right)=\cos 2 \theta=\cos \left(\frac{\pi}{2}-2 A\right)=\sin 2 A
\)

Example 2.56: Prove that \((\cos A-\cos B)^2+(\sin A-\sin B)^2=4 \sin ^2 \frac{A-B}{2}\).

Solution:
\(
\begin{aligned}
\text { L.H.S. } & =(\cos A-\cos B)^2+(\sin A-\sin B)^2 \\
& =\cos ^2 A+\cos ^2 B-2 \cos A \cos B+\sin ^2 A+\sin ^2 B-2 \sin A \sin B
\end{aligned}
\)
\(
\begin{aligned}
& =\left(\cos ^2 A+\sin ^2 A\right)+\left(\cos ^2 B+\sin ^2 B\right)-2(\cos A \cos B+\sin A \sin B) \\
& =2-2 \cos (A-B)=2(1-\cos (A-B))=4 \sin ^2 \frac{A-B}{2}
\end{aligned}
\)

Example 2.57: If \(\sin A=\frac{3}{5}\) and \(0^{\circ}<A<90^{\circ}\), find the values of \(\sin 2 A, \cos 2 A, \tan 2 A\), and \(\sin 4 A\).

Solution: Given \(\sin A=\frac{3}{5}\) and \(A\) is an acute angle.
\(
\therefore \quad \cos A=\frac{4}{5}
\)
\([\because A\) is acute \(]\)
and \(\tan A=\frac{3}{4}\)
Now, \(\sin 2 A=2 \sin A \cos A=2 \times \frac{3}{5} \times \frac{4}{5}=\frac{24}{25}\)
\(
\begin{aligned}
& \cos 2 A=1-2 \sin ^2 A=1-2 \times \frac{9}{25}=\frac{7}{25} \\
& \tan 2 A=\frac{2 \tan A}{1-\tan ^2 A}=\frac{2 \times \frac{3}{4}}{1-\left(\frac{3}{4}\right)^2}=\frac{\frac{6}{4}}{1-\frac{9}{16}}=\frac{\frac{6}{4}}{\frac{7}{16}}=\frac{6}{4} \times \frac{16}{7}=\frac{24}{7} \\
& \sin 4 A=2 \sin 2 A \cos 2 A=2 \times \frac{24}{25} \times \frac{7}{25}=\frac{336}{625} .
\end{aligned}
\)

Example 2.58: If \(\tan \alpha=\frac{1}{7}, \sin \beta=\frac{1}{\sqrt{10}}\), prove that \(\alpha+2 \beta=\frac{\pi}{4}\), where \(0<\alpha<\frac{\pi}{2}\) and \(0<\beta<\frac{\pi}{2}\).

Solution: \(\tan (\alpha+2 \beta)=\frac{\tan \alpha+\tan 2 \beta}{1-\tan \alpha \tan 2 \beta}=\frac{\frac{1}{7}+\tan 2 \beta}{1-\frac{1}{7} \tan 2 \beta} \dots(i)\)
\(
\text { Now, } \tan 2 \beta=\frac{2 \tan \beta}{1-\tan ^2 \beta}=\frac{2 \times \frac{1}{3}}{1-\frac{1}{9}}=\frac{3}{4}[\tan \beta>0 \text { as } 0<\beta<\pi / 2]
\)
Substituting the value of \(\tan 2 \beta\) in Eq. (i), we get
\(
\tan (\alpha+2 \beta)=\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7} \times \frac{3}{4}}=\frac{25}{25}=1
\)
Now, \(0<\alpha<\frac{\pi}{2}\) and \(0<\beta<\frac{\pi}{2}\)
\(
\begin{aligned}
& \therefore \quad 0<2 \beta<\pi, \text { but } \tan 2 \beta=\frac{3}{4}>0 \\
& \Rightarrow \quad 0<2 \beta<\frac{\pi}{2}
\end{aligned}
\)
Hence, \(0<\alpha+2 \beta<\pi\).
In the interval \((0, \pi), \tan \theta\) takes value 1 at \(\pi / 4\) only
\(
\therefore \quad \alpha+2 \beta=\frac{\pi}{4}
\)

Example 2.59: Show that \(\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 \theta}}}=2 \cos \theta, \quad 0<\theta<\pi / 16\).

Solution: \(\text { L.H.S. }=\sqrt{2+\sqrt{2+\sqrt{2(1+\cos 8 \theta)}}}\)
\(
\begin{aligned}
&=\sqrt{2+\sqrt{2+\sqrt{2\left(2 \cos ^2 4 \theta\right)}}}\\
&\left[\because 1+\cos 8 \theta=2 \cos ^2 \frac{8 \theta}{2}\right]
\end{aligned}
\)
\(
\begin{aligned}
& =\sqrt{2+\sqrt{2+\sqrt{\left(4 \cos ^2 4 \theta\right)}}} \\
& =\sqrt{2+\sqrt{2+2 \cos 4 \theta}} \\
& =\sqrt{2+\sqrt{2(1+\cos 4 \theta)}} \\
& =\sqrt{2+\sqrt{2\left(2 \cos ^2 2 \theta\right)}} \left[\because 1+\cos 4 \theta=2 \cos ^2 2 \theta\right]
\end{aligned}
\)
\(
=\sqrt{2+2 \cos 2 \theta}=\sqrt{2(1+\cos 2 \theta)}=\sqrt{2\left(2 \cos ^2 \theta\right)}=2 \cos \theta=\text { R.H.S. }
\)

Example 2.60: Prove that \(\frac{\sec 8 \theta-1}{\sec 4 \theta-1}=\frac{\tan 8 \theta}{\tan 2 \theta}\).

Solution: L.H.S. \(=\frac{\sec 8 \theta-1}{\sec 4 \theta-1}=\frac{\frac{1}{\cos 8 \theta}-1}{\frac{1}{\cos 4 \theta}-1}=\frac{1-\cos 8 \theta}{\cos 8 \theta} \times \frac{\cos 4 \theta}{1-\cos 4 \theta}\)
\(
=\frac{2 \sin ^2 4 \theta}{\cos 8 \theta} \times \frac{\cos 4 \theta}{2 \sin ^2 2 \theta}
\)
\(
\left[\begin{array}{l}
\because 1-\cos 8 \theta=2 \sin ^2 \frac{8 \theta}{2}=2 \sin ^2 4 \theta \\
\text { and } 1-\cos 4 \theta=2 \sin ^2 \frac{4 \theta}{2}=2 \sin ^2 2 \theta
\end{array}\right]
\)
\(
\begin{aligned}
& =\frac{(2 \sin 4 \theta \cos 4 \theta)}{\cos 8 \theta} \times \frac{\sin 4 \theta}{2 \sin ^2 2 \theta} \\
& =\left(\frac{2 \sin 4 \theta \cos 4 \theta}{\cos 8 \theta}\right) \times\left(\frac{2 \sin 2 \theta \cos 2 \theta}{2 \sin ^2 2 \theta}\right) \\
& =\left(\frac{\sin 2(4 \theta)}{\cos 8 \theta}\right) \times\left(\frac{\cos 2 \theta}{\sin 2 \theta}\right)=\left(\frac{\sin 8 \theta}{\cos 8 \theta}\right) \times\left(\frac{\cos 2 \theta}{\sin 2 \theta}\right) \\
& =\tan 8 \theta \cot 2 \theta=\frac{\tan 8 \theta}{\tan 2 \theta}=\text { R.H.S. }
\end{aligned}
\)

Example 2.61: Show that \(\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}=4\).

Solution: L.H.S. \(=\frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}}=\frac{\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}\)
\(
\begin{aligned}
& =\frac{2\left(\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ}\right)}{\sin 20^{\circ} \cos 20^{\circ}} \\
& =\frac{2\left(\sin 60^{\circ} \cos 20^{\circ}-\cos 60^{\circ} \sin 20^{\circ}\right)}{\sin 20^{\circ} \cos 20^{\circ}} \\
& =\frac{2 \sin \left(60^{\circ}-20^{\circ}\right)}{\sin 20^{\circ} \cos 20^{\circ}}=\frac{2 \sin 40^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}} \\
& =\frac{4 \sin 40^{\circ}}{2 \sin 20^{\circ} \cos 20^{\circ}}=\frac{4 \sin 40^{\circ}}{\sin 40^{\circ}}=4
\end{aligned}
\)

Example 2.62: Prove that \(\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)=\frac{1}{8}\).

Solution:
We have \(\cos \frac{7 \pi}{8}=\cos \left(\pi-\frac{\pi}{8}\right)=-\cos \frac{\pi}{8}\) and \(\cos \frac{5 \pi}{8}=\cos \left(\pi-\frac{3 \pi}{8}\right)=-\cos \frac{3 \pi}{8}\)
\(
\therefore \quad \text { L.H.S. }=\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1-\cos \frac{3 \pi}{8}\right)\left(1-\cos \frac{\pi}{8}\right)
\)
\(
\begin{aligned}
& =\left(1-\cos ^2 \frac{\pi}{8}\right)\left(1-\cos ^2 \frac{3 \pi}{8}\right) \\
& =\sin ^2 \frac{\pi}{8} \sin ^2 \frac{3 \pi}{8} \\
& =\frac{1}{4}\left(2 \sin ^2 \frac{\pi}{8}\right)\left(2 \sin ^2 \frac{3 \pi}{8}\right) \\
& =\frac{1}{4}\left[\left(1-\cos \frac{\pi}{4}\right)\left(1-\cos \frac{3 \pi}{4}\right)\right] \left[\because 1-\cos \theta=2 \sin ^2 \frac{\theta}{2}\right]
\end{aligned}
\)
\(
=\frac{1}{4} \cdot\left[\left(1-\frac{1}{\sqrt{2}}\right)\left(1+\frac{1}{\sqrt{2}}\right)\right]=\frac{1}{4}\left(1-\frac{1}{2}\right)=\frac{1}{8}=\text { R.H.S. }
\)

Example 2.63: Prove that \(\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\cos ^4 \frac{7 \pi}{8}=\frac{3}{2}\).

Solution:
We have \(\frac{7 \pi}{8}=\pi-\frac{\pi}{8}\) and \(\frac{5 \pi}{8}=\pi-\frac{3 \pi}{8}\)
\(
\begin{aligned}
& \Rightarrow \cos \frac{7 \pi}{8}=-\cos \frac{\pi}{8} \text { and } \cos \frac{5 \pi}{8}=-\cos \frac{3 \pi}{8} \\
& \Rightarrow \cos ^4 \frac{7 \pi}{8}=\cos ^4 \frac{\pi}{8} \text { and } \cos ^4 \frac{5 \pi}{8}=\cos ^4 \frac{3 \pi}{8} \\
& \therefore \text { L.H.S. }=2 \cos ^4 \frac{\pi}{8}+2 \cos ^4 \frac{3 \pi}{8}
\end{aligned}
\)
\(
\begin{aligned}
& =2\left[\left(\cos ^2 \frac{\pi}{8}\right)^2+\left(\cos ^2 \frac{3 \pi}{8}\right)^2\right] \\
& =2\left\{\frac{1+\cos \frac{\pi}{4}}{2}\right\}^2+\left\{\frac{1+\cos \frac{3 \pi}{4}}{2}\right\}^2 \left[\because \frac{1+\cos 2 \theta}{2}=\cos ^2 \theta\right]
\end{aligned}
\)
\(
\begin{aligned}
& =\frac{1}{2}\left\{\left(1+\cos \frac{\pi}{4}\right)^2+\left(1+\cos \frac{3 \pi}{4}\right)^2\right\}^2 \\
& =\frac{1}{2}\left\{\left(1+\frac{1}{\sqrt{2}}\right)^2+\left(1-\frac{1}{\sqrt{2}}\right)^2\right\} \\
& =\frac{1}{2}\left\{\left(1+\frac{1}{2}+\sqrt{2}\right)+\left(1+\frac{1}{2}-\sqrt{2}\right)\right\}=\frac{3}{2}=\text { R.H.S. }
\end{aligned}
\)

Example 2.64: If \(\pi<x<2 \pi\), prove that \(\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}=\cot \left(\frac{x}{2}+\frac{\pi}{4}\right)\).

Solution:
\(
\text { L.H.S. }=\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}=\frac{\sqrt{2 \cos ^2 \frac{x}{2}}+\sqrt{2 \sin ^2 \frac{x}{2}}}{\sqrt{2 \cos ^2 \frac{x}{2}}-\sqrt{2 \sin ^2 \frac{x}{2}}}
\)
\(
\begin{aligned}
& =\frac{\sqrt{2}\left|\cos \frac{x}{2}\right|+\sqrt{2}\left|\sin \frac{x}{2}\right|}{\sqrt{2}\left|\cos \frac{x}{2}\right|-\sqrt{2}\left|\sin \frac{x}{2}\right|} \\
& =\frac{\left|\cos \frac{x}{2}\right|+\left|\sin \frac{x}{2}\right|}{\left|\cos \frac{x}{2}\right|-\left|\sin \frac{x}{2}\right|} \\
& =\frac{-\cos \frac{x}{2}+\sin \frac{x}{2}}{-\cos \frac{x}{2}-\sin \frac{x}{2}} \left[\because \pi<x<2 \pi, \therefore \frac{\pi}{2}<\frac{x}{2}<\pi\right]
\end{aligned}
\)
\(
\begin{aligned}
&\Rightarrow \cos x / 2 \text { is negative and } \sin x / 2 \text { is positive. }\\
&=\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}=\frac{\cot \frac{x}{2}-1}{\cot \frac{x}{2}+1} \text { [dividing numerator and denominator by } \sin x / 2 \text { ] }
\end{aligned}
\)
\(
=\cot \left(\frac{x}{2}+\frac{\pi}{4}\right)=\text { R.H.S. }
\)

Example 2.65: If \(\sin \alpha+\sin \beta=a\) and \(\cos \alpha+\cos \beta=b\), prove that \(\tan \frac{a-\beta}{2}= \pm \sqrt{\frac{4-a^2-b^2}{a^2+b^2}}\).

Solution:
Given, \(\sin \alpha+\sin \beta=a \dots(i)\)
and \(\cos \alpha+\cos \beta=b \dots(ii)\)
\(
\begin{aligned}
& \text { Now }(\cos \alpha+\cos \beta)^2+(\sin \alpha+\sin \beta)^2=b^2+a^2 \\
& \Rightarrow \cos ^2 \alpha+\cos ^2 \beta+2 \cos \alpha \cos \beta+\sin ^2 \alpha+\sin ^2 \beta+2 \sin \alpha \sin \beta=b^2+a^2 \\
& \Rightarrow\left(\cos ^2 \alpha+\sin ^2 \alpha\right)+\left(\cos ^2 \beta+\sin ^2 \beta\right)+2(\cos \alpha \cos \beta+\sin \alpha \sin \beta)=a^2+b^2 \\
& \Rightarrow 2+2 \cos (\alpha-\beta)=a^2+b^2
\end{aligned}
\)
\(
\text { or } \cos (\alpha-\beta)=\frac{a^2+b^2-2}{2}
\)
\(
\begin{aligned}
&\text { Now, } \tan \frac{\alpha-\beta}{2}= \pm \sqrt{\frac{1-\cos (\alpha-\beta)}{1+\cos (\alpha-\beta)}}\\
&=4 \pm \sqrt{\frac{1-\frac{a^2+b^2-2}{2}}{1+\frac{a^2+b^2-2}{2}}}= \pm \sqrt{\frac{4-a^2-b^2}{a^2+b^2}}
\end{aligned}
\)

Example 2.66: If \(\tan \frac{\theta}{2}=\sqrt{\frac{a-b}{a+b}} \tan \frac{\varphi}{2}\), prove that \(\cos \alpha=\frac{a \cos \varphi+b}{a+b \cos \varphi}\).

Solution: Given, \(\tan \frac{\theta}{2}=\sqrt{\frac{a-b}{a+b}} \tan \frac{\varphi}{2} \dots(i)\)
\(
\begin{aligned}
&\text { Now, } \cos \theta=\frac{1-\tan ^2 \frac{\theta}{2}}{1+\tan ^2 \frac{\theta}{2}}=\frac{1-\frac{a-b}{a+b} \tan ^2 \frac{\varphi}{2}}{1+\frac{a-b}{a+b} \tan ^2 \frac{\varphi}{2}}\\
&=\frac{1-\frac{a-b}{a+b} \frac{\sin ^2 \frac{\varphi}{2}}{\cos ^2 \frac{\varphi}{2}}}{1+\frac{a-b}{a+b} \frac{\sin ^2 \frac{\varphi}{2}}{\cos ^2 \frac{\varphi}{2}}}
\end{aligned}
\)
\(
\begin{aligned}
& =\frac{(a+b) \cos ^2 \frac{\varphi}{2}-(a-b) \sin ^2 \frac{\varphi}{2}}{(a+b) \cos ^2 \frac{\varphi}{2}+(a-b) \sin ^2 \frac{\varphi}{2}} \\
& =\frac{a\left(\cos ^2 \frac{\varphi}{2}-\sin ^2 \frac{\varphi}{2}\right)+b\left(\cos ^2 \frac{\varphi}{2}+\sin ^2 \frac{\varphi}{2}\right)}{a\left(\cos ^2 \frac{\varphi}{2}+\sin ^2 \frac{\varphi}{2}\right)+b\left(\cos ^2 \frac{\varphi}{2}-\sin ^2 \frac{\varphi}{2}\right)} \\
& =\frac{a \cos \varphi+b}{a+b \cos \varphi}
\end{aligned}
\)

Example 2.67: If \(\cos \theta=\cos \alpha \cos \beta\), prove that \(\tan \frac{\theta+\alpha}{2} \tan \frac{\theta-\alpha}{2}=\tan ^2 \frac{\beta}{2}\).

Solution: Given, \(\cos \theta=\cos \alpha \cos \beta\), we have \(\cos \beta=\frac{\cos \theta}{\cos \alpha}\)
Now, \(\tan ^2 \frac{\beta}{2}=\frac{1-\cos \beta}{1+\cos \beta}=\frac{1-\frac{\cos \theta}{\cos \alpha}}{1+\frac{\cos \theta}{\cos \alpha}}\)
\(
\begin{aligned}
& =\frac{\cos \alpha-\cos \theta}{\cos \alpha+\cos \theta} \\
& =\frac{2 \sin \frac{\alpha+\theta}{2} \sin \frac{\theta-\alpha}{2}}{2 \cos \frac{\theta+\alpha}{2} \cos \frac{\theta-\alpha}{2}}=\tan \frac{\theta+\alpha}{2} \tan \frac{\theta-\alpha}{2}
\end{aligned}
\)

VALUES OF TRIGONOMETRIC RATIOS OF STANDARD ANGLES

Value of \(\sin 15^{\circ}, \cos 15^{\circ}, \sin 75^{\circ}, \cos 75^{\circ}, \tan 15^{\circ}, \tan 75^{\circ}\):
\(
\sin 15^{\circ}=\sin \left(45^{\circ}-30^{\circ}\right)=\sin 45^{\circ} \cos 30^{\circ}-\sin 30^{\circ} \cos 45^{\circ}=\frac{1}{\sqrt{2}} \frac{\sqrt{3}}{2}-\frac{1}{2} \frac{1}{\sqrt{2}}=\frac{\sqrt{3}-1}{2 \sqrt{2}}
\)
Also, \(\sin 15^{\circ}=\cos 75^{\circ}=-\cos 105^{\circ}\)
Similarly, we can prove that \(\cos 15^{\circ}=\frac{\sqrt{3}+1}{2 \sqrt{2}}\)
Also, \(\cos 15^{\circ}=\sin 75^{\circ}=\sin 105^{\circ}\)
\(
\tan 15^{\circ}=\tan \left(60^{\circ}-45^{\circ}\right)=\frac{\tan 60^{\circ}-\tan 45^{\circ}}{1+\tan 60^{\circ} \tan 45^{\circ}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}
\)
\(
\tan 75^{\circ}=\tan \left(60^{\circ}+45^{\circ}\right)=\frac{\tan 60^{\circ}+\tan 45^{\circ}}{1-\tan 60^{\circ} \tan 45^{\circ}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}=2+\sqrt{3}
\)

Value of \(\sin 18^{\circ}, \cos 18^{\circ}\):
Let \(\theta=18^{\circ}\), then \(5 \theta=90^{\circ}\)
\(
\begin{aligned}
& \Rightarrow \quad 2 \theta+3 \theta=90^{\circ} \\
& \Rightarrow \quad 2 \theta=90^{\circ}-3 \theta \\
& \Rightarrow \quad \sin 2 \theta=\sin \left(90^{\circ}-3 \theta\right) \\
& \Rightarrow \quad \sin 2 \theta=\cos 3 \theta \\
& \Rightarrow \quad 2 \sin \theta \cos \theta=4 \cos ^3 \theta-3 \cos \theta \\
& \Rightarrow \quad 2 \sin \theta=4 \cos ^2 \theta-3 \text { [dividing by } \cos \theta \text { ] }\\
& \Rightarrow \quad 2 \sin \theta=4\left(1-\sin ^2 \theta\right)-3=1-4 \sin ^2 \theta \\
& \Rightarrow \quad 4 \sin ^2 \theta+2 \sin \theta-1=0 \\
& \Rightarrow \quad \sin \theta=\frac{-2 \pm \sqrt{4+16}}{8}=\frac{-2 \pm 2 \sqrt{5}}{8}=\frac{-1 \pm \sqrt{5}}{4} \\
& \because \quad \theta=18^{\circ}
\end{aligned}
\)
\(\therefore \sin \theta=\sin 18^{\circ}>0\), for \(18^{\circ}\) lies in the first quadrant.
\(
\therefore \quad \sin \theta \text {, i.e., } \sin 18^{\circ}=\frac{\sqrt{5}-1}{4}
\)

Value of \(\cos \mathbf{1 8}^{\boldsymbol{\circ}}\):
\(
\begin{aligned}
& \cos ^2 18^{\circ}=1-\sin ^2 18^{\circ}=1-\left(\frac{\sqrt{5}-1}{4}\right)^2=1-\frac{5+1-2 \sqrt{5}}{16}=\frac{10+2 \sqrt{5}}{16} \\
& \Rightarrow \cos 18^{\circ}=\frac{1}{4} \sqrt{10+2 \sqrt{5}} \left[\because \cos 18^{\circ}>0\right]
\end{aligned}
\)

Value of \(\cos 36^{\circ}, \sin 36^{\circ}\):
\(
\begin{aligned}
&\cos 36^{\circ}=1-2 \sin ^2 18^{\circ}=1-2\left(\frac{\sqrt{5}-1}{4}\right)^2=\frac{\sqrt{5}+1}{4}\\
&\text { Value of } \sin \mathbf{3 6}^{\circ} \text { : }\\
&\begin{aligned}
& \sin ^2 36^{\circ}=1-\cos ^2 36^{\circ}=1-\left(\frac{\sqrt{5}+1}{4}\right)^2=1-\frac{6+2 \sqrt{5}}{16}=\frac{16-6-2 \sqrt{5}}{16}=\frac{10-2 \sqrt{5}}{16} \\
& \therefore \sin 36^{\circ}=\frac{1}{4} \sqrt{10-2 \sqrt{5}} \left[\because \sin 36^{\circ}>0\right]
\end{aligned}
\end{aligned}
\)
Note:
\(
\begin{aligned}
& \sin 54^{\circ}=\sin \left(90^{\circ}-36^{\circ}\right)=\cos 36^{\circ}=\frac{\sqrt{5}+1}{4} \\
& \cos 54^{\circ}=\cos \left(90^{\circ}-36^{\circ}\right)=\sin 36^{\circ}=\frac{1}{4}(\sqrt{10-2 \sqrt{5}})
\end{aligned}
\)

Value of \(\tan 7 \frac{1^{\circ}}{2}, \cot 7 \frac{1}{2}^{\circ}\):
Let \(\theta=7 \frac{1}{2}^{\circ}\), then \(2 \theta=15^{\circ}\)
\(
\tan \theta=\frac{1-\cos 2 \theta}{\sin 2 \theta} \left[\because 1-\cos 2 \theta=2 \sin ^2 \theta \text { and } \sin 2 \theta=2 \sin \theta \cos \theta\right]
\)
\(
=\frac{1-\cos 15^{\circ}}{\sin 15^{\circ}}=\frac{1-\frac{\sqrt{3}+1}{2 \sqrt{2}}}{\frac{\sqrt{3}-1}{2 \sqrt{2}}}=\frac{2 \sqrt{2}-\sqrt{3}-1}{\sqrt{3}-1}=(\sqrt{3}-\sqrt{2})(\sqrt{2}-1)
\)

Value of \(\cot 82 \frac{1^{\circ}}{2}\):
\(
\cot 82 \frac{1}{2}^{\circ}=\cot \left(90^{\circ}-7 \frac{1}{2}^{\circ}\right)=\tan 7 \frac{1}{2}^{\circ}=(\sqrt{3}-\sqrt{2})(\sqrt{2}-1)
\)

Value of \(\cot 7 \frac{1}{2}^{\circ}\):
\(
\text { Let } \theta=7 \frac{1^{\circ}}{2} \text {, then } 2 \theta=15^{\circ}
\)
Now, \(\cot \theta=\frac{1+\cos 2 \theta}{\sin 2 \theta}=\frac{1+\cos 15^{\circ}}{\sin 15^{\circ}}=\frac{1+\frac{\sqrt{3}+1}{2 \sqrt{2}}}{\frac{\sqrt{3}-1}{2 \sqrt{2}}}=\frac{2 \sqrt{2}+\sqrt{3}+1}{\sqrt{3}-1}=(\sqrt{3}+\sqrt{2})(\sqrt{2}+1)\)

Value of \(\tan 82 \frac{1}{2}^{\circ}\):
\(
\tan 82 \frac{1^{\circ}}{2}=\tan \left(90^{\circ}-7 \frac{1}{2}^{\circ}\right)=\cot 7 \frac{1}{2}^{\circ}=(\sqrt{3}+\sqrt{2})(\sqrt{2}+1)
\)

Summary Table

\(
\begin{array}{|l|l|l|l|l|l|l|l|}
\hline & 7.5^{\circ} & 15^{\circ} & 18^{\circ} & 22.5^{\circ} & 36^{\circ} & 67.5^{\circ} & 75^{\circ} \\
\hline \sin & \frac{\sqrt{8-2 \sqrt{6}-2 \sqrt{2}}}{4} & \frac{\sqrt{3}-1}{2 \sqrt{2}} & \frac{\sqrt{5}-1}{4} & \frac{\sqrt{2-\sqrt{2}}}{2} & \frac{\sqrt{10-2 \sqrt{5}}}{4} & \frac{\sqrt{2+\sqrt{2}}}{2} & \frac{\sqrt{3}+1}{2 \sqrt{2}} \\
\hline \cos & \frac{\sqrt{8+2 \sqrt{6}+2 \sqrt{2}}}{4} & \frac{\sqrt{3}+1}{2 \sqrt{2}} & \frac{\sqrt{10+2 \sqrt{5}}}{4} & \frac{\sqrt{2+\sqrt{2}}}{2} & \frac{\sqrt{5}+1}{4} & \frac{\sqrt{2-\sqrt{2}}}{2} & \frac{\sqrt{3}-1}{2 \sqrt{2}} \\
\hline \tan & (\sqrt{3}-\sqrt{2})(\sqrt{2}-1) & 2-\sqrt{3} & \frac{\sqrt{10+2 \sqrt{5}}}{4} & \sqrt{2}-1 & \sqrt{5-2 \sqrt{5}} & \sqrt{2}+1 & 2+\sqrt{3} \\
\hline \cot & (\sqrt{3}+\sqrt{2})(\sqrt{2}+1) & 2+\sqrt{3} & \sqrt{(5+2 \sqrt{5}}) & \sqrt{2}+1 & \sqrt{\left(1+\frac{2}{\sqrt{5}}\right)} & \sqrt{2}-1 & 2-\sqrt{3} \\
\hline
\end{array}
\)

Example 2.68: Find the angle \(\theta\) whose cosine is equal to its tangent.

Solution:
\(
\begin{aligned}
& \text { Given, } \cos \theta=\tan \theta \quad \Rightarrow \quad \cos ^2 \theta=\sin \theta \\
& \Rightarrow \quad 1-\sin ^2 \theta=\sin \theta \text { or } \sin ^2 \theta+\sin \theta-1=0 \\
& \Rightarrow \quad \sin \theta=\frac{-1 \pm \sqrt{5}}{2}=2 \frac{\sqrt{5}-1}{4}=2 \sin 18^{\circ} \\
& \Rightarrow \quad \theta=\sin ^{-1}\left(2 \sin 18^{\circ}\right)
\end{aligned}
\)

Example 2.69: Find the value of \(\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}\).

Solution:
\(
\begin{aligned}
& \cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ} \\
& =\left(\cos 12^{\circ}+\cos 132^{\circ}\right)+\left(\cos 84^{\circ}+\cos 156^{\circ}\right) \\
& =2 \cos \left(\frac{12^{\circ}+132^{\circ}}{2}\right) \cos \left(\frac{132^{\circ}-12^{\circ}}{2}\right)+2 \cos \left(\frac{84^{\circ}+156^{\circ}}{2}\right) \cos \left(\frac{156^{\circ}-84^{\circ}}{2}\right) \\
& =2 \cos 72^{\circ} \cos 60^{\circ}+2 \cos 120^{\circ} \cos 36^{\circ} \\
& =2 \sin 18^{\circ} \cos 60^{\circ}+2 \cos 120^{\circ} \cos 36^{\circ} \\
& =2\left(\frac{\sqrt{5}-1}{4}\right) \frac{1}{2}+2\left(-\frac{1}{2}\right)\left(\frac{\sqrt{5}+1}{4}\right)=-\frac{1}{2}
\end{aligned}
\)

Example 2.70: Prove that \(\cos 36^{\circ} \cos 72^{\circ} \cos 108^{\circ} \cos 144^{\circ}=\frac{1}{16}\).

Solution:
\(
\begin{aligned}
& \cos 36^{\circ} \cos 72^{\circ} \cos 108^{\circ} \cos 144^{\circ} \\
& =\cos 36^{\circ} \sin 18^{\circ}\left(-\sin 18^{\circ}\right)\left(-\cos 36^{\circ}\right) \\
& =\cos ^2 36^{\circ} \sin ^2 18^{\circ}=\left(\frac{\sqrt{5}+1}{4}\right)^2\left(\frac{\sqrt{5}-1}{4}\right)^2 \\
& =\left[\left(\frac{\sqrt{5}+1}{4}\right)\left(\frac{\sqrt{5}-1}{4}\right)\right]^2=\frac{1}{16}
\end{aligned}
\)

SUM OF SINES OR COSINES OF \(N\) ANGLES IN A.P.

\(\sin \alpha+\sin (\alpha+\alpha \beta)+\sin (\alpha+2 \beta)+\cdots+\sin (\alpha+\overline{n-1} \beta)=\frac{\sin \frac{n \beta}{2}}{\sin \frac{\beta}{2}} \times \sin \left[\alpha+(n-1) \frac{\beta}{2}\right]\)

Proof: Let \(S=\sin \alpha+\sin (\alpha+\beta)+\sin (\alpha+2 \beta)+\cdots+\sin (\alpha+\overline{n-1} \beta)\)
Here angles are in A.P. and common difference of angles \(=\beta\)
Multiplying both sides by \(2 \sin \frac{\beta}{2}\), we get
\(
2 S \sin \frac{\beta}{2}=2 \sin \alpha \sin \frac{\beta}{2}+2 \sin (\alpha+\beta) \sin \frac{\beta}{2}+\cdots+2 \sin (\alpha+\overline{n-1} \beta) \sin \frac{\beta}{2} \dots(i)
\)
\(
\begin{aligned}
& \text { Now, } 2 \sin \alpha \sin \frac{\beta}{2}=\cos \left(\alpha-\frac{\beta}{2}\right)-\cos \left(\alpha+\frac{\beta}{2}\right) \\
& 2 \sin (\alpha+\beta) \sin \frac{\beta}{2}=\cos \left(\alpha+\frac{\beta}{2}\right)-\cos \left(\alpha+\frac{3 \beta}{2}\right) \\
& 2 \sin (\alpha+2 \beta) \sin \frac{\beta}{2}=\cos \left(\alpha+\frac{3 \beta}{2}\right)-\cos \left(\alpha+\frac{5 \beta}{2}\right) \\
& \vdots \\
\end{aligned}
\)
\(
\frac{2 \sin (\alpha+\overline{n-1} \beta) \sin \frac{\beta}{2}=\cos \left[\alpha+(2 n-3) \frac{\beta}{2}\right]-\cos \left[\alpha(2 n-1) \frac{\beta}{2}\right]}{\text { Adding, we get R.H.S. of .Eq. }(i)=\cos \left(\alpha-\frac{\beta}{2}\right)-\cos \left[\alpha+(2 n-1) \frac{\beta}{2}\right]} .
\)
\(
\begin{aligned}
& \text { or } 2 \sin \frac{\beta}{2} S=2 \sin \left[\alpha+(n-1) \frac{\beta}{2}\right] \sin \frac{n \beta}{2} \\
& \Rightarrow S=\frac{\sin \frac{n \beta}{2}}{\sin \frac{\beta}{2}} \sin \left[\alpha+(n-1) \frac{\beta}{2}\right]
\end{aligned}
\)
In the above result replacing \(\alpha\) by \(\pi / 2+\alpha\), we get
\(
\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\cdots+\cos (\alpha+\overline{n-1} \beta)=\frac{\sin \frac{n \beta}{2}}{\sin \frac{\beta}{2}} \cos \left[\alpha+(n-1) \frac{\beta}{2}\right]
\)

Example 2.71: Find the value of \(\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}\).

Solution:
\(
\begin{aligned}
S & =\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7} \\
& =\frac{\sin \left(3 \frac{\pi}{7}\right)}{\sin \left(\frac{\pi}{7}\right)} \cos \left(\frac{\pi}{7}+\frac{3 \pi}{7}\right) \\
& =\frac{2 \sin \left(\frac{3 \pi}{7}\right) \cos \left(\frac{4 \pi}{7}\right)}{2 \sin \left(\frac{2 \pi}{7}\right)}=\frac{\sin \left(\frac{7 \pi}{7}\right)-\sin \left(\frac{\pi}{7}\right)}{2 \sin \left(\frac{2 \pi}{7}\right)}=-\frac{1}{2}
\end{aligned}
\)

Example 2.72: Prove that \(\sin \theta+\sin 3 \theta+\sin 5 \theta+\cdots+\sin (2 n-1) \theta=\frac{\sin ^2 n \theta}{\sin \theta}\).

Solution:
\(
\begin{aligned}
\sin \theta+\sin 3 \theta+\sin 5 \theta+\cdots+\sin (2 n-1) \theta & =\frac{\sin \left[n\left(\frac{2 \theta}{2}\right)\right]}{\sin \left(\frac{2 \theta}{2}\right)} \sin \left(\frac{\theta+(2 n-1) \theta}{2}\right) \\
& =\frac{\sin ^2 n \theta}{\sin \theta}
\end{aligned}
\)

CONDITIONAL IDENTITIES

Some Standard Identities in Triangle

Case-I: \(\tan A+\tan B+\tan C=\tan A \tan B \tan C\)

Proof:
In \(\triangle A B C\), we have \(A+B+C=\pi\)
\(
\begin{aligned}
& \Rightarrow A+B=\pi-C \\
& \Rightarrow \tan (A+B)=\tan (\pi-C) \\
& \Rightarrow \frac{\tan A+\tan B}{1-\tan A \tan B}=-\tan C \\
& \Rightarrow \tan A+\tan B=-\tan C+\tan A \tan B \tan C \\
& \Rightarrow \tan A+\tan B+\tan C=\tan A \tan B \tan C
\end{aligned}
\)

Case-II: \(\tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{C}{2} \tan \frac{B}{2}+\tan \frac{C}{2} \tan \frac{A}{2}=1\)

Proof:
Since \(A+B+C=\pi\), we have \(\frac{A}{2}+\frac{B}{2}=\frac{\pi}{2}-\frac{C}{2}\)
\(
\begin{aligned}
& \Rightarrow \tan \left(\frac{A}{2}+\frac{B}{2}\right)=\tan \left(\frac{\pi}{2}-\frac{C}{2}\right)=\cot \frac{C}{2} \\
& \Rightarrow \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}=\frac{1}{\tan \frac{C}{2}} \\
& \Rightarrow \tan \frac{A}{2} \tan \frac{C}{2}+\tan \frac{B}{2} \tan \frac{C}{2}=1-\tan \frac{A}{2} \tan \frac{B}{2} \\
& \Rightarrow \tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}=1
\end{aligned}
\)

Case-III: \(\sin 2 A+\sin 2 B+\sin 2 C=4 \sin A \sin B \sin C\)

Proof:
\(
\begin{aligned}
(\sin 2 A+\sin 2 B)+\sin 2 C & =2 \sin (A+B) \cos (A-B)+\sin 2 C \\
& =2 \sin (\pi-C) \cos (A-B)+\sin 2 C \\
& =2 \sin C \cos (A-B)+2 \sin C \cos C \\
& =2 \sin C[\cos (A-B)+\cos C] \\
& =2 \sin C[\cos (A-B)+\cos \{\pi-(A+B)\}] \\
& =2 \sin C[\cos (A-B)-\cos (A+B)] \\
& =2 \sin C \times 2 \sin A \sin B=4 \sin A \sin B \sin C
\end{aligned}
\)

Case-IV: \(\cos 2 A+\cos 2 B+\cos 2 C=-1-4 \cos A \cos B \cos C\)

Proof:
\(
\begin{aligned}
& (\cos 2 A+\cos 2 B)+\cos 2 C \\
& =2 \cos (A+B) \cos (A-B)+2 \cos ^2 C-1 \\
& =2 \cos (\pi-C) \cos (A-B)+2 \cos ^2 C-1 \\
& =-2 \cos C \cos (A-B)+2 \cos ^2 C-1 \\
& =-2 \cos C[\cos (A-B)-\cos C]-1 \\
& =-2 \cos C[\cos (A-B)-\cos \{\pi-(A+B)\}]-1 \\
& =-2 \cos C[\cos (A-B)+\cos (A+B)]-1 \\
& =-1-4 \cos A \cos B \cos C
\end{aligned}
\)

Case-V: \(\cos A+\cos B+\cos C=1+4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\)

Proof:
\(
\begin{aligned}
& (\cos A+\cos B)+\cos C-1 \\
& =2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}+\cos C-1 \\
& =2 \cos \left(\frac{\pi}{2}-\frac{C}{2}\right) \cos \frac{A-B}{2}+\cos C-1 \\
& =2 \sin \frac{C}{2} \cos \frac{A-B}{2}+1-2 \sin ^2 \frac{C}{2}-1 \\
& =2 \sin \frac{C}{2} \cos \frac{A-B}{2}-2 \sin ^2 \frac{C}{2} \\
& =2 \sin \frac{C}{2}\left[\cos \frac{A-B}{2}-\sin \frac{C}{2}\right] \\
& =2 \cos \frac{C}{2}\left[\cos \frac{A-B}{2}-\sin \left(\frac{\pi}{2}-\frac{A+B}{2}\right)\right] \\
& =2 \sin \frac{C}{2}\left[\cos \frac{A-B}{2}-\cos \frac{A+B}{2}\right] \\
& =2 \sin \frac{C}{2}\left(2 \sin \frac{A}{2} \sin \frac{B}{2}\right)=4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}
\end{aligned}
\)

Case-VI: \(\sin A+\sin B+\sin C=4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}\)

Proof:
\(
\begin{aligned}
& (\sin A+\sin B)+\sin C \\
& =2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}+\sin C \\
& =2 \sin \left(\frac{\pi}{2}-\frac{C}{2}\right) \cos \frac{A-B}{2}+\sin C \\
& =2 \cos \frac{C}{2} \cos \frac{A-B}{2}+2 \sin \frac{C}{2} \cos \frac{C}{2} .
\end{aligned}
\)
\(
\begin{aligned}
& =2 \cos \frac{C}{2}\left[\cos \frac{A-B}{2}+\sin \frac{C}{2}\right] \\
& =2 \cos \frac{C}{2}\left[\cos \frac{A-B}{2}+\sin \left(\frac{\pi}{2}-\frac{A+B}{2}\right)\right] \\
& =2 \cos \frac{C}{2}\left[\cos \frac{A-B}{2}+\cos \frac{A+B}{2}\right] \\
& =4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}
\end{aligned}
\)
\(
\begin{aligned}
&\text { Note: }\\
&\tan A+\tan B+\tan C=\tan A \tan B \tan C \text { is true for } A+B+C=n \pi \text {, where } n \in N \text {. }
\end{aligned}
\)

Example 2.73: If \(A+B+C=180^{\circ}\), prove that \(\cos ^2 A+\cos ^2 B+\cos ^2 C=1-2 \cos A \cos B \cos C\).

Solution:
\(
\begin{aligned}
\cos ^2 A+\cos ^2 B+\cos ^2 C & =\frac{1+\cos 2 A}{2}+\frac{1+\cos 2 B}{2}+\frac{1+\cos 2 C}{2} \\
& =\frac{1}{2}(\cos 2 A+\cos 2 B+\cos 2 C)+\frac{3}{2} \\
& =\frac{1}{2}(-1-4 \cos A \cos B \cos C)+\frac{3}{2} \\
& =1-2 \cos A \cos B \cos C .
\end{aligned}
\)

Example 2.74: Prove that in triangle \(A B C, \cos ^2 A+\cos ^2 B-\cos ^2 C=1-2 \sin A \sin B \cos C\).

Solution:
\(
\begin{aligned}
\cos ^2 A+\cos ^2 B-\cos ^2 C & =\cos ^2 A+\sin ^2 C-\sin ^2 B \\
& =\cos ^2 A+\sin (C+B) \sin (C-B) \\
& =1-\sin ^2 A+\sin A \sin (C-B) \\
& =1-\sin A[\sin A-\sin (C-B)] \\
& =1-\sin A[\sin (B+C)-\sin (C-B)] \\
& =1-2 \sin A \sin B \cos C
\end{aligned}
\)

Example 2.75: In triangle \(A B C\), prove that
\(
\sin (B+C-A)+\sin (C+A-B)+\sin (A+B-C)=4 \sin A \sin B \sin C .
\)

Solution:
\(\begin{aligned} \sin (B+C-A)+\sin (C+A-B)+\sin (A+B-C) & =\sin (\pi-2 A)+\sin (\pi-2 B)+\sin (\pi-2 C) \\ & =\sin 2 A+\sin 2 B+\sin 2 C=4 \sin A \sin B \sin C\end{aligned}\)

Example 2.76: If \(x+y+z=x y z\), prove that \(\frac{2 x}{1-x^2}+\frac{2 y}{1-y^2}+\frac{2 z}{1-z^2}=\frac{2 x}{1-x^2} \frac{2 y}{1-y^2} \frac{2 z}{1-z^2}\).

Solution: \(\text { Let } x=\tan A, y=\tan B, z=\tan C\)
Now \(x+y+z=x y z\)
\(\Rightarrow \tan A+\tan B+\tan C=\tan A \tan B \tan C\)
\(\Rightarrow A+B+C=n \pi\)
\(\Rightarrow 2 A+2 B+2 C=2 n \pi\)
\(\Rightarrow \tan 2 A+\tan 2 B+\tan 2 C=\tan 2 A \tan 2 B \tan 2 C\)
\(\Rightarrow \frac{2 \tan A}{1-\tan ^2 A}+\frac{2 \tan B}{1-\tan ^2 B}+\frac{2 \tan C}{1-\tan ^2 C}=\frac{2 \tan A}{1-\tan ^2 A} \frac{2 \tan B}{1-\tan ^2 B} \frac{2 \tan C}{1-\tan ^2 C}\)
\(\Rightarrow \frac{2 x}{1-x^2}+\frac{2 y}{1-y^2}+\frac{2 z}{1-z^2}=\frac{2 x}{1-x^2}+\frac{2 y}{1-y^2} \frac{2 z}{1-z^2}\)

Example 2.77: If \(A+B+C=\pi\), prove that \(\sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}-\sin ^2 \frac{C}{2}=1-2 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}\).

Solution:
\(
\begin{aligned}
\sin ^2 \frac{A}{2}-\sin ^2 \frac{C}{2}+\sin ^2 \frac{B}{2} & =\sin \left(\frac{A+C}{2}\right) \sin \left(\frac{A-C}{2}\right)+1-\cos ^2 \frac{B}{2} \\
& =\cos \left(\frac{B}{2}\right) \sin \left(\frac{A-C}{2}\right)-\cos ^2 \frac{B}{2}+1 \\
& =\cos \left(\frac{B}{2}\right)\left[\sin \left(\frac{A-C}{2}\right)-\cos \frac{B}{2}\right]+1 \\
& =\cos \left(\frac{B}{2}\right)\left[\sin \left(\frac{A-C}{2}\right)-\sin \left(\frac{A+C}{2}\right)\right]+1 \\
& =1-2 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}
\end{aligned}
\)

Example 2.78: The product of the sines of the angles of a triangle is \(p\) and the product of their cosines is \(q\). Show that the tangents of the angles are the roots of the equation \(q x^2-p x^2+(1+q)\) \(x-p=0\).

Solution: From the question, \(\sin A \sin B \sin C=p\) and \(\cos A \cos B \cos C=q\)
\(\therefore \tan A \tan B \tan C=\frac{p}{q} \dots(i)\)
Also, \(\tan A+\tan B+\tan C=\tan A \tan B \tan C=\frac{p}{q} \dots(ii)\)
\(
\begin{aligned}
& \text { Now, } \tan A \tan B+\tan B \tan C+\tan C \tan A \\
& =\frac{\sin A \sin B \cos C+\sin B \sin C \cos A+\sin C \sin A \cos B}{\cos A \cos B \cos C} \\
& \left.=\frac{1}{2 q}\left[\left(\sin ^2 A+\sin ^2 B-\sin ^2 C\right)+\left(\sin ^2 B+\sin ^2 C \sin ^2 A\right)+\sin ^2 C+\sin ^2 A-\sin ^2 B\right)\right] \\
& \quad\left[\because A+B+C=\pi \text { and } 2 \sin A \sin B \cos C=\sin ^2 A+\sin ^2 B-\sin ^2 C\right]
\end{aligned}
\)
\(
=\frac{1}{2 q}\left[\sin ^2 A+\sin ^2 B+\sin ^2 C\right]=\frac{1}{4 q}[3-(\cos 2 A+\cos 2 B+\cos 2 C)]=\frac{1}{q}[1+\cos A \cos B \cos C]=\frac{1}{q}(1+q)
\)
The equation whose roots are \(\tan A, \tan B \tan C\) will be given by
\(
x^3-(\tan A+\tan B+\tan C) x^2+(\tan A \tan B+\tan B \tan C+\tan C \tan A) x-\tan A \tan B \tan C=0
\)
or \(x^3-\frac{p}{q} x^2+\frac{1+q}{q} x-\frac{p}{q}=0\), or \(q x^3-p x^2+(1+q) x-p=0\)

SOME IMPORTANT RESULTS AND THEIR APPLICATIONS

Case-I:  \(\cos A \cos (60-A) \cos (60+A)=\frac{1}{4} \cos 3 A\)

Proof: We have
\(
\begin{aligned}
\text { L.H.S. } & =\cos A \cos (60-A) \cos (60+A) \\
& =\cos A\left(\cos ^2 60^{\circ}-\sin ^2 A\right) \\
& =\cos A\left(\frac{1}{4}-\sin ^2 A\right)=\cos A\left(\frac{1}{4}-\left(1-\cos ^2 A\right)\right)=\cos A\left(-\frac{3}{4}+\cos ^2 A\right) \\
& =\frac{1}{4} \cos A\left(-3+4 \cos ^2 A\right)=\frac{1}{4}\left(4 \cos ^3 A-3 \cos A\right) \\
& =\frac{1}{4} \cos 3 A=\text { R.H.S }
\end{aligned}
\)

Case-II: \(\sin A \sin (60-A) \sin (60+A)=\frac{1}{4} \sin 3 A\)

Proof: We have
\(
\begin{aligned}
\text { L.H.S. } & =\sin A \sin (60-A) \sin (60+A) \\
& =\sin A\left(\sin ^2 60^{\circ}-\sin ^2 A\right) \left[\because \sin (A+B) \sin (A-B)=\sin ^2 A-\sin ^2 B\right] \\
& =\sin A\left(\frac{3}{4}-\sin ^2 A\right)=\frac{1}{4} \sin A\left(3-4 \sin ^2 A\right) \\
& =\frac{1}{4}\left(3 \sin A-4 \sin ^3 A\right) \\
& =\frac{1}{4} \sin 3 A=\text { R.H.S }
\end{aligned}
\)

Case-III: \(\tan \alpha \tan \left(60^{\circ}-\alpha\right) \tan \left(60^{\circ}+\alpha\right)=\tan 3 \alpha\)

Proof: 
\(
\begin{aligned}
& \tan \left(60^{\circ}-\alpha\right)=\frac{\sqrt{3}-\tan \alpha}{1+\sqrt{3} \tan \alpha} \\
& \tan \left(60^{\circ}+\alpha\right)=\frac{\sqrt{3}+\tan \alpha}{1-\sqrt{3} \tan \alpha}
\end{aligned}
\)
\(
\mathrm{LHS}=\tan \alpha \cdot \frac{(\sqrt{3}-\tan \alpha)(\sqrt{3}+\tan \alpha)}{(1+\sqrt{3} \tan \alpha)(1-\sqrt{3} \tan \alpha)}
\)
\(
\tan \alpha \cdot \frac{(\sqrt{3})^2-(\tan \alpha)^2}{(1)^2-(\sqrt{3} \tan \alpha)^2}
\)
\(
\frac{3 \tan \alpha-\tan ^3 \alpha}{1-3 \tan ^2 \alpha}
\)
\(
=\tan 3 \alpha
\)

Example 2.79: Prove that \(\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}=\frac{1}{16}\).

Solution:
\(
\begin{aligned}
\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ} \cos 60^{\circ} & =\cos 20^{\circ} \cos \left(60^{\circ}-20^{\circ}\right) \cos \left(60^{\circ}+20^{\circ}\right) \cos 60^{\circ} \\
& =\frac{1}{4} \cos \left(3 \times 20^{\circ}\right) \cos 60^{\circ}=\frac{1}{4} \cos ^2 60^{\circ}=\frac{1}{4} \times \frac{1}{4}=\frac{1}{16}
\end{aligned}
\)

Example 2.80: Prove that \(\sin 10^{\circ} \sin 30^{\circ} \sin 50^{\circ} \sin 70^{\circ}=\frac{1}{16}\).

Solution:
\(
\begin{aligned}
\sin 10^{\circ} \sin 30^{\circ} \sin 50^{\circ} \sin 70^{\circ} & =\sin 10^{\circ} \sin \left(60^{\circ}-10^{\circ}\right) \sin \left(60^{\circ}+10^{\circ}\right) \sin 30^{\circ} \\
& =\frac{1}{4} \sin \left(3 \times 10^{\circ}\right) \sin 30^{\circ}=\frac{1}{4} \sin ^2 30^{\circ}=\frac{1}{16}
\end{aligned}
\)

Example 2.81: Prove that \(\tan 20^{\circ} \tan 40^{\circ} \tan 80^{\circ}=\tan 60^{\circ}\).

Solution:
\(
\begin{aligned}
& \tan 20^{\circ} \tan 40^{\circ} \tan 80^{\circ} \\
& =\tan 20^{\circ} \tan \left(60^{\circ}-20^{\circ}\right) \tan \left(60^{\circ}+20^{\circ}\right)=\tan \left(3 \times 20^{\circ}\right)=\tan 60^{\circ}
\end{aligned}
\)

Case-IV: \(\cos A \cos 2 A \cos 2^2 A \cos 2^3 A \cdots \cos 2^{n-1} A=\frac{\sin 2^n A}{2^n \sin A}\)

Proof: \(\text { L.H.S. }=\cos A \cos 2 A \cos 2^2 A \cos 2^3 A \cdots \cos 2^{n-1} A\)
\(
\begin{aligned}
& =\frac{1}{2 \sin A}\left[(2 \sin A \cos A) \cos 2 A \cos 2^2 A \cos 2^3 A \cdots \cos 2^{n-1} A\right] \\
& =\frac{1}{2 \sin A}\left[\left(\sin 2 A \cos 2 A \cos 2^2 A \cos 2^3 A \cdots \cos 2^{n-1} A\right]\right. \\
& =\frac{1}{2^2 \sin A}\left[(2 \sin 2 A \cos 2 A) \cos 2^2 A \cos 2^3 A \cdots \cos 2^{n-1} A\right] \\
& =\frac{1}{2^2 \sin A}\left[\sin 2(2 A) \cdot \cos 2^2 A \cos 2^3 A \cdots \cos 2^{n-1} A\right] \\
& =\frac{1}{2^3 \sin A}\left[\left(2 \sin 2^2 A \cos 2^2 A\right) \cos 2^3 A \cdots \cos 2^{n-1} A\right]
\end{aligned}
\)
\(
\begin{aligned}
& =\frac{1}{2^3 \sin A}\left[\sin \left(2 \times 2^2 A\right) \cos 2^3 A \cdots \cos 2^{n-1} A\right] \\
& =\frac{1}{2^3 \sin A}\left[\left(\sin 2^3 A \cos 2^3 A \cos 2^4 A \cdots \cos 2^{n-1} A\right]\right. \\
& \cdots \\
& \cdots \\
& =\frac{1}{2^{n-1} \sin A}\left[\sin 2^{n-1} A \cos 2^{n-1} A\right] \\
& =\frac{1}{2^n \sin A}\left[2 \sin 2^{n-1} A \cos 2^{n-1} A\right] \\
& =\frac{1}{2^n \sin A} \sin \left(2 \times 2^{n-1} A\right) \\
& =\frac{1}{2^n \sin A} \sin 2^n A=\text { R.H.S. }
\end{aligned}
\)

Example 2.82: If \(\theta=\frac{\pi}{2^n+1}\), show that \(\cos \theta \cos 2 \theta ~2 \cos 2^2 \theta \cdots \cos 2^{n-1} \theta=\frac{1}{2^n}\).

Solution: In the above result, put \(\theta=\frac{\pi}{2^n+1}\)
\(
\begin{aligned}
\text { R.H.S. }=\frac{\sin 2^n \theta}{2^n \sin \theta}=\frac{\sin \left(\frac{\pi}{2^n+1}\right) 2^n}{2^n \sin \left(\frac{\pi}{2^n+1}\right)} & =\frac{\sin \left(\frac{2^n+1-1}{2^n+1}\right) \pi}{2^n \sin \left(\frac{\pi}{2^n+1}\right)} \\
& =\frac{\sin \left(\pi-\frac{\pi}{2^n+1}\right)}{2^n \sin \left(\frac{\pi}{2^n+1}\right)} \\
& =\frac{\sin \left(\frac{\pi}{2^n+1}\right)}{2^n \sin \left(\frac{\pi}{2^n+1}\right)} \\
& =\frac{1}{2^n}
\end{aligned}
\)

Example 2.83: Prove that \(\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{14 \pi}{15}=\frac{1}{16}\).

Solution: We have
\(
\text { L.H.S. }=\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \left(\pi-\frac{\pi}{15}\right)
\)
\(
\begin{aligned}
& =\left(\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15}\right)\left(-\cos \frac{\pi}{15}\right) \\
& =-\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \\
& =-\cos A \cos 2 A \cos 2^2 A \cos 2^3 A, \text { where } A=\pi / 15 \\
& =-\left[\frac{\sin 2^4 A}{2^4 \sin A}\right]=-\frac{\sin 16 A}{2^4 \sin A} \\
& =-\frac{\sin (15 A+A)}{16 \sin A}=\frac{-\sin (\pi+A)}{16 \sin A} [\because 15 A=\pi]\\
& =\frac{\sin A}{16 \sin A}=\frac{1}{16}=\frac{1}{16}=\text { R.H.S. }
\end{aligned}
\)

Example 2.84: Prove that \(\sin 6^{\circ} \sin 42^{\circ} \sin 66^{\circ} \sin 78^{\circ}=\frac{1}{16}\).

Solution:
\(
\begin{aligned}
& \sin 6^{\circ} \sin 42^{\circ} \sin 66^{\circ} \sin 78^{\circ} \\
& =\sin 6^{\circ} \cos 48^{\circ} \cos 24^{\circ} \cos 12^{\circ} \\
& =\sin 6^{\circ} \frac{2^3 \sin 12^{\circ} \cos 12^{\circ} \cos 24^{\circ} \cos 48^{\circ}}{2^3 \sin 12^{\circ}} \\
& =\sin 6^{\circ} \frac{\sin 96^{\circ}}{2^3 \sin 12^{\circ}} \\
& =\frac{2 \sin 6^{\circ} \cos 6^{\circ}}{2^4 \sin 12^{\circ}}=\frac{\sin 12^{\circ}}{2^4 \sin 12^{\circ}}=\frac{1}{16}
\end{aligned}
\)

Example 2.85: In \(\triangle A B C, \tan A+\tan B+\tan C \geq 3 \sqrt{3}\), where \(A, B, C\) are acute angles.

Solution: In \(\triangle A B C\),
\(\tan A+\tan B+\tan C=\tan A \tan B \tan C\)
Also, \(\frac{\tan A+\tan B+\tan C}{3} \geq \sqrt[3]{\tan A \tan B \tan C} \text { [since A.M. } \geq \text { G.M.] }\)
\(
\begin{aligned}
& \Rightarrow \tan A \tan B \tan C \geq \sqrt[3]{\tan A \tan B \tan C} \\
& \Rightarrow \tan ^2 A \tan ^2 B \tan ^2 C \geq 27 \\
& \Rightarrow \tan A \tan B \tan C \geq 3 \sqrt{3} \text { [cubing both sides] }\\
& \Rightarrow \tan A+\tan B+\tan C \geq 3 \sqrt{3}
\end{aligned}
\)

Example 2.86: In \(\triangle A B C\), prove that \(\cos A+\cos B+\cos C \leq 3 / 2\).

Solution:
\(
\begin{aligned}
& \text { Let } \cos A+\cos B+\cos C=x \\
& \Rightarrow 2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)+1-2 \sin ^2 \frac{C}{2}=x \\
& \Rightarrow 2 \sin \frac{C}{2} \cos \left(\frac{A-B}{2}\right)+1-2 \sin ^2 \frac{C}{2}=x \\
& \Rightarrow 2 \sin ^2 \frac{C}{2}-2 \sin \frac{C}{2} \cos \left(\frac{A-B}{2}\right)+x-1=0
\end{aligned}
\)
This is quadratic in \(\sin C / 2\), which is real. So, discriminant \(D \geq 0\).
\(
\begin{aligned}
& 4 \cos ^2\left(\frac{A-B}{2}\right)-4 \times 2(x-1) \geq 0 \\
& \Rightarrow 2(x-1) \leq \cos ^2\left(\frac{A-B}{2}\right) \\
& \Rightarrow 2(x-1) \leq 1 \\
& \Rightarrow x \leq 3 / 2
\end{aligned}
\)
Thus, \(\cos A+\cos B+\cos C \leq 3 / 2\)

Note: Since \(\cos A+\cos B+\cos C=1+4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\)
We have \(\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \leq \frac{1}{8}\).
Students are advised to remember this as a standard result.

Example 2.87: Find the least value of \(\sec A+\sec B+\sec C\) in an acute angle triangle.

Solution: In an acute angle triangle, \(\sec A, \sec B\) and \(\sec C\) are positive. Now A.M. \(\geq\) H.M.
\(
\Rightarrow \quad \frac{\sec A+\sec B+\sec C}{3} \geq \frac{3}{\cos A+\cos B+\cos C}
\)
But in \(\triangle A B C, \cos A+\cos B+\cos C \leq 3 / 2\)
\(
\begin{aligned}
& \Rightarrow \quad \frac{\sec A+\sec B+\sec C}{3} \geq 2 \\
& \Rightarrow \quad \sec A+\sec B+\sec C \geq 6
\end{aligned}
\)

EXERCISES

Q1. Are the set of angles \(\alpha\) and \(\beta\) given by \(\alpha=\left(2 n+\frac{1}{2}\right), \pi \pm A\) and \(\beta=m \pi+(-1)^m\left(\frac{\pi}{2}-A\right)[latex] same, where [latex]n, m \in I\)?

Solution: Let \(m=2 k\), i.e., \(m\) is even where \(k \in I\)
Now, \(\beta=2 k \pi+\frac{\pi}{2}-A=\left(2 k+\frac{1}{2}\right) \pi-A \dots(i)\)
If \(m=2 k+1\), i.e., \(m\) is odd, then
\(
\beta=(2 k+1) \pi-\left(\frac{\pi}{2}-A\right)=\left(2 k+\frac{1}{2}\right) \pi+A \dots(ii)
\)
From Eqs. (i) and (ii), \(\beta\) can be expressed as
\(
\beta=\left(2 k+\frac{1}{2}\right) \pi \pm A, k \in I
\)
which is same as \(\alpha\).

Q2. If \(A B C\) is a triangle and \(\tan \frac{A}{2}, \tan \frac{B}{2}, \tan \frac{C}{2}\) are in H.P., then find the minimum value of \(\cot B / 2\).

Solution:
\(
\begin{aligned}
& A+B+C=\pi \\
& \Rightarrow \frac{A}{2}+\frac{B}{2}=\frac{\pi}{2}-\frac{C}{2} \\
& \Rightarrow \cot \left(\frac{A}{2}+\frac{B}{2}\right)=\cot \left(\frac{\pi}{2}-\frac{C}{2}\right) \\
& \Rightarrow \frac{\cot \frac{A}{2} \cot \frac{B}{2}-1}{\cot \frac{A}{2}+\cot \frac{B}{2}}=\tan \frac{C}{2}=\frac{1}{\cot \frac{C}{2}} \\
& \Rightarrow \cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}=\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2} \dots(i)
\end{aligned}
\)
But \(\tan \frac{A}{2}, \tan \frac{B}{2}, \tan \frac{C}{2}\) are in H.P.
\(\therefore \cot \frac{A}{2}, \cot \frac{B}{2}, \cot \frac{C}{2}\) are in A.P.
So, \(\cot \frac{A}{2}+\cot \frac{C}{2}=2 \cot \frac{B}{2}\).
Hence, Eq. (i) becomes \(\cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}=3 \cot \frac{B}{2} \Rightarrow \cot \frac{A}{2} \cot \frac{C}{2}=3\)
\(\Rightarrow\) G.M. of \(\cot \frac{A}{2}\) and \(\cot \frac{C}{2}=\sqrt{\cot \frac{A}{2} \cot \frac{C}{2}}=\sqrt{3}\)
and A.M. of \(\cot \frac{A}{2}\) and \(\cot \frac{C}{2}=\frac{\cot \frac{A}{2}+\cot \frac{C}{2}}{2}=\cot \frac{B}{2}\)
But A.M. \(\geq\) G.M.
\(
\Rightarrow \quad \cot \frac{B}{2} \geq \sqrt{3}
\)
Therefore, the minimum value of \(\cot B / 2\) is \(\sqrt{3}\).

Q3. Find the sum of the series \(\operatorname{cosec} \theta+\operatorname{cosec} 2 \theta+\operatorname{cosec} 4 \theta+\cdots\) to \(n\) terms.

Solution:
\(
\begin{aligned}
& \operatorname{cosec} \theta=\frac{1}{\sin \theta} \\
& \quad=\frac{1}{\sin \theta} \frac{\sin \theta / 2}{\sin \theta / 2}=\frac{\sin \left(\theta-\frac{\theta}{2}\right)}{\sin \theta \sin \left(\frac{\theta}{2}\right)}=\frac{\sin \theta \cos \frac{\theta}{2}-\cos \theta \sin \frac{\theta}{2}}{\sin \theta \sin \frac{\theta}{2}} \\
& \therefore \operatorname{cosec} \theta=\cot \frac{\theta}{2}-\cot \theta \\
& \text { Similarly, } \operatorname{cosec} 2 \theta=\cot \theta-\cot 2 \theta \\
& \operatorname{cosec} 4 \theta=\cot 2 \theta-\cot 4 \theta \\
& \vdots \\
& \text { cosec } 2^{n-1} \theta=\cot 2^{n-2} \theta-\cot 2^{n-1} \theta \\
& \text { Therefore, sum }=\cot \frac{\theta}{2}-\cot 2^{n-1} \theta
\end{aligned}
\)

Q4. In \(\triangle A B C\), if \(\sin ^3 \theta=\sin (A-\theta) \sin (B-\theta) \sin (C-\theta)\), prove that \(\cot \theta=\cot A+\cot B+\cot C\).

Solution:
\(
\begin{aligned}
& \text { Let } \cot \theta=\cot A+\cot B+\cot C \\
& \Rightarrow \cot \theta-\cot A=\cot B+\cot C \\
& \Rightarrow \frac{\sin (A-\theta)}{\sin A \sin \theta}=\frac{\sin (B+C)}{\sin B \sin C} \\
& \Rightarrow \sin (A-\theta)=\frac{\sin ^2 A \sin \theta}{\sin B \sin C} \dots(i)
\end{aligned}
\)
Similarly, \(\sin (B-\theta)=\frac{\sin ^2 B \sin \theta}{\sin A \sin C} \dots(ii)\)
and \(\sin (C-\theta)=\frac{\sin ^2 C \sin \theta}{\sin A \sin B} \dots(iii)\)
By multiplying corresponding sides of Eqs. (i), (ii), and (iii), we have \(\sin ^3 \theta=\sin (A-\theta) \sin (B-\theta) \sin (C-\theta)\)

Q5. In triangle \(A B C\), prove that \(\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2} \leq \frac{3}{2}\). Hence, deduce that \(\cos \frac{\pi+A}{4} \cos \frac{\pi+B}{4} \cos \frac{\pi+C}{4} \leq \frac{1}{8}\).

Solution:
Let \(\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2}=k\)
or \(2 \sin \frac{A+B}{4} \cos \frac{A-B}{4}+\cos \frac{A+B}{2}=k\)
\(
\Rightarrow \quad 2 \sin ^2 \frac{A+B}{4}-2 \cos \frac{A-B}{4} \sin \frac{A+B}{4}+k-1=0
\)
\(
\begin{aligned}
& \text { Since } \sin \frac{A+B}{4} \text { is real, } 4 \cos ^2 \frac{A-B}{4}-8(k-1) \geq 0 \\
& \Rightarrow \quad 2(k-1) \leq \cos ^2 \frac{A-B}{4} \leq 1 \Rightarrow k \leq 3 / 2
\end{aligned}
\)
\(
\begin{aligned}
& \text { Hence, } 2 \sin \frac{A+B}{4}\left[\cos \frac{A-B}{4}-\sin \frac{\pi-C}{4}\right] \leq \frac{1}{2} \\
& \Rightarrow 2 \sin \frac{A+B}{4}\left[\cos \frac{A-B}{4}-\cos \frac{\pi+C}{4}\right] \leq \frac{1}{2} \\
& \Rightarrow 4 \sin \frac{A+B}{4} \sin \frac{\pi+C+A-B}{8} \sin \frac{\pi+C-A+B}{8} \leq \frac{1}{2} \\
& \Rightarrow 4 \sin \frac{\pi-C}{4} \sin \frac{\pi-B}{4} \sin \frac{\pi-A}{4} \leq \frac{1}{2} \Rightarrow \cos \frac{\pi+C}{4} \cos \frac{\pi+B}{4} \cos \frac{\pi+A}{4} \leq \frac{1}{8}
\end{aligned}
\)

Q6. If \(\frac{x}{\tan (\theta+\alpha)}=\frac{y}{\tan (\theta+\beta)}=\frac{z}{\tan (\theta+\gamma)}\), then show that \(\sum \frac{x+y}{x-y} \sin ^2(\alpha-\beta)=0\).

Solution:
Here \(\frac{x}{y}=\frac{\tan (\theta+\alpha)}{\tan (\theta+\beta)}\). By componendo and dividendo, we get
\(
\begin{array}{r}
\frac{x+y}{x-y}=\frac{\tan (\theta+\alpha)+\tan (\theta+\beta)}{\tan (\theta+\alpha)-\tan (\theta+\beta)}=\frac{\sin (2 \theta+\alpha+\beta)}{\sin (\alpha-\beta)} \\
\therefore \frac{x+y}{x-y} \sin ^2(\alpha-\beta)=\sin (2 \theta+\alpha+\beta) \sin (\alpha-\beta) \\
=\frac{1}{2}[\cos 2(\theta+\beta)-\cos 2(\theta+\alpha)] \dots(i)
\end{array}
\)
Similarly, \(\frac{y+z}{y-z} \sin ^2(\beta-\gamma)=\frac{1}{2}[\cos 2(\theta+\gamma)-\cos 2(\theta+\beta)] \dots(ii)\)
and \(\frac{z+x}{z-x} \sin ^2(\gamma-\alpha)=\frac{1}{2}[\cos 2(\theta+\alpha)-\cos 2(\theta+\gamma)] \dots(iii)\)
Adding Eqs. (i), (ii), and (iii), we get L.H.S. \(=0\).

Q7. If \(\tan 6 \theta=p / q\), find the value of \(\frac{1}{2}(p \operatorname{cosec} 2 \theta-q \sec 2 \theta)\) in terms of \(p\) and \(q\).

Solution:
\(
\begin{aligned}
&\text { Here, we have } \tan 6 \theta=p / q\\
&\begin{aligned}
& \Rightarrow \frac{\sin 6 \theta}{\cos 6 \theta}=\frac{p}{q} \quad \Rightarrow \frac{p}{\sin 6 \theta}=\frac{q}{\cos 6 \theta}=\frac{\sqrt{p^2+q^2}}{\sqrt{1}}=\sqrt{p^2+q^2}=k \text { (say) } \\
& \text { Now } y=\frac{1}{2}(p \operatorname{cosec} \theta-q \sec 2 \theta)=\frac{1}{2}\left(\frac{p}{\sin 2 \theta}-\frac{q}{\cos 2 \theta}\right) \\
& \Rightarrow y=\frac{1}{2}\left[\frac{p \cos 2 \theta-q \sin 2 \theta}{\sin 2 \theta \cos 2 \theta}\right] \\
& \qquad=\left[\frac{2 k \sin 6 \theta \cos 2 \theta-2 k \cos 6 \theta \sin 2 \theta}{4 \sin 2 \theta \cos 2 \theta}\right]=k \frac{\sin (6 \theta-2 \theta)}{\sin 4 \theta}=k=\sqrt{p^2+q^2}
\end{aligned}
\end{aligned}
\)

Q8. If \(0<\alpha<\pi / 2\) and \(\sin \alpha+\cos \alpha+\tan \alpha+\cot \alpha+\sec \alpha+\operatorname{cosec} \alpha=7\), then prove that \(\sin 2 \alpha\) is a root of the equation \(x^2-44 x-36=0\).

Solution:
\(
\begin{aligned}
& \sin \alpha+\cos \alpha+(\tan \alpha+\cot \alpha)+(\sec \alpha+\operatorname{cosec} \alpha)=7 \\
& \Rightarrow \quad(\sin \alpha+\cos \alpha)+\frac{1}{\sin \alpha \cos \alpha}+\frac{\sin \alpha+\cos \alpha}{\sin \alpha \cos \alpha}=7 \\
& \Rightarrow \quad(\sin \alpha+\cos \alpha)\left(1+\frac{1}{\sin \alpha \cos \alpha}\right)=7-\frac{1}{\sin \alpha \cos \alpha} \\
& \Rightarrow \quad(1+\sin 2 \alpha)\left(1+\frac{4}{\sin 2 \alpha}+\frac{4}{\sin ^2 2 \alpha}\right)=49-\frac{28}{\sin 2 \alpha}+\frac{4}{\sin ^2 2 \alpha}
\end{aligned}
\)
\(
\begin{aligned}
&\text { Let } \sin 2 \alpha=x\\
&\begin{aligned}
& \Rightarrow(1+x)\left(1+\frac{4}{x}+\frac{4}{x^2}\right)=49-\frac{28}{x}+\frac{4}{x^2} \quad \Rightarrow(1+x)\left(x^2+4 x+4\right)=49 x^2-28 x+4 \\
& \Rightarrow x^3-44 x^2-36 x=0 \\
& \Rightarrow x^2-44 x-36=0 .(\text { as } x=\sin 2 \alpha \neq 0)
\end{aligned}
\end{aligned}
\)

Q9. Prove that \(1+\cot \theta \leq \cot \frac{\theta}{2}\) for \(0<\theta<\pi\). Find \(\theta\) when the equality sign holds.

Solution:
We have
\(
\begin{aligned}
1+\cot \theta-\cot \frac{\theta}{2} & =1+\frac{\cot ^2 \frac{\theta}{2}-1}{2 \cot \frac{\theta}{2}}-\cot \frac{\theta}{2} \\
& =\frac{2 \cot \frac{\theta}{2}+\cot ^2 \frac{\theta}{2}-1-2 \cot ^2 \frac{\theta}{2}}{2 \cot \frac{\theta}{2}} \\
& =\frac{-\left(\cot \frac{\theta}{2}-1\right)^2}{2 \cot \frac{\theta}{2}} \leq 0 \text { for } 0<\theta<\pi
\end{aligned}
\)
\(
\begin{aligned}
&\Rightarrow \quad 1+\cot \theta \leq \cot \frac{\theta}{2}\\
&\text { Equality holds when } \cot \frac{\theta}{2}-1=0 \Rightarrow \theta=\frac{\pi}{2} \text {. }
\end{aligned}
\)

Q10. Show that \(2^{\sin x}+2^{\cos x} \geq 2^{1-1 / \sqrt{2}}\).

Solution:
\(
\begin{aligned}
&\text { Since A.M. of two positive quantities } \geq \text { their G.M., we have }\\
&\begin{aligned}
& \frac{2^{\sin x}+2^{\cos x}}{2} \geq \sqrt{2^{\sin x} 2^{\cos x}}=\sqrt{2^{\sin x+\cos x}}=\sqrt{2^{\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)}} \geq \sqrt{2^{-\sqrt{2}}} \\
& \Rightarrow \quad 2^{\sin x}+2^{\cos x} \geq 2 \times 2^{\frac{-1}{\sqrt{2}}}=2^{1-\frac{1}{\sqrt{2}}}
\end{aligned}
\end{aligned}
\)

Q11. If \(A, B\) and \(C\) are the angles of a triangle, show that \(\tan ^2 \frac{A}{2}+\tan ^2 \frac{B}{2}+\tan ^2 \frac{C}{2} \geq 1\).

Solution:
\(
\begin{aligned}
&\text { We have } \frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi}{2} \text {, so that }\\
&\begin{aligned}
& \tan \left(\frac{A}{2}+\frac{B}{2}\right)=\tan \left(\frac{\pi}{2}-\frac{C}{2}\right) \\
& \Rightarrow \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}=\frac{1}{\tan \frac{C}{2}} \\
& \Rightarrow \tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}=1
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
& \text { Now } \tan ^2 \frac{A}{2}+\tan ^2 \frac{B}{2}+\tan ^2 \frac{C}{2}-1 \\
& =\frac{1}{2}\left[\sum 2 \tan ^2 \frac{A}{2}-\sum 2 \tan \frac{A}{2} \tan \frac{B}{2}\right] \\
& =\frac{1}{2}\left[\left(\tan \frac{A}{2}-\tan \frac{B}{2}\right)^2+\left(\tan \frac{B}{2}-\tan \frac{C}{2}\right)^2+\left(\tan \frac{C}{2}-\tan \frac{A}{2}\right)^2\right] \geq 0
\end{aligned}
\)

Q12. Let \(A, B, C\) be three angles such that \(A=\pi / 4\) and \(\tan B \tan C=p\). Find all possible values of \(p\) such that \(A, B, C\) are the angles of a triangle.

Solution:
\(
\begin{aligned}
& A+B+C=\pi \\
& \Rightarrow B+C=\frac{3 \pi}{4} \Rightarrow 0<B, C<\frac{3 \pi}{4} \text { Also } \tan B \tan C=p \\
& \Rightarrow \frac{\sin B \sin C}{\cos B \cos C}=\frac{p}{1} \\
& \Rightarrow \frac{\cos B \cos C-\sin B \sin C}{\cos B \cos C+\sin B \sin C}=\frac{1-p}{1+p} \\
& \Rightarrow \frac{\cos (B+C)}{\cos (B-C)}=\frac{1-p}{1+p} \\
& \Rightarrow \frac{1+p}{\sqrt{2}(p-1)}=\cos (B-C) \dots(i)
\end{aligned}
\)
Since \(B\) or \(C\) can vary from 0 to \(3 \pi / 4\), we get
\(
0 \leq B-C<\frac{3 \pi}{4} \Rightarrow-\frac{1}{\sqrt{2}}<\cos (B-C) \leq 1 .
\)
Equation (i) will now lead to \(-\frac{1}{\sqrt{2}}<\frac{p+1}{\sqrt{2}(p-1)} \leq 1\)
For \(0<1+\frac{p+1}{p-1} \quad \Rightarrow \frac{2 p}{(p-1)}>0 \quad \Rightarrow p<0\) or \(p>1 \dots(ii)\)
\(
\begin{aligned}
&\begin{aligned}
& \text { Also } \frac{p+1-\sqrt{2}(p-1)}{\sqrt{2}(p-1)} \leq 0 \\
& \Rightarrow \frac{\left(p-(\sqrt{2}+1)^2\right)}{(p-1)} \geq 0 \\
& \Rightarrow p<1 \text { or } p \geq(\sqrt{2}+1)^2 \dots(iii)
\end{aligned}\\
&\text { Combining Eqs. (ii) and (iii), we get } p<0 \text { or } p \geq(\sqrt{2}+1)^2 \text {. }
\end{aligned}
\)

Q13. Eliminate \(x\) from the equations, \(\sin (a+x)=2 b\) and \(\sin (a-x)=2 c\).

Solution:
\(
\begin{aligned}
& \text { Adding } \sin (a+x)+\sin (a-x)=2(b+c) \\
& \Rightarrow 2 \sin a \cos x=2(b+c) \\
& \Rightarrow \cos x=\frac{b+c}{\sin a} \dots(i)
\end{aligned}
\)
\(
\begin{aligned}
&\text { Subtracting, we get }\\
&\begin{aligned}
& \sin (a+x)-\sin (a-x)=2(b-c) \\
& \Rightarrow 2 \cos a \sin x=2(b-c) \\
& \Rightarrow \sin x=\frac{b-c}{\cos a} \dots(ii)
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&\text { Squaring and adding Eq. (i) and Eq. (ii), we get }\\
&\frac{(b+c)^2}{\sin ^2 a}+\frac{(b-c)^2}{\cos ^2 a}=1
\end{aligned}
\)

Q14. If \(\tan \beta=\frac{n \sin \alpha \cos \alpha}{1-n \sin ^2 \alpha}\), prove that \(\tan (\alpha-\beta)=(1-n) \tan \alpha\).

Solution:
\(
\begin{aligned}
\tan \beta & =\frac{n \sin \alpha \cos \alpha}{1-n \sin ^2 \alpha} \\
& =\frac{\frac{n \sin \alpha \cos \alpha}{\cos ^2 \alpha}}{\frac{1}{\cos ^2 \alpha}-\frac{n \sin ^2 \alpha}{\cos ^2 \alpha}} \text { [dividing numerator and denominator by } \cos ^2 \alpha \text { ] }
\end{aligned}
\)
\(
=\frac{n \tan \alpha}{\sec ^2 \alpha-n \tan ^2 \alpha}=\frac{n \tan \alpha}{1+\tan ^2 \alpha-n \tan ^2 \alpha}=\frac{n \tan \alpha}{1+(1-n) \tan ^2 \alpha} \dots(i)
\)
\(
\text { Now, L.H.S. } \begin{aligned}
= & \tan (\alpha-\beta)=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \tan \beta} \\
= & \frac{\tan \alpha-\frac{n \tan \alpha}{1+(1-n) \tan ^2 \alpha}}{1+\tan \alpha \frac{n \tan \alpha}{1+(1-n) \tan ^2 \alpha}} \text { [From Eq. (i)] }
\end{aligned}
\)
\(
=\frac{\tan \alpha+(1-n) \tan ^3 \alpha-n \tan \alpha}{1+(1-n) \tan ^2 \alpha+n \tan ^2 \alpha}=\quad \frac{(1-n) \tan \alpha+(1-n) \tan ^3 \alpha}{1+\tan ^2 \alpha}=(1-n) \tan \alpha
\)

Q15. Show that \(\frac{\sin x}{\cos 3 x}+\frac{\sin 3 x}{\cos 9 x}+\frac{\cos 9 x}{\cos 27 x}=\frac{1}{2}[\tan 27 x-\tan x]\).

Solution: L.H.S. contains \(x, 3 x, 9 x\) and \(27 x\), whereas R.H.S contains \(27 x\) and \(x\) only. So, we will manipulate terms as shown below
\(
\begin{aligned}
\text { R.H.S. } & =\frac{1}{2}[\tan 27 x-\tan x] \\
& =\frac{1}{2}[(\tan 27 x-\tan 9 x)+(\tan 9 x-\tan 3 x)+(\tan 3 x-\tan x)] \\
& =\frac{1}{2}\left[\left(\frac{\sin 27 x}{\cos 27 x}-\frac{\sin 9 x}{\cos 9 x}\right)+\left(\frac{\sin 9 x}{\cos 9 x}-\frac{\sin 3 x}{\cos 3 x}\right)+\left(\frac{\sin 3 x}{\cos 3 x}-\frac{\sin x}{\cos x}\right)\right] \\
& =\frac{1}{2}\left[\frac{\sin (27 x-9 x)}{\cos 27 x \cos 9 x}+\frac{\sin (9 x-3 x)}{\cos 9 x \cos 3 x}+\frac{\sin (3 x-x)}{\cos 3 x \cos x}\right] \\
& =\frac{1}{2}\left[\frac{\sin 18 x}{\cos 27 x \cos 9 x}+\frac{\sin 6 x}{\cos 9 x \cos 3 x}+\frac{\sin 2 x}{\cos 3 x \cos x}\right]
\end{aligned}
\)
\(
\begin{aligned}
& =\frac{1}{2}\left[\frac{2 \sin 9 x \cos 9 x}{\cos 27 x \cos 9 x}+\frac{2 \sin 3 x \cos 3 x}{\cos 9 x \cos 3 x}+\frac{2 \sin x \cos x}{\cos 3 x \cos x}\right] \\
& =\frac{\sin 9 x}{\cos 27 x}+\frac{\sin 3 x}{\cos 9 x}+\frac{\sin x}{\cos 3 x}=\frac{\sin x}{\cos 3 x}+\frac{\sin 3 x}{\cos 9 x}+\frac{\sin 9 x}{\cos 27 x}=\text { L.H.S. }
\end{aligned}
\)

Q16. Prove that \(\frac{2 \cos 2^n \theta+1}{2 \cos \theta+1}=(2 \cos \theta-1)(2 \cos 2 \theta-1)\left(2 \cos 2^2 \theta-1\right) \cdots\left(2 \cos 2^{n-1} \theta-1\right)\).

Solution:
We have to prove \(\frac{2 \cos 2^n \theta+1}{2 \cos \theta+1}=(2 \cos \theta-1)(2 \cos 2 \theta-1)\left(2 \cos 2^2 \theta-1\right) \ldots\left(2 \cos 2^{n-1} \theta-1\right)\)
or \(2 \cos 2^n \theta+1=[(2 \cos \theta+1)(2 \cos \theta-1)](2 \cos 2 \theta-1)\left(2 \cos 2^2 \theta-1\right) \ldots\left(2 \cos 2^{n-1} \theta-1\right)\)
Now \([(2 \cos \theta+1)(2 \cos \theta-1)](2 \cos 2 \theta-1)\left(2 \cos 2^2 \theta-1\right) \ldots\left(2 \cos 2^{n-1} \theta-1\right)\)
\(=\left(4 \cos ^2 \theta-1\right)(2 \cos 2 \theta-1)\left(2 \cos 2^2 \theta-1 \ldots\left(2 \cos 2^{n-1} \theta-1\right)\right.\)
\(
=(2 \cos 2 \theta+1)(2 \cos 2 \theta-1)\left(2 \cos 2^2 \theta-1\right) \ldots\left(2 \cos 2^{n-1} \theta-1\right) \quad\left[\text { using } \cos 2 \theta=2 \cos ^2 \theta-1\right]
\)
\(
\begin{aligned}
& =\left(4 \cos ^2 2 \theta-1\right)\left(2 \cos 2^2 \theta-1\right) \ldots\left(2 \cos 2^{n-1} \theta-1\right) \\
& =\left(2 \cos ^2 \theta+1\right)\left(2 \cos 2^2 \theta-1\right) \ldots\left(2 \cos 2^{n-1} \theta-1\right) \\
& =\left(4 \cos ^2 2^2 \theta-1\right)\left(2 \cos 2^3 \theta-1\right) \ldots\left(2 \cos 2^{n-1} \theta-1\right) \\
& \vdots \\
& =\left(2 \cos 2^{n-1} \theta+1\right)\left(2 \cos 2^{n-1} \theta-1\right) \\
& =4 \cos ^2 2^{n-1} \theta-1 \\
& =2 \cos 2^n \theta+1
\end{aligned}
\)

Q17. Prove that \(\frac{\tan 2^n \theta}{\tan \theta}=(1+\sec 2 \theta)\left(1+\sec 2^2 \theta\right)\left(1+\sec 2^3 \theta\right) \cdots\left(1+\sec 2^n \theta\right)\).

Solution:
We have to prove that \(\frac{\tan 2^n \theta}{\tan \theta}=(1+\sec 2 \theta)\left(1+\sec 2^2 \theta\right)\left(1+\sec 2^3 \theta\right) \ldots\left(1+\sec 2^n \theta\right)\)
or \(\tan 2^n \theta=\tan \theta(1+\sec 2 \theta)\left(1+\sec 2^2 \theta\right)\left(1+\sec 2^3 \theta\right) \ldots\left(1+\sec 2^n \theta\right)\)
Now \(\tan \theta(1+\sec 2 \theta)\left(1+\sec 2^2 \theta\right)\left(1+\sec 2^3 \theta\right) \ldots\left(1+\sec 2^n \theta\right)\)
\(
\begin{aligned}
& =\tan \theta\left(\frac{1+\cos 2 \theta}{\cos 2 \theta}\right)\left(1+\sec 2^2 \theta\right)\left(1+\sec 2^3 \theta\right) \ldots\left(1+\sec 2^n \theta\right) \\
& =\frac{\sin \theta}{\cos \theta}\left(\frac{2 \cos 2 \theta}{\cos 2 \theta}\right)\left(1+\sec 2^2 \theta\right)\left(1+\sec 2^3 \theta\right) \ldots\left(1+\sec 2^n \theta\right) \\
& =(\tan 2 \theta)\left(1+\sec 2^2 \theta\right)\left(1+\sec 2^3 \theta\right) \ldots\left(1+\sec 2^n \theta\right) \\
& =(\tan 2 \theta)\left(\frac{1+\cos 2^2 \theta}{\cos 2^2 \theta}\right)\left(1+\sec 2^3 \theta\right) \ldots\left(1+\sec 2^n \theta\right) \\
& =(\tan 2 \theta)\left(\frac{2 \cos 2}{\cos 2^2 \theta}\right)\left(1+\sec 2^3 \theta\right) \ldots\left(1+\sec 2^n \theta\right) \\
& =\left(\tan 2^2 \theta\right)\left(1+\sec 2^3 \theta\right) \ldots\left(1+\sec 2^n \theta\right) \\
& \vdots \\
& =\tan 2^{n-1} \theta\left(1+\sec 2^n \theta\right) \\
& =\tan 2^{n-1} \theta\left(\frac{1+\cos 2^n \theta}{\cos 2^n \theta}\right) \\
& =\tan 2^{n-1} \theta\left(\frac{2 \cos 2^{n-1} \theta}{\cos 2^n \theta}\right) \\
& =\tan 2^n \theta
\end{aligned}
\)