Summary
0 of 55 Questions completed
Questions:
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading…
You must sign in or sign up to start the quiz.
You must first complete the following:
0 of 55 Questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 point(s), (0)
Earned Point(s): 0 of 0, (0)
0 Essay(s) Pending (Possible Point(s): 0)
\(\text { If } \tan \theta=\frac{-4}{3}, \text { then } \sin \theta \text { is }\)
Correct choice is b. Since \(\tan \theta=-\frac{4}{3}\) is negative, \(\theta\) lies either in second quadrant or in fourth quadrant. Thus \(\sin \theta=\frac{4}{5}\) if \(\theta\) lies in the second quadrant or \(\sin \theta=-\frac{4}{5}\), if \(\theta\) lies in the fourth quadrant.
If \(\sin \theta\) and \(\cos \theta\) are the roots of the equation \(a x^{2}-b x+c=0\), then \(a\), \(b\) and \(c\) satisfy the relation.
The correct choice is (b). Given that \(\sin \theta\) and \(\cos \theta\) are the roots of the equation \(a x^{2}-b x+c=0\), so \(\sin \theta+\cos \theta=\frac{b}{a}\) and \(\sin \theta \cos \theta=\frac{c}{a}\)
Using the identity \((\sin \theta+\cos \theta)^{2}=\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta\), we have \(\frac{b^{2}}{a^{2}}=1+\frac{2 c}{a}\) or \(a^{2}-b^{2}+2 a c=0\)
The greatest value of \(\sin x \cos x\) is
(d) is the correct choice, since
\(
\sin x \cos x=\frac{1}{2} \sin 2 x \leq \frac{1}{2}, \text { since }|\sin 2 x| \leq 1 .
\)
The value of \(\sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}\) is
Correct choice is (c). Indeed \(\sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}\).
\(
\begin{aligned}
&=\frac{\sqrt{3}}{2} \sin 20^{\circ} \sin \left(60^{\circ}-20^{\circ}\right) \sin \left(60^{\circ}+20^{\circ}\right)\left(\operatorname{since} \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right) \\
&=\frac{\sqrt{3}}{2} \sin 20^{\circ}\left[\sin ^{2} 60^{\circ}-\sin ^{2} 20^{\circ}\right] \\
&=\frac{\sqrt{3}}{2} \sin 20^{\circ}\left[\frac{3}{4}-\sin ^{2} 20^{\circ}\right] \\
&=\frac{\sqrt{3}}{2} \times \frac{1}{4}\left[3 \sin 20^{\circ}-4 \sin ^{3} 20^{\circ}\right] \\
&=\frac{\sqrt{3}}{2} \times \frac{1}{4}\left(\sin 60^{\circ}\right) \\
&=\frac{\sqrt{3}}{2} \times \frac{1}{4} \times \frac{\sqrt{3}}{2}=\frac{3}{16}
\end{aligned}
\)
The value of \(\cos \frac{\pi}{5} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{5} \cos \frac{8 \pi}{5}\) is
(d) is the correct answer. We have
\(
\begin{aligned}
&\cos \frac{\pi}{5} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{5} \cos \frac{8 \pi}{5} \\
&=\frac{1}{2 \sin \frac{\pi}{5}} 2 \sin \frac{\pi}{5} \cos \frac{\pi}{5} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{5} \cos \frac{8 \pi}{5} \\
&=\frac{1}{2 \sin \frac{\pi}{5}} \sin \frac{2 \pi}{5} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{5} \cos \frac{8 \pi}{5} \\
&=\frac{1}{4 \sin \frac{\pi}{5}} \sin \frac{4 \pi}{5} \cos \frac{4 \pi}{5} \cos \frac{8 \pi}{5}
\end{aligned}
\)
\(\begin{aligned}
&=\frac{1}{8 \sin \frac{\pi}{5}} \sin \frac{8 \pi}{5} \cos \frac{8 \pi}{5} \\
&=\frac{\sin \frac{16 \pi}{5}}{16 \sin \frac{\pi}{5}}=\frac{\sin \left(3 \pi+\frac{\pi}{5}\right)}{16 \sin \frac{\pi}{5}} \\
&=\frac{-\sin \frac{\pi}{5}}{16 \sin \frac{\pi}{5}} \\
&=-\frac{1}{16}
\end{aligned}\)
If \(3 \tan \left(\theta-15^{\circ}\right)=\tan \left(\theta+15^{\circ}\right), 0^{\circ}<\theta<90^{\circ}\), then \(\theta=\) ___
(a) Given that \(3 \tan \left(\theta-15^{\circ}\right)=\tan \left(\theta+15^{\circ}\right)\) which can be rewritten as
\(
\frac{\tan \left(\theta+15^{\circ}\right)}{\tan \left(\theta-15^{\circ}\right)}=\frac{3}{1} .
\)
Applying componendo and Dividendo; we get \(\frac{\tan \left(\theta+15^{\circ}\right)+\tan \left(\theta-15^{\circ}\right)}{\tan \left(\theta+15^{\circ}\right)-\tan \left(\theta-15^{\circ}\right)}=2\)
\(
\begin{aligned}
&\Rightarrow \frac{\sin \left(\theta+15^{\circ}\right) \cos \left(\theta-15^{\circ}\right)+\sin \left(\theta-15^{\circ}\right) \cos \left(\theta+15^{\circ}\right)}{\sin \left(\theta+15^{\circ}\right) \cos \left(\theta-15^{\circ}\right)-\sin \left(\theta-15^{\circ}\right) \cos \left(\theta+15^{\circ}\right)}=2 \\
&\Rightarrow \frac{\sin 2 \theta}{\sin 30^{\circ}}=2 \quad \text { i.e., } \quad \sin 2 \theta=1 \\
\end{aligned}
\)
giving \(\theta=\frac{\pi}{4}\)
State whether the following statement is True or False.
\(
\text { “The inequality } 2^{\sin \theta}+2^{\cos \theta} \geq 2^{1-\frac{1}{\sqrt{2}}} \text { holds for all real values of } \theta”
\)
(b) True. Since \(2^{\sin \theta}\) and \(2^{\cos \theta}\) are positive real numbers, so A.M. (Arithmetic Mean) of these two numbers is greater or equal to their G.M. (Geometric Mean) and hence
\(
\begin{aligned}
&\frac{2^{\sin \theta}+2^{\cos \theta}}{2} \geq \sqrt{2^{\sin \theta} \times 2^{\cos \theta}}=\sqrt{2^{\sin \theta+\cos \theta}} \\
&\geq 2^{\frac{\sin \theta+\cos \theta}{2}}=2^{\frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}} \sin \theta+\frac{1}{\sqrt{2}} \cos \theta\right)} \\
&\geq 2^{\frac{1}{\sqrt{2}} \sin \left(\frac{\pi}{4}+\theta\right)}
\end{aligned}
\)
Since,
\(
\begin{aligned}
&-1 \leq \sin \left(\frac{\pi}{4}+\theta\right) \leq 1 \text {, we have } \\
&\frac{2^{\sin \theta}+2^{\cos \theta}}{2} \geq 2^{\frac{-1}{\sqrt{2}}} \Rightarrow 2^{\sin \theta}+2^{\cos \theta} \geq 2^{1-\frac{1}{\sqrt{2}}}
\end{aligned}
\)
Match each item given under the column C1 to its correct answer given under column C2
C1 C2
(a)\((1-\cos x) / \sin x\) (i) \(\cot ^{2} x / 2\)
(b) \((1+\cos x) /(1-\cos x)\) (ii) \(\cot x / 2\)
\((c)(1+\cos x) / \sin x\) (iii) \(|\cos x+\sin x|\)
(d) \(\sqrt{ }(1+\sin 2 x)\) (iv) \(\tan x / 2\)
(a) \(\frac{1-\cos x}{\sin x}=\frac{2 \sin ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}=\tan \frac{x}{2}\). Hence (a) matches with (iv) denoted by (a) \(\leftrightarrow\) (iv)
(b) \(\frac{1+\cos x}{1-\cos x}=\frac{2 \sin ^{2} \frac{x}{2}}{2 \sin ^{2} \frac{x}{2}}=\cot ^{2} \frac{x}{2}\). Hence (b) matches with (i) i.e., (b) \(\leftrightarrow\) (i)
(c) \(\frac{1+\cos x}{\sin x}=\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}=\cot \frac{x}{2}\).
Hence (c) matches with (ii) i.e., (c) \(\leftrightarrow\) (ii)
(d) \(\sqrt{1+\sin 2 x}=\sqrt{\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x}\)
\(
\begin{aligned}
&=\sqrt{(\sin x+\cos x)^{2}} \\
&=|(\sin x+\cos x)| \text {. Hence (d) matches with (iii), i.e., (d) } \leftrightarrow \text { (iii) }
\end{aligned}
\)
If \(\sin \theta+\operatorname{cosec} \theta=2\), then \(\sin ^{2} \theta+\operatorname{cosec}^{2} \theta\) is equal to
(c)
\(\sin \theta+\operatorname{cosec} \theta=2\)
Squaring L.H.S and R.H.S
We get,
\(
\begin{aligned}
&\Rightarrow(\sin \theta+\operatorname{cosec} \theta)^{2}=2^{2} \\
&\Rightarrow \sin ^{2} \theta+\operatorname{cosec}^{2} \theta+2 \sin \theta \operatorname{cosec} \theta=4 \\
&\Rightarrow \sin ^{2} \theta+\operatorname{cosec}^{2} \theta+2 \sin \theta\left(\frac{1}{\sin \theta}\right)=4 \\
&\Rightarrow \sin ^{2} \theta+\operatorname{cosec}^{2} \theta+2=4 \\
&\Rightarrow \sin ^{2} \theta+\operatorname{cosec}^{2} \theta=2
\end{aligned}
\)
If \(f(x)=\cos ^{2} x+\sec ^{2} x\), then
(d) Given that: \(f(x)=\cos ^{2} x+\sec ^{2} x\)
We know that \(A M \geq G M\)
\(
\begin{aligned}
&\Rightarrow \frac{\cos ^{2} x+\sec ^{2} x}{2} \geq \sqrt{\cos ^{2} x \cdot \sec ^{2} x} \\
&\Rightarrow \frac{\cos ^{2} x+\sec ^{2}}{2} \geq 1 \\
&\Rightarrow \cos ^{2} x+\sec ^{2} x \geq 2 \\
&\Rightarrow f(x) \geq 2
\end{aligned}
\)
If \(\tan \theta=\frac{1}{2}\) and \(\tan \phi=\frac{1}{3}\), then the value of \(\theta+\phi\) is
(d) According to the questlon,
\(\tan \theta=\frac{1}{2}\) and \(\tan \phi=\frac{1}{3}\)
We know that,
\(
\begin{aligned}
&\tan (\theta+\phi)=\frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi} \\
&=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}=\frac{\frac{5}{6}}{\frac{5}{6}}=1 \\
&\Rightarrow \tan (\theta+\phi)=\tan \frac{\pi}{4} \\
&\Rightarrow \theta+\phi=\frac{\pi}{4}
\end{aligned}
\)
Which of the following is not correct?
We know that,
a) \(\sin \theta=-1 / 5\) is correct since \(\sin \theta \in[-1,1]\)
b) \(\cos \theta=1\) is correct since \(\cos \theta \in[-1,1]\)
c) \(\sec \theta=1 / 2\)
\(\Rightarrow(1 / \cos \theta)=1 / 2\)
\(\Rightarrow \cos \theta=2\) is incorrect since \(\cos \theta \in[-1,1]\)
d) \(\tan \theta=20\) is correct since \(\tan \theta \in \mathrm{R}\).
Thus, optlon (c) \(\sec \theta=1 / 2\) Is the correct answer.
The value of \(\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}\) is
(b) Let, \(\mathrm{N}=\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots . \tan 89^{\circ}\)
\(N=\left[\tan 1^{\circ} \tan 2^{\circ} \ldots \tan 44^{\circ}\right] \tan 45^{\circ}\left[\tan \left(90^{\circ}-44^{\circ}\right) \cdot \tan \right.\) \(\left.\left(90^{\circ}-43^{\circ}\right) \ldots \tan \left(90^{\circ}-1^{\circ}\right)\right]\)
\(N=\left[\tan 1^{\circ} \tan 2^{\circ} \ldots \tan 44^{\circ}\right] \cdot \tan 45^{\circ} .\left[\cot 44^{\circ} \cdot \cot 43^{\circ} \ldots\right.\) \(\left.\cot 1^{\circ}\right]\)
We know, \(\tan \varnothing \times \cot \varnothing=1\) and \(\tan 45^{\circ}=1\)
Therefore, simplifying \(N\) further, we get \(N=1\)
The value of \(\frac{1-\tan ^{2} 15^{\circ}}{1+\tan ^{2} 15^{\circ}}\) is
According to the question,
Let \(\theta=15^{\circ} \Rightarrow 2 \theta=30^{\circ}\)
Now, since we know that,
\(
\begin{aligned}
&\cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta} \\
&\Rightarrow \cos 30^{\circ}=\frac{1-\tan ^{2} 15^{\circ}}{1+\tan ^{2} 15^{\circ}} \\
&\Rightarrow \frac{\sqrt{3}}{2}=\frac{1-\tan ^{2} 15^{\circ}}{1+\tan ^{2} 15^{\circ}}
\end{aligned}
\)
Thus, optlon (c) \(\sqrt{3} / 2\) Is the correct answer.
The value of \(\cos 1^{\circ} \cos 2^{\circ} \cos 3^{\circ} \ldots \cos 179^{\circ}\) is
(b) We have:
\(
\begin{aligned}
&\cos 1^{\circ} \cdot \cos 2^{\circ} \cdot \cos 3^{\circ} \ldots \ldots \ldots \ldots \cdot \cos 179^{\circ} \\
&=\cos 1^{\circ} \cdot \cos 2^{\circ} \cdot \cos 3^{\circ} \ldots \ldots \ldots . . \cos 90^{\circ} \ldots \ldots \ldots \ldots . \cos 179^{\circ} \\
&\therefore \cos 1^{\circ} \cdot \cos 2^{\circ} \cdot \cos 3^{\circ} \ldots \ldots \ldots \ldots . . \cos 179^{\circ}=0 \text { As }\left[\cos 90^{\circ}=0\right]
\end{aligned}
\)
If \(\tan \theta=3\) and \(\theta\) lies in third quadrant, then the value of \(\sin \theta\) is
(c) According to the question,
Given that, \(\tan \theta=3\) and \(\theta\) lies in third quadrant
\(
\Rightarrow \cot \theta=1 / 3
\)
We know that,
\(
\begin{aligned}
&\operatorname{cosec}^{2} \theta=1+\cot ^{2} \theta \\
&=1+\left(\frac{1}{3}\right)^{2}=1+\frac{1}{9}=\frac{10}{9} \\
&\Rightarrow \sin ^{2} \theta=\frac{9}{10} \\
&\Rightarrow \sin \theta=\pm \frac{3}{\sqrt{10}} \\
&\Rightarrow \sin \theta=-\frac{3}{\sqrt{10}}
\end{aligned}
\)
since \(\theta\) lies in third quadrant.
Thus, optlon (c) \(-3 / \sqrt{10}\) Is the correct answer.
The value of \(\tan 75^{\circ}-\cot 75^{\circ}\) is equal to
(a) The given expression is \(\tan 75^{\circ}-\cot 75^{\circ}\)
\(
\begin{aligned}
&\tan 75^{\circ}-\cot 75^{\circ}=\tan 75^{\circ}-\cot \left(90-15^{\circ}\right) \\
&=\tan 75^{\circ}-\tan 15^{\circ} \\
&=\frac{\sin 75^{\circ}}{\cos 75^{\circ}}-\frac{\sin 15^{\circ}}{\cos 15^{\circ}} \\
&=\frac{\sin 75^{\circ} \cos 15^{\circ}-\cos 75^{\circ} \sin 15^{\circ}}{\cos 75^{\circ} \cos 15^{\circ}} \\
&=\frac{\sin \left(75^{\circ}-15^{\circ}\right)}{\frac{1}{2} \times 2 \cos 75^{\circ} \cos 15^{\circ}} \\
&=\frac{2 \sin 60^{\circ}}{\cos \left(75^{\circ}+15^{\circ}\right)+\cos \left(75^{\circ}-15^{\circ}\right)} \\
&=\frac{2 \times \frac{\sqrt{3}}{2}}{\cos 90^{\circ}+\cos 60^{\circ}} \\
&=\frac{\sqrt{3}}{0+\frac{1}{2}} \\
&=2 \sqrt{3}
\end{aligned}
\)
Which of the following is correct?
(b) We know that, 1 radian \(=180^{\circ} / \pi=57^{\circ} 30^{\prime}\) approx.
\(57^{\circ}\) lies between 0 and 90 degrees and since in first quadrant \(\sin \theta\) increases when \(\theta\) increases.
\(
\Rightarrow \sin 1^{\circ}<\sin 1
\)
If \(\tan \alpha=\frac{m}{m+1}, \tan \beta=\frac{1}{2 m+1}\), then \(\alpha+\beta\) is equal to
(d) As given, \(\tan \alpha=\frac{m}{m+1}\)
and \(\tan \beta=\frac{1}{2 m+1}\)
\(
\begin{aligned}
&\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \\
&=\frac{\frac{m}{m+1}+\frac{1}{2 m+1}}{1-\frac{m}{m+1} \times \frac{1}{2 m+1}}=\frac{m(2 m+1)+(m+1)}{(m+1)(2 m+1)-m} \\
&=\frac{2 m^{2}+2 m+1}{2 m^{2}+2 m+1}=1 \\
&\text { So, } \alpha+\beta=\frac{\pi}{4}
\end{aligned}
\)
The minimum value of \(3 \cos x+4 \sin x+8\) is
(d) Let \(y=3 \cos x+4 \sin x+8\)
\(
\Rightarrow y-8=3 \cos x+4 \sin x
\)
We know that, minimum value of \(A \cos \theta+B \sin \theta=-\sqrt{ }\left(A^{2}+B^{2}\right)\)
\(\Rightarrow y-8=-\sqrt{ }\left(3^{2}+4^{2}\right)=-\sqrt{25}=-5\)
\(\Rightarrow \mathrm{y}=-5+8\)
\(=3\)
The value of \(\tan 3 \mathrm{~A}-\tan 2 \mathrm{~A}-\tan \mathrm{A}\) is equal to
(a) \(\tan 3 \mathrm{~A}=\tan (2 \mathrm{~A}+\mathrm{A})\)
\(
\begin{aligned}
&\tan 3 A=\frac{\tan 2 A+\tan A}{1-\tan 2 A \cdot \tan A} \\
&\Rightarrow \tan 3 A(1-\tan 2 A \cdot \tan A)=\tan 2 A+\tan A \\
&\Rightarrow \tan 3 A-\tan 3 A \cdot \tan 2 A \cdot \tan A=\tan 2 A+\tan A \\
&\Rightarrow \tan 3 A-\tan 2 A-\tan A=\tan 3 A \cdot \tan 2 A \cdot \tan A
\end{aligned}
\)
The value of \(\sin \left(45^{\circ}+\theta\right)-\cos \left(45^{\circ}-\theta\right)\) is
(d) We know that,
\(
\sin (90-\theta)=\cos \theta
\)
So,
\(
\begin{aligned}
&\sin \left(45^{\circ}+\theta\right)=\cos \left[90-\left(45^{\circ}+\theta\right)\right]=\cos \left(45^{\circ}-\theta\right) \\
&\therefore \sin \left(45^{\circ}+\theta\right)-\cos \left(45^{\circ}-\theta\right) \\
&=\cos \left(45^{\circ}-\theta\right)-\cos \left(45^{\circ}-\theta\right) \\
&=0
\end{aligned}
\)
The value of \(\cot \left(\frac{\pi}{4}+\theta\right) \cot \left(\frac{\pi}{4}-\theta\right)\) is
(c) \(
\begin{aligned}
&\text { Given, } \cot \left(\frac{\pi}{4}+\theta\right) \cot \left(\frac{\pi}{4}-\theta\right) \\
&=\frac{\cot \frac{\pi}{4} \cot \theta-1}{\cot \frac{\pi}{4}+\cot \theta} \times \frac{\cot \frac{\pi}{4} \cot \theta+1}{\cot \theta-\cot \frac{\pi}{4}} \\
&=\frac{\cot \theta-1}{1+\cot \theta} \times \frac{1+\cot \theta}{\cot \theta-1}=1
\end{aligned}
\)
\(\cos 2 \theta \cos 2 \phi+\sin ^{2}(\theta-\phi)-\sin ^{2}(\theta+\phi)\) is equal to
\(\left[\right.\) Hint: Use \(\left.\sin ^{2} A-\sin ^{2} B=\sin (A+B) \sin (A-B)\right]\)
(b) Given that: \(\cos 2 \theta \cos 2 \Phi+\sin ^{2}(\theta-\Phi)-\sin ^{2}(\theta+\Phi)\)
\(
\begin{aligned}
&\cos 2 \theta \cos 2 \Phi+\sin ^{2}(\theta-\Phi)-\sin ^{2}(\theta+\Phi) \\
&=\cos 2 \theta \cos 2 \Phi+\sin (\theta-\Phi+\theta+\Phi) \cdot \sin (\theta-\Phi-\theta-\Phi) \ldots \ldots\left[\because \sin ^{2} A\right. \\
&\left.-\sin ^{2} B=\sin (A+B) \cdot \sin (A-B)\right] \\
&=\cos 2 \theta \cos 2 \Phi+\sin 2 \theta \cdot \sin (-2 \Phi) \\
&=\cos 2 \theta \cos 2 \Phi-\sin 2 \theta \sin 2 \Phi \ldots \ldots .[\because \sin (-\theta)=-\sin \theta] \\
&=\cos (2 \theta+2 \Phi)=\cos 2(\theta+\Phi)
\end{aligned}
\)
The value of \(\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}\) is
(c) \(
\begin{aligned}
&\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ} \\
&=\left(\cos 12^{\circ}+\cos 132^{\circ}\right)+\left(\cos 84^{\circ}+\cos 156^{\circ}\right) \\
&=2 \cos 72^{\circ} \cos 60^{\circ}+2 \cos 120^{\circ} \cos 36^{\circ} \\
&=\cos 72^{\circ}-\cos 36^{\circ}=\sin 18^{\circ}-\cos 36^{\circ} \\
&=-1 / 2
\end{aligned}
\)
If \(\tan A=\frac{1}{2}, \tan B=\frac{1}{3}\), then \(\tan (2 A+B)\) is equal to
(c) Given, \(\tan A=\frac{1}{2}, \tan B=\frac{1}{3} \quad \ldots(i)\)
Now, \(\tan (2 A+B)=\frac{\tan 2 A+\tan B}{1-\tan 2 A \tan B}\)
\(
\begin{aligned}
&=\frac{\frac{2 \tan A}{1-\tan ^{2} A}+\tan B}{1-\frac{2 \tan A}{1-\tan ^{2} A} \times \tan B} \\
&=\frac{\frac{4}{3}+\frac{1}{3}}{1-\frac{4}{3} \times \frac{1}{3}}=3
\end{aligned}
\)
The value of \(\sin \frac{\pi}{10} \sin \frac{13 \pi}{10}\) is
(c)
\(
\begin{aligned}
\sin \frac{\pi}{10} \sin \frac{13 \pi}{10} &=\sin \frac{\pi}{10} \sin \left(\pi+\frac{3 \pi}{10}\right)=-\sin \frac{\pi}{10} \sin \frac{3 \pi}{10} \\
&=-\sin 18^{\circ} \sin 54^{\circ}=-\sin 18^{\circ} \cos 36^{\circ} \\
&=-\left(\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt{5}+1}{4}\right)=\frac{1-5}{16}=-\frac{1}{4}
\end{aligned}
\)
The value of \(\sin 50^{\circ}-\sin 70^{\circ}+\sin 10^{\circ}\) is equal to
(b)
\(
\begin{aligned}
&\sin 50^{\circ}-\sin 70^{\circ}+\sin 10^{\circ} \\
&=2 \sin \left(\frac{50^{\circ}-70^{\circ}}{2}\right) \cos \left(\frac{50^{\circ}+70^{\circ}}{2}\right)+\sin 10^{\circ}\left[\because \sin A-\sin B=2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)\right. \\
&=2 \sin \left(-10^{\circ}\right) \cos 60^{\circ}+\sin 10^{\circ} \\
&=2 \times \frac{1}{2} \sin \left(-10^{\circ}\right)+\sin 10^{\circ} \\
&=-\sin 10^{\circ}+\sin 10^{\circ} \\
&=0
\end{aligned}
\)
If \(\sin \theta+\cos \theta=1\), then the value of \(\sin 2 \theta\) is equal to
(c) Given that: \(\sin \theta+\cos \theta=1\)
\(
\begin{aligned}
&\Rightarrow(\sin \theta+\cos \theta)^{2}=(1)^{2} \\
&\Rightarrow \sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta=1 \\
&\Rightarrow 1+\sin 2 \theta=1 \\
&\Rightarrow \sin 2 \theta=1-1=0
\end{aligned}
\)
If \(\alpha+\beta=\frac{\pi}{4}\), then the value of \((1+\tan \alpha)(1+\tan \beta)\) is
(b)
\(
\begin{aligned}
&\text { Given that: } \alpha+\beta=\frac{\pi}{4} \\
&\Rightarrow \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=1 \\
&\Rightarrow \tan \alpha+\tan \beta=1-\tan \alpha \tan \beta \\
&\Rightarrow \tan \alpha+\tan \beta+\tan \alpha \tan \beta=1 \\
&\Rightarrow 1+\tan \alpha+\tan \beta+\tan \alpha \tan \beta=1+1 \\
&\Rightarrow 1(1+\tan \alpha)+\tan \beta(1+\tan \alpha)=2 \\
&\Rightarrow(1+\tan \alpha)(1+\tan \beta)=2
\end{aligned}
\)
If \(\sin \theta=\frac{-4}{5}\) and \(\theta\) lies in third quadrant then the value of \(\cos \frac{\theta}{2}\) is
(c) Given that, \(\sin \theta=-\frac{4}{5}\) and \(\theta\) lies in the 3rd quadrant.
\(
\begin{aligned}
&\Rightarrow \cos \theta=-\sqrt{1-\frac{16}{25}}=-\frac{3}{5} \\
&\text { Now, } \cos \frac{\theta}{2}=\pm \sqrt{\frac{1+\cos \theta}{2}} \\
&=\pm \sqrt{\frac{1-\frac{3}{5}}{2}} \\
&=\pm \sqrt{\frac{1}{5}}
\end{aligned}
\)
But we take \(\cos \frac{\theta}{2}=-\frac{1}{\sqrt{5}}\). Since, if \(\theta\) lies in 3rd quadrant, then \(\frac{\theta}{2}\) will be in 2nd quadrant.
Number of solutions of the equation \(\tan x+\sec x=2 \cos x\) lying in the interval \([0,2 \pi]\) is
(c) Given equation is \(\tan x+\sec x=2 \cos x\)
\(
\begin{aligned}
&\Rightarrow \frac{\sin x}{\cos x}+\frac{1}{\cos x}=2 \cos x \\
&\Rightarrow 1+\sin x=2 \cos ^{2} x \\
&\Rightarrow 1+\sin x=2\left(1-\sin ^{2} x\right) \\
&\Rightarrow 2 \sin ^{2} x+2 \sin x-\sin x-1=0 \\
&\Rightarrow 2 \sin x(\sin x+1)-1(\sin x+1)=0 \\
&\Rightarrow(2 \sin x-1)(\sin x+1)=0 \\
&\Rightarrow \text { either } \sin x=\frac{1}{2} \text { or } \sin x=-1 \\
&\Rightarrow \text { either } x=\frac{\pi}{6}, \frac{5 \pi}{6} \in[0, \pi] \text { or } x=\frac{3 \pi}{2}
\end{aligned}
\)
But \(x=\frac{3 \pi}{2}\) can not be possible.
\(\therefore\) Number of solutions are 2 .
The value of \(\sin \frac{\pi}{18}+\sin \frac{\pi}{9}+\sin \frac{2 \pi}{9}+\sin \frac{5 \pi}{18}\) is given by
(a)
\(
\begin{aligned}
&\text { Given expression, } \sin \frac{\pi}{18}+\sin \frac{\pi}{9}+\sin \frac{2 \pi}{9}+\sin \frac{5 \pi}{18} \\
&=\sin 10^{\circ}+\sin 20^{\circ}+\sin 40^{\circ}+\sin 50^{\circ} \\
&=\sin 50^{\circ}+\sin 10^{\circ}+\sin 40^{\circ}+\sin 20^{\circ} \\
&=\sin 130^{\circ}+\sin 10^{\circ}+\sin 140^{\circ}+\sin 20^{\circ} \\
&=2 \sin 70^{\circ} \cos 60^{\circ}+2 \sin 80^{\circ} \cdot \cos 60^{\circ} \quad\left[\because \sin x+\sin y=2 \sin \frac{x+y}{2}\right. \\
&\left.\cdot \cos \frac{x-y}{2}\right] \\
&=2 \cdot \frac{1}{2} \sin 70^{\circ}+2 \cdot \frac{1}{2} \sin 80^{\circ} \quad\left[\because \cos 60^{\circ}=\frac{1}{2}\right] \\
&=\sin 70^{\circ}+\sin 80^{\circ}=\sin \frac{7 \pi}{18}+\sin \frac{4 \pi}{9}
\end{aligned}
\)
If \(\mathrm{A}\) lies in the second quadrant and \(3 \tan \mathrm{A}+4=0\), then the value of \(2 \cot \mathrm{A}-5 \cos \mathrm{A}+\sin \mathrm{A}\) is equal to
(b) We have, 3 tan \(A+4=0, A\) lies in second quadrant.
\(
\Rightarrow \tan \mathrm{A}=-\frac{4}{3}
\)
\(\cot \mathrm{A}=-\frac{3}{4}, \sin \mathrm{A}=\frac{4}{5}\) and \(\cos \mathrm{A}=-\frac{3}{5}\),
since A lie in second quadrant.
Now, \(2 \cot A-5 \cos A+\sin A\)
\(
=2 \times-\frac{3}{4}-5 \times-\frac{3}{5}+\frac{4}{5}
\)
\(
=-\frac{3}{2}+3+\frac{4}{5}
\)
\(
=\frac{-15+30+8}{10}
\)
\(
=\frac{23}{10}
\)
The value of \(\cos ^{2} 48^{\circ}-\sin ^{2} 12^{\circ}\) is
(a)
\(
\begin{aligned}
&\cos ^{2} 48^{\circ}-\sin ^{2} 12^{\circ} \\
&=\cos \left(48^{\circ}+12^{\circ}\right) \cos \left(48^{\circ}-12^{\circ}\right)\left[\cos (A+B) \cos (A-B)=\cos ^{2} A-\sin ^{2} B\right] \\
&=\cos 60^{\circ} \cos 36^{\circ} \\
&=\frac{1}{2} \times\left(\frac{\sqrt{5}+1}{4}\right) \\
&=\frac{\sqrt{5}+1}{8}
\end{aligned}
\)
If \(\tan \alpha=\frac{1}{7}, \tan \beta=\frac{1}{3}\), then \(\cos 2 \alpha\) is equal to
(b) We have, \(\tan \alpha=\frac{1}{7}\)
\(\tan \beta=\frac{1}{3}\)
Now, \(\cos 2 \alpha=\frac{1-\tan ^{2} \alpha}{1+\tan ^{2} \alpha}\)
\(
=\frac{1-\frac{1}{40}}{1+\frac{1}{49}}=\frac{48}{50} \quad \ldots(i)
\)
Now, \(\sin 2 \beta=\frac{2 \tan \beta}{1+\tan ^{2} \beta}\)
\(
=\frac{2 \times \frac{1}{3}}{1+\frac{1}{9}}=\frac{2}{3} \times \frac{9}{10}=\frac{3}{5} \quad \ldots(i i)
\)
Also, \(\sin 4 \beta=2 \sin 2 \beta \cos 2 \beta=2 \cdot \frac{3}{5}\left(\frac{1-\tan ^{2} \beta}{1+\tan ^{2} \beta}\right)\)
(using \((i i)\) )
\(
=\frac{6}{5}\left(\frac{1-\frac{1}{9}}{1+\frac{1}{9}}\right)=\frac{6}{5} \times \frac{8}{9} \times \frac{9}{10}=\frac{48}{50} \quad \ldots(\text { iii })
\)
Hence, \(\cos 2 \alpha=\sin 4 \beta\) (from (i) and (iii))
If \(\tan \theta=\frac{a}{b}\), then \(b \cos 2 \theta+a \sin 2 \theta\) is equal to
Correct option is (b)
\(
\begin{aligned}
&\tan \theta=\frac{\mathrm{a}}{\mathrm{b}} \\
&\Rightarrow 1+\tan ^{2} \theta=\sec ^{2} \theta \Rightarrow 1+\left(\frac{\mathrm{a}}{\mathrm{b}}\right)^{2}=\sec ^{2} \theta \Rightarrow \frac{\mathrm{a}^{2}+\mathrm{b}^{2}}{\mathrm{~b}^{2}}=\sec ^{2} \theta \Rightarrow \cos ^{2} \theta= \\
&\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}+\mathrm{b}^{2}} \Rightarrow \\
&\cos \theta=\sqrt{\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}+\mathrm{b}^{2}}}
\end{aligned}
\)
we know that, \(\sin \theta=\sqrt{1-\frac{b^{2}}{a^{2}+b^{2}}}=\sqrt{\frac{a^{2}}{a^{2}+b^{2}}}\)
Now, \(\mathrm{b} \cos 2 \theta+a \sin 2 \theta=b\left(\cos ^{2} \theta-\sin ^{2} \theta\right)+a(2 \sin \theta \cos \theta)=b\left(\frac{b^{2}}{a^{2}+b^{2}}-\right.\)
\(
\begin{aligned}
&\left.\frac{a^{2}}{a^{2}+b^{2}}\right)+a\left(2 \frac{a b}{a^{2}+b^{2}}\right) \\
&=\frac{1}{b^{2}+a^{2}}\left[b^{3}-a^{2} b+2 a^{2} b\right]=\frac{b^{3}+a^{2} b}{b^{2}+a^{2}}=\frac{b\left(b^{2}+a^{2}\right)}{b^{2}+a^{2}}=b
\end{aligned}
\)
If for real values of \(x, \cos \theta=x+\frac{1}{x}\), then
(d) Given that, \(\cos \theta=x+1 / x\)
\(
\begin{aligned}
&\cos \theta=\left(x^{2}+1\right) / x \\
&\Rightarrow x^{2}+1=x \cos \theta \\
&\Rightarrow x^{2}-\cos \theta x+1=0
\end{aligned}
\)
We know that, \(b^{2}-4 a c \geq 0\)
\(
\begin{aligned}
&\Rightarrow(-\cos \theta)^{2}-4 \times 1 \times 1 \geq 0 \\
&\Rightarrow \cos ^{2} \theta-4 \geq 0 \\
&\Rightarrow \cos ^{2} \theta \geq 4 \\
&\Rightarrow \cos \theta \geq \pm 2
\end{aligned}
\)
But \(-1 \leq \cos \theta \leq 1\)
So, no value of \(\theta\) is possible
\(
\text { The value of } \frac{\sin 50^{\circ}}{\sin 130^{\circ}} \text { is }
\)___
(b)
\(
\begin{aligned}
&\frac{\sin 50^{\circ}}{\sin 130^{\circ}}=\frac{\sin 50^{\circ}}{\sin \left(180^{\circ}-50^{\circ}\right)} \\
&=\frac{\sin 50^{\circ}}{\sin 50^{\circ}}
\end{aligned}
\)
\(
\text { If } k=\sin \left(\frac{\pi}{18}\right) \sin \left(\frac{5 \pi}{18}\right) \sin \left(\frac{7 \pi}{18}\right) \text {, then the numerical value of } k \text { is }
\)___
(c)
\(
\begin{aligned}
&\mathrm{k}=\sin (\pi / 8) \sin (5 \pi / 18) \sin (7 \pi / 18) \\
&=\sin 10^{\circ} \sin {50^{\circ}} \sin {70^{\circ}} \\
&=\frac{1}{2}\left[\cos 40^{\circ}-\cos 60^{\circ}\right] \sin 70^{\circ} \\
&=\frac{1}{2} \cos 40^{\circ} \sin 70^{\circ}-\frac{1}{4} \sin 70^{\circ} \\
&=\frac{1}{4}\left[\sin 110^{\circ}+\sin 30^{\circ}\right]-\frac{1}{4} \sin 70^{\circ} \\
&=\frac{1}{4} \sin \left(180^{\circ}-70^{\circ}\right)+\frac{1}{4} \times \frac{1}{2}-\frac{1}{4} \sin 70^{\circ}=\frac{1}{8}
\end{aligned}
\)
\(\text { If } \tan \mathrm{A}=\frac{1-\cos \mathrm{B}}{\sin \mathrm{B}} \text {, then } \tan 2 \mathrm{~A}=\)___
(c) Given that, \(\tan A=\frac{1-\cos B}{\sin B}\)
Now, \(\tan 2 \mathrm{~A}=\frac{2 \tan \mathrm{A}}{1-\tan ^{2} \mathrm{~A}}\)
\(
=\frac{2 \times \frac{1-\cos B}{\sin B}}{1-\left(\frac{1-\cos B}{\sin B}\right)^{2}}
\)
\(
=\frac{2 \times \frac{2 \sin ^{2} \frac{B}{2}}{2 \sin \frac{B}{2} \cos \frac{B}{2}}}{1-\left(\frac{2 \sin ^{2} \frac{B}{2}}{2 \sin \frac{B}{2} \cos \frac{B}{2}}\right)^{2}}
\)
\(
=\frac{2 \times \frac{\sin \frac{B}{2}}{\cos \frac{B}{2}}}{1-\left(\frac{\sin \frac{B}{2}}{\cos \frac{B}{2}}\right)^{2}}
\)
\(
=\frac{2 \tan \frac{B}{2}}{1-\tan ^{2} \frac{B}{2}}
\)
\(=\tan \mathrm{B}\)
\(
\Rightarrow \tan 2 A=\tan B
\)
If \(\sin x+\cos x=a\), then
(i) \(\sin ^{6} x+\cos ^{6} x=\)
(ii) \(|\sin x-\cos x|=\)
(a) Given that, \(\sin x+\cos x=a\)
On squaring both sides, we get
\(\begin{aligned} &(\sin x+\cos x)^{2}=(a)^{2} \\ \Rightarrow & \sin ^{2} x+\cos ^{2} x+2 \sin x \cos x=a^{2} \end{aligned}\)
\(\Rightarrow \quad \sin x \cdot \cos x=\frac{1}{2}\left(a^{2}-1\right)\)
\((i) \sin ^{6} x+\cos ^{6} x=\left(\sin ^{2} x\right)^{3}+\left(\cos ^{2} x\right)^{3}\)
\(=\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x\right)\) \(=\sin ^{4} x+\cos ^{4} x-\frac{1}{4}\left(a^{2}-1\right)^{2}\)
\(=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x-\frac{1}{4}\left(a^{2}-1\right)^{2}\)
\(=1-2 \cdot \frac{1}{4}\left(a^{2}-1\right)^{2}-\frac{1}{4}\left(a^{2}-1\right)^{2}=\frac{1}{4}\left[4-3\left(a^{2}-1\right)^{2}\right]\)
\((i i)|\sin x-\cos x|=\sqrt{(\sin x-\cos x)^{2}}\)
\(=\sqrt{\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x}\)
\(=\sqrt{1-2 \frac{1}{2}\left(a^{2}-1\right)}=\sqrt{1-a^{2}+1}=\sqrt{2-a^{2}}\)
In a triangle \(\mathrm{ABC}\) with \(\angle \mathrm{C}=90^{\circ}\) the equation whose roots are \(\tan \mathrm{A}\) and \(\tan \mathrm{B}\) is ___ .
(b) Given that, a \(\triangle \mathrm{ABC}\) with \(\angle \mathrm{C}=90^{\circ}\) and an equation whose roots are tan \(\mathrm{A}\) and tan \(\mathrm{B}\) \(\Rightarrow x^{2}-(\tan A+\tan B) x+\tan A \cdot \tan B=0(\mathrm{i})\)
Now, \(A+B=90^{\circ}\left[\because \angle C=90^{\circ}\right]\)
\(
\begin{aligned}
&\Rightarrow \tan (A+B)=\tan 90^{\circ} \\
&\Rightarrow \frac{\tan A+\tan B}{1-\tan A \tan B}=\frac{1}{0} \\
&\Rightarrow 1-\tan A \tan B=0 \\
&\Rightarrow \tan A \tan B=1 \text { (ii) }
\end{aligned}
\)
Now,
\(
\tan A+\tan B=\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}
\)
\(
\begin{aligned}
&=\frac{\sin A \cos B+\sin B \cos A}{\cos A \cos B} \\
&=\frac{\sin (A+B)}{\cos A \cos B}=\frac{\sin 90^{\circ}}{\cos A \cos \left(90^{\circ}-A\right)}
\end{aligned}
\)
\(
=\frac{1}{\cos A \sin A}=\frac{2}{2 \cos A \sin A}=\frac{2}{\sin 2 A}
\)
\(
\Rightarrow \tan A+\tan B=\frac{2}{\sin 2 A} \text { (iii) }
\)
Substituting the values from (II) and (III) In equatlon (I), we get
\(
x^{2}-\left(\frac{2}{\sin 2 A}\right) x+1=0
\)
\(3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4\left(\sin ^{6} x+\cos ^{6} x\right)=\)___.
(c)
\(
\begin{aligned}
&(\sin x-\cos x)^{4} \\
&=\left[(\sin x-\cos x)^{2}\right]^{2} \\
&=\left(\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x\right)^{2} \\
&=(1-2 \sin x \cos x)^{2} \\
&=1-4 \sin x \cos x+4 \sin ^{2} x \cos ^{2} x \\
&(\sin x+\cos x)^{2} \\
&=\sin ^{2} x+\cos ^{2} x+2 \sin _{x} \cos _{x} \\
&=1+2 \sin {x} \cos {x} \\
&\sin ^{6} x+\cos ^{6} x \\
&=\left(\sin ^{2} x\right)^{3}+\left(\cos ^{2} x\right)^{3} \\
&=\left(\sin ^{2} x+\cos ^{2} x\right)^{3}-3 \sin ^{2} x \cos ^{2} x\left(\sin ^{2} x+\cos ^{2} x\right) \ldots\left[\because a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b)\right] \\
&=1^{3}-3 \sin ^{2} x \cos ^{2} x(1) \\
&=1-3 \sin ^{2} x \cos ^{2} x \\
& 3(\sin x-\cos x)^4+4\left(\sin ^6 x+\cos ^6 x\right)+6[\sin x+\cos x]^2 \\
& =3-12 \sin x \cos x+12 \sin ^2 x \cos ^2 x+4 \\
& -12 \sin ^2 x \cos ^2 x+6+12 \sin x \cos x \\
& =13
\end{aligned}
\)
\(
\text { Given } x>0 \text {, the values of } f(x)=-3 \cos \sqrt{3+x+x^{2}} \text { lie in the interval }
\)___.
(a) Given that: \(f(x)=-3 \cos \sqrt{3+x+x^{2}}\)
Put \(\sqrt{3+x+x^{2}}=y\)
\(
\begin{aligned}
&\therefore f(x)=-3 \cos y \\
&\because-1 \leq \cos y \leq 1 \\
&3 \geq-3 \cos y \geq-3 \\
&\Rightarrow-3 \leq-3 \cos y \leq 3 \\
&\therefore-3 \leq-3 \cos \sqrt{3+x+x^{2}} \leq 3, x>0
\end{aligned}
\)
The maximum distance of a point on the graph of the function \(y=\sqrt{3} \sin x+\cos x\) from \(x\)-axis is ____.
(d) Given that \(y=\sqrt{3} \sin x+\cos x \ldots \ldots\) (i)
\(\therefore\) The maximum distance from a point on the graph of equation (i) from \(\mathrm{x}\)-axis
\(
\begin{aligned}
&=\sqrt{(\sqrt{3})^{2}+(1)^{2}} \\
&=\sqrt{3+1} \\
&=2 .
\end{aligned}
\)
State whether the following statement is True or False.
\(
\text { If } \tan \mathrm{A}=\frac{1-\cos \mathrm{B}}{\sin \mathrm{B}} \text {, then } \tan 2 \mathrm{~A}=\tan \mathrm{B}
\)
(a) True
\(
\begin{aligned}
&\tan A=\frac{1-\cos B}{\sin B} \\
&\Rightarrow \tan A=\frac{2 \sin ^{2}\left(\frac{B}{2}\right)}{2 \sin \left(\frac{B}{2}\right) \cos \left(\frac{B}{2}\right)} \\
&\Rightarrow \tan A=\tan \left(\frac{B}{2}\right) \\
&\text { Now, } \tan 2 A=\frac{2 \tan A}{1-\tan ^{2} A}=\frac{2 \tan \left(\frac{B}{2}\right)}{1-\tan ^{2}\left(\frac{B}{2}\right)}=\tan B \\
&\therefore \tan 2 A=\tan B
\end{aligned}
\)
State whether the following statement is True or False.
\(
\text { The equality } \sin \mathrm{A}+\sin 2 \mathrm{~A}+\sin 3 \mathrm{~A}=3 \text { holds for some real value of } \mathrm{A} \text {. }
\)
(b) We know that, maximum value of \(\sin A\) is 1 .
Given that, \(\sin A+\sin 2 A+\sin 3 A=3\)
It is possible when \(\sin A=\sin 2 A=\sin 3 A=1\)
But \(\sin 2 \mathrm{~A}\) and \(\sin 3 \mathrm{~A}\) is not equal to 1 .
Hence, the statement Is false.
State whether the following statement is True or False.
\(\sin 10^{\circ} \text { is greater than } \cos 10^{\circ} .\)
(b) Let, \(\sin 10^{\circ}>\cos 10^{\circ}\)
\(
\begin{aligned}
&\Rightarrow \sin 10^{\circ}>\cos \left(90^{\circ}-80^{\circ}\right) \\
&\Rightarrow \sin 10^{\circ}>\sin 80^{\circ}
\end{aligned}
\)
Which is not possible since value of sin increases with increase in \(\theta\).
Hence, statement Is false.
State whether the following statement is True or False.
\(
\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}=\frac{1}{16}
\)
(a)
\(
\begin{aligned}
& L H S=\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15} \text { , On dividing and multiplying by } 2 \sin \frac{2 \pi}{15}, \text { we get } \\
& =\frac{1}{2 \sin \frac{2 \pi}{15}} \times\left(2 \sin \frac{2 \pi}{15} \times \cos \frac{2 \pi}{15}\right) \times \cos \frac{4 \pi}{15} \times \cos \frac{8 \pi}{15} \times \cos \frac{16 \pi}{15} \\
& =\frac{1}{2 \times 2 \sin \frac{2 \pi}{15}} \times\left(2 \sin \frac{4 \pi}{15} \times \cos \frac{4 \pi}{15}\right) \times \cos \frac{8 \pi}{15} \times \cos \frac{16 \pi}{15} \\
& =\frac{1}{2 \times 4 \sin \frac{2 \pi}{15}}\left(2 \sin \frac{8 \pi}{15} \times \cos \frac{8 \pi}{15}\right) \times \cos \frac{16 \pi}{15} \\
& =\frac{1}{2 \times 8 \sin \frac{2 \pi}{15}}\left(2 \sin \frac{16 \pi}{15} \times \cos \frac{16 \pi}{15}\right) \\
& =\frac{1}{16 \sin \frac{2 \pi}{15}}\left(\sin \frac{32 \pi}{15}\right) \\
& =-\frac{1}{16 \sin \frac{2 \pi}{15}}\left(\sin 2 \pi-\frac{32 \pi}{15}\right)[\because \sin (2 \pi-\theta)=-\sin \theta] \\
& =-\frac{1}{16 \sin \frac{2 \pi}{15}} \sin \left(-\frac{2 \pi}{15}\right) \\
& =\frac{1}{16}=R H S
\end{aligned}
\)
State whether the following statement is True or False.
One value of \(\theta\) which satisfies the equation \(\sin ^{4} \theta-2 \sin ^{2} \theta-1\) lies between 0 and \(2 \pi\).
(b) We have, \(\sin ^{4} \theta-2 \sin ^{2} \theta-1=0\)
Now,
\(
\sin ^{2} \theta=\frac{-(-2) \pm \sqrt{(-2)^{2}-4 \times 1 \times(-1)}}{2 \times 1}
\)
\(
\sin ^{2} \theta=\frac{2 \pm \sqrt{4+4}}{2}=\frac{2 \pm \sqrt{8}}{2}=\frac{2 \pm 2 \sqrt{2}}{2}
\)
\(
\begin{aligned}
&\Rightarrow \sin ^{2} \theta=1 \pm \sqrt{2} \\
&\Rightarrow \sin ^{2} \theta=1+\sqrt{2} \text { or } 1-\sqrt{2}
\end{aligned}
\)
We know that, \(-1 \leq \sin \theta \leq 1\)
\(
\Rightarrow \sin ^{2} \theta \leq 1
\)
but \(\sin ^{2} \theta=1+\sqrt{2}\) or \(1-\sqrt{2}\)
Which is not possible.
Hence, the above statement Is false.
State whether the following statement is True or False.
\(
\text { If } \operatorname{cosec} x=1+\cot x \text { then } x=2 n \pi, 2 n \pi+\frac{\pi}{2}
\)
(a) We have, \(\operatorname{cosec} x=1+\cot x\)
\(
\Rightarrow \frac{1}{\sin x}=1+\frac{\cos x}{\sin x}
\)
\(
\Rightarrow \frac{1}{\sin x}=\frac{\sin x+\cos x}{\sin x}
\)
\(
\Rightarrow \sin x+\cos x=1
\)
\(
\Rightarrow \frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x=\frac{1}{\sqrt{2}}
\)
\(
\Rightarrow \sin \frac{\pi}{4} \sin x+\cos \frac{\pi}{4} \cos x=\frac{1}{\sqrt{2}}
\)
\(
\Rightarrow \cos \left(x-\frac{\pi}{4}\right)=\cos \frac{\pi}{4}
\)
\(
x=2 n \pi+\frac{\pi}{4}+\frac{\pi}{4}=2 n \pi+\frac{\pi}{2}
\)
\(
\text { or } x=2 n \pi+\frac{\pi}{4}-\frac{\pi}{4}
\)
\(
=2 n \pi
\)
Hence, statement Is true.
State whether the following statement is True or False.
\(
\text { If } \tan \theta+\tan 2 \theta+\sqrt{3} \tan \theta \tan 2 \theta=\sqrt{3} \text {, then } \theta=\frac{n \pi}{3}+\frac{\pi}{9}
\)
(a)
\(
\begin{aligned}
&\tan \theta+\tan 2 \theta+\sqrt{ } 3 \tan \theta \cdot \tan 2 \theta=\sqrt{ } 3 \\
&=>\tan \theta+\tan 2 \theta=\sqrt{ } 3-\sqrt{ } 3 \tan \theta \cdot \tan 2 \theta \\
&=>\tan \theta+\tan 2 \theta=\sqrt{ } 3[1-\tan \theta \tan 2 \theta] \\
&=>(\tan \theta+\tan 2 \theta) /(1-\tan \theta \cdot \tan 2 \theta)=\sqrt{ } 3 \\
&=>\tan (\theta+2 \theta)=\tan \pi / 3 \\
&=>\tan 3 \theta=\tan \pi / 3 \\
&=>3 \theta=n \pi+\pi / 3, n \in Z \\
&=>\theta=n \pi / 3+\pi / 9, n \in Z \\
&\therefore \text { Solution set }=\left\{\frac{n \pi}{3}+\frac{\pi}{9}: n \in Z\right\}
\end{aligned}
\)
State whether the following statement is True or False.
\(
\text { If } \tan (\pi \cos \theta)=\cot (\pi \sin \theta) \text {, then } \cos \left(\theta-\frac{\pi}{4}\right)=\pm \frac{1}{2 \sqrt{2}}
\)
(a) We have, \(\tan (\pi \cos \theta)=\cot (\pi \sin \theta)\)
\(
\Rightarrow \tan (\pi \cos \theta)=\tan \left(\frac{\pi}{2}-\pi \sin \theta\right)
\)
\(\Rightarrow \pi \cos \theta=\frac{\pi}{2}-\pi \sin \theta\)
\(
\Rightarrow \pi \cos \theta+\pi \sin \theta=\frac{\pi}{2}
\)
\(
\Rightarrow \cos \theta+\sin \theta=\frac{1}{2}
\)
\(
\Rightarrow \cos \frac{\pi}{4} \cos \theta+\sin \frac{\pi}{4} \sin \theta=\frac{1}{2 \sqrt{2}}
\)
\(
\left[\because \cos \left(\theta-\frac{\pi}{4}\right) \text { or } \cos \left(\frac{\pi}{4}-\theta\right)\right]
\)
\(
\Rightarrow \cos \left(\theta-\frac{\pi}{4}\right)=\pm \frac{1}{2 \sqrt{2}}
\)
State whether the following statement is True or False.
In the following match each item given under the column \(\mathrm{C}_{1}\) to its correct answer given under the column \(\mathrm{C}_{2}\):
C1 C2
(a) \(\sin (x+y) \sin (x-y)^{2}\) (i) \(\cos ^{2} x-\sin ^{2} y\)
(b) \(\cos (x+y) \cos (x-y)\) (ii) \(\frac{1-\tan \theta}{1+\tan \theta}\)
(c) \(\cot \left(\frac{\pi}{4}+\theta\right)\) (iii) \(\frac{1+\tan \theta}{1-\tan \theta}\)
(d) \(\tan \left(\frac{\pi}{4}+\theta\right)\) (iv) \(\sin ^{2} x-\sin ^{2} y\)
(b) False, Correct match is shown below.
(a) \(\leftrightarrow\) (iv)
(b) \(\leftrightarrow\) (i)
(c) \(\leftrightarrow\) (ii)
(d) \(\leftrightarrow\) (iii)
You cannot copy content of this page