Quiz

Hint

\(
\begin{gathered}
2 \cos \theta-3=-5 \\
2 \cos \theta=-2 \\
\cos \theta=-1 \\
\theta=\pi
\end{gathered}
\)

Hint

If we rewrite the right side, we can write the equation in terms of cosine:
\(
\begin{aligned}
3 \cos \theta+3 & =2 \sin ^2 \theta \\
3 \cos \theta+3 & =2\left(1-\cos ^2 \theta\right) \\
3 \cos \theta+3 & =2-2 \cos ^2 \theta \\
2 \cos ^2 \theta+3 \cos \theta+1 & =0 \\
(2 \cos \theta+1)(\cos \theta+1) & =0 \\
2 \cos \theta+1 & =0 \\
\cos \theta & =-\frac{1}{2} \\
\theta & =\frac{2 \pi}{3}, \frac{4 \pi}{3} \\
\cos \theta+1 & =0 \\
\cos \theta & =-1 \\
\theta & =\pi
\end{aligned}
\)
Our solutions are \(\theta=\frac{2 \pi}{3}, \frac{4 \pi}{3}, \pi\).

Hint

\(370^{\circ}\) is greater than \(360^{\circ}\) so let us subtract \(360^{\circ}\)
\(
\begin{gathered}
370^{\circ}-360^{\circ}=10^{\circ} \\
\cos \left(370^{\circ}\right)=\cos \left(10^{\circ}\right)=\mathbf{0 . 9 8 5} \text { (to } 3 \text { decimal places) }
\end{gathered}
\)

Hint

\(
m \angle B+155^{\circ}=180^{\circ}
\)

\(
m \angle B=25^{\circ}
\)

Hint

\(
\begin{aligned}
\sin \left(35^{\circ}\right) & =\frac{\text { Opposite }}{\text { Hypotenuse }} \\
& =\frac{2.8}{4.9} \\
& =\mathbf{0 . 5 7}
\end{aligned}
\)

Hint

To solve \(m\), use the sine ratio.
\(\operatorname{Sin} 72.3^{\circ}=m / 315\)
\(0.953=\mathrm{m} / 315\)
\(m=315 \times 0.953\)
\(m=300.195 \mathrm{mtr}\)
The man is \(300.195 \mathrm{mtr}\) above the ground.

Hint

Let \(x\) be the angle of elevation of the sun, then
\(
\begin{aligned}
& \tan x=55 / 23=2.391 \\
& x=\tan ^{-1}(2.391)
\end{aligned}
\)
or \(x=67.30\) degrees

Hint

From the given,
\(
\begin{aligned}
& P Q=12 \mathrm{~cm} \\
& \mathrm{PR}=13 \mathrm{~cm}
\end{aligned}
\)
In the right triangle \(P Q R, Q\) is right angle.
By Pythagoras theorem,
\(
\begin{aligned}
& \mathrm{PR}^2=\mathrm{PQ}^2+\mathrm{QR}^2 \\
& \mathrm{QR}^2=(13)^2-(12)^2 \\
& =169-144 \\
& =25 \\
& \mathrm{QR}=5 \mathrm{~cm} \\
& \tan \mathrm{P}=\mathrm{QR} / \mathrm{PQ}=5 / 12 \\
& \cot \mathrm{R}=\mathrm{QR} / \mathrm{PQ}=5 / 12
\end{aligned}
\)
So, \(\tan \mathrm{P}-\cot \mathrm{R}=(5 / 12)-(5 / 12)=0\)

Hint

\(
\begin{aligned}
& \left(\sin ^4 \theta-\cos ^4 \theta+1\right) \operatorname{cosec}^2 \theta \\
& =\left[\left(\sin ^2 \theta-\cos ^2 \theta\right)\left(\sin ^2 \theta+\cos ^2 \theta\right)+1\right] \operatorname{cosec}^2 \theta \\
& \text { Using the identity } \sin ^2 A+\cos ^2 A=1 \\
& =\left(\sin ^2 \theta-\cos ^2 \theta+1\right) \operatorname{cosec}^2 \theta \\
& =\left[\sin ^2 \theta-\left(1-\sin ^2 \theta\right)+1\right] \operatorname{cosec}^2 \theta \\
& =2 \sin ^2 \theta \operatorname{cosec}^2 \theta \\
& =2 \sin ^2 \theta\left(1 / \sin ^2 \theta\right) \\
& =2
\end{aligned}
\)

Hint

Given,
\(\sin 3 A=\cos \left(A-26^{\circ}\right) ; 3 A\) is an acute angle
\(
\begin{aligned}
& \cos \left(90^{\circ}-3 A\right)=\cos \left(A-26^{\circ}\right)\left\{\text { since } \cos \left(90^{\circ}-A\right)=\sin A\right\} \\
& \Rightarrow 90^{\circ}-3 A=A-26 \\
& \Rightarrow 3 A+A=90^{\circ}+26^{\circ} \\
& \Rightarrow 4 A=116^{\circ} \\
& \Rightarrow A=116^{\circ} / 4 \\
& \Rightarrow A=29^{\circ}
\end{aligned}
\)

Hint

Given,
\(
\tan \theta+\sec \theta=l \text {….(i) }
\)
We know that,
\(
\begin{aligned}
& \sec ^2 \theta-\tan ^2 \theta=1 \\
& (\sec \theta-\tan \theta)(\sec \theta+\tan \theta)=1 \\
& (\sec \theta-\tan \theta) l=1\{\text { from (i) }\} \\
& \sec \theta-\tan \theta=1 / l \ldots \text { (ii) } \\
& \text { Adding (i) and (ii), } \\
& \tan \theta+\sec \theta+\sec \theta-\tan \theta=l+(1 / l) \\
& 2 \sec \theta=\left(l^2+1\right) l \\
& \sec \theta=\left(l^2+1\right) / 2 l
\end{aligned}
\)

Hint

Given,
\(a \sin \theta+b \cos \theta=c\)
Squaring on both sides,
\((a \sin \theta+b \cos \theta)^2=c^2\)
\(a^2 \sin ^2 \theta+b^2 \cos ^2 \theta+2 a b \sin \theta \cos \theta=c^2\)
Using the identity \(\sin ^2 A+\cos ^2 A=1\),
\(a^2\left(1-\cos ^2 \theta\right)+b^2\left(1-\sin ^2 \theta\right)+2 a b \sin \theta \cos \theta=c^2\)
\(a^2-a^2 \cos ^2 \theta+b^2-b^2 \sin ^2 \theta+2 a b \sin \theta \cos \theta=c^2\)
\(a^2+b^2-c^2=a^2 \cos ^2 \theta+b^2 \sin ^2 \theta-2 a b \sin \theta \cos \theta\)
\(a^2+b^2-c^2=(a \cos \theta-b \sin \theta)^2\)
\(\Rightarrow a \cos \theta-b \sin \theta=\sqrt{ }\left(a^2+b^2-c^2\right)\)

Hint

We first calculate distance \(d\) using the time and speed ( 1 minute \(=1 / 60\) hour)
\(
d=600 \times (1 / 60)=10 \text { miles }
\)
We next express the tangent of the given angles of elevation as follows
\(
\tan \left(20^{\circ}\right)=h /(d+x)
\)
and
\(
\tan \left(60^{\circ}\right)=\mathrm{h} / \mathrm{x}
\)
Eliminate \(x\) in the two equations above to find a relationship between \(\mathrm{h}\) and \(\mathrm{d}\)
\(
\mathrm{h}=\mathrm{d} /\left[1 / \tan \left(20^{\circ}\right)-1 / \tan \left(60^{\circ}\right)\right]
\)
\(=4.6\) miles (rounded to 2 decimal places)

Hint

First, we can notice that \(\pi-\frac{\pi}{6}=\frac{5 \pi}{6}\) and so the terminal line for \(\frac{5 \pi}{6}\) will form an angle of \(\frac{\pi}{6}\) with the negative \(x\)-axis in the second quadrant and we’ll have the following unit circle for this problem.

The coordinates of the line representing \(\frac{5 \pi}{6}\) will be the same as the coordinates of the line representing \(\frac{\pi}{6}\) except that the \(x\) coordinate will now be negative. So, our new coordinates will then be \(\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\) and so the answer is,
\(
\cos \left(\frac{5 \pi}{6}\right)=-\frac{\sqrt{3}}{2}
\)

Hint

First we can notice that \(-\pi-\frac{\pi}{3}=-\frac{4 \pi}{3}\) and so (remembering that negative angles are rotated clockwise) we can see that the terminal line for \(-\frac{4 \pi}{3}\) will form an angle of \(\frac{\pi}{3}\) with the negative \(x\)-axis in the second quadrant and we’ll have the following unit circle for this problem. The coordinates of the line representing \(-\frac{4 \pi}{3}\) will be the same as the coordinates of the line representing \(\frac{\pi}{3}\) except that the \(x\) coordinate will now be negative. So, our new coordinates will then be \(\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\) and so the answer is,
\(
\sin \left(-\frac{4 \pi}{3}\right)=\frac{\sqrt{3}}{2}
\)

Hint

\(x^2+x^2=12\) and solve for x.

Hint

First, sketch the scenario. The path taken by Shane forms a right-angled triangle. The distance from the starting point forms the hypotenuse.

\(
x=\sqrt{3^2+6^2}=6.71 \mathrm{~m}
\)

Hint

\(\mathrm{OE}\) is the radius of the circle, which is \(12 \mathrm{~cm}\)
\(
\begin{aligned}
& O P^2+P E^2=O E^2 \\
& 6^2+P E^2=12^2 \\
& P E=\sqrt{12^2-6^2} \\
& E F=2 \times P E=20.78 \mathrm{~cm}
\end{aligned}
\)

Hint

Let the point above the ground \(=x\)
As given in the question,
Length of the foot ladder \(=35\)
Base of the ladder \(=21\) feet
Let the equation be \(a^2+b^2=c^2\)
\(
(21)^2+(x)^2=(35)^2
\)
\(441+x=1225\)
Subtracting by 441 on both sides
\(
x^2=784
\)
Squaring and both sides
\(
x=\sqrt{ } 784=28
\)

Hint

\(
\begin{aligned}
15^2+20^2 & =r^2 \\
225+400 & =r^2 \\
\sqrt{625} & =\sqrt{r^2} \\
25 & =r
\end{aligned}
\)

Hint

\(
\text { Draw a diagram, letting } x \text { be the distance covered by the car travelling eastbound. }
\)

By Pythagorean theorem (since the directions east and north make a right angle) we have:
\(
\begin{aligned}
& 15^2+x^2=(x+5)^2 \\
& 225+x^2=x^2+10 x+25 \\
& 225-25=10 x \\
& 200=10 x \\
& x=20
\end{aligned}
\)
Hence, the eastbound car has travelled 20 miles.

Hint

Because this forms a right triangle, with the 650 feet of string being the hypotenuse, use the pythagorean theorem, \(a^2+b^2=c^2\). A is \(600 \mathrm{ft}\) and \(c\) is \(650 \mathrm{ft}\).
Set up the equation:
\(
\begin{aligned}
& 600^2+x^2=650^2 \\
& 360000+x^2=422500 \\
& x^2=62500 \\
& x=250
\end{aligned}
\)
The kite was 250 feet high off the ground.

Hint

In the right triangle above, according to the Pythagorean theorem, we have
\(
(x-17)^2+x^2=25^2
\)
\(
x=-7 \text { or } x=24
\)
Because the base of a triangle can never be a negative value, we can ignore \(x=-7\).
So, the value \(x\) is 24 .
Then,
\(
x-17=24-17=7
\)
So, the base and height of the right triangle are 24 inches and 7 inches respectively.

Hint

\(
\begin{aligned}
& -1 \leq \sin x \leq 1 \\
& -4 \leq 4 \sin x \leq 4 \\
& p=4+1=5
\end{aligned}
\)

Hint

\(
\begin{aligned}
& a=3 \mathrm{~km} \\
& \alpha=5^{\circ} \\
& \tan \alpha=x / a \\
& x=a \cdot \tan \alpha=a \cdot \tan 5^{\circ}=3 \cdot \tan 5^{\circ}=3 \cdot 0.087489=0.262=0.2625 \mathrm{~km}
\end{aligned}
\)

Hint

To solve this problem, first set up a diagram that shows all of the info given in the problem. 

Next, we need to interpret which side length corresponds to the shadow of the building, which is what the problem is asking us to find. Is it the hypotenuse, or the base of the triangle? Think about when you look at a shadow. When you see a shadow, you are seeing it on something else, like the ground, the sidewalk, or another object. We see the shadow on the ground, which corresponds to the base of our triangle, so that is what we’ll be solving for. We’ll call this base \(b\).

Next, think about which trig functions relate our known angle, \(22^{\circ}\), to the base (or adjacent) and the opposite sides of the triangle. If you thought tangent (or cotangent), you are correct! We know that \(\tan x=\frac{\text { opposite }}{\text { adjacent }}\) and \(\cot x=\frac{\text { adjacent }}{\text { opposite }}\). For simplicity’s sake, we’ll use tangent to solve this problem. We have:
\(
\begin{aligned}
& \tan 22^{\circ}=\frac{60}{b} \\
& b=\frac{60}{\tan 22^{\circ}} \\
& b=\frac{60}{.4} \text { (Use a calculator and round to two places to find that } \tan 22^{\circ}=.40 \text { ) } \\
& b=150 \text { meters }
\end{aligned}
\)
Therefore the shadow cast by the building is 150 meters long.

Hint

To solve this problem, we need to create a diagram, but in order to create that diagram, we need to understand the vocabulary that is being used in this question. The following diagram clarifies the difference between an angle of depression (an angle that looks downward; relevant to our problem) and the angle of elevation (an angle that looks upward; relevant to other problems, but not this specific one.) Imagine that the top of the blue altitude line is the top of the lighthouse, the green line labelled GroundHorizon is sea level, and point B is where the boat is.

Merging together the given info and this diagram, we know that the angle of depression is \(19^{\circ}\) and and the altitude (blue line) is 105 meters. While the blue line is drawn on the left hand side in the diagram, we can assume is it is the same as the right hand side. Next, we need to think of the trig function that relates the given angle, the given side, and the side we want to solve for. The altitude or blue line is opposite the known angle, and we want to find the distance between the boat (point B) and the top of the lighthouse. That means that we want to determine the length of the hypotenuse, or red line labelled SlantRange. The sine function relates opposite and hypotenuse, so we’ll use that here. We get:
\(\sin 19^{\circ}=\frac{105}{d}\) (where \(d\) is the distance between the top of the lighthouse and the boat)
\(d=\frac{105}{\sin 19^{\circ}}\)
\(d=\frac{105}{.33}\) (using a calculator in degree mode and rounding to two digits, we get that \(\sin 19^{\circ}=.33\) )
\(d=318.18\) meters

Hint

As with other trig problems, begin with a sketch of a diagram of the given and sought-after information.

Angelina and her car start at the bottom left of the diagram. The road she is driving on is the hypotenuse of our triangle, and the angle of the road relative to flat ground is \(22^{\circ}\). Because we want to find the change in height (also called elevation), we want to determine the difference between her ending and starting heights, which is labelled \(\mathrm{x}\) in the diagram. Next, consider which trig function relates together an angle and the sides opposite and hypotenuse relative to it; the correct one is sine. Then, set up:
\(
\begin{aligned}
& \sin 22^{\circ}=\frac{x}{4000} \\
& 4000 \cdot \sin 22^{\circ}=x \\
& 4000 \cdot .37=x \text { (using a calculator in degree mode and rounding to two decimals we get that } \sin 22^{\circ}=.37 \text { ) } \\
& 1480=x
\end{aligned}
\)
Therefore the change in height between Angelina’s starting and ending points is 1480 meters.

Hint

To solve this problem, let’s start by drawing a diagram of the two buildings, the distance in between them, and the angle between the tops of the two buildings. Then, label in the given lengths and angle. 

We are being asked to find the height of the taller building, but this diagram does not provide a triangle that has as one of its sides the entire height of the larger (rightmost and blue) building. However, we can instead find the distance \(\overline{\mathrm{HI}}\), and then add that to the 40 foot height of the shorter building to find the entire height of the taller building. Start by finding \(\overline{\mathrm{HI}}\) :
\(
\begin{aligned}
& \tan 48^{\circ}=\frac{\overline{\mathrm{HI}}}{50} \\
& 50 \cdot \tan 48^{\circ}=\overline{\mathrm{HI}} \\
& 50 \cdot 1.11=\overline{\mathrm{HI}} \\
& 55.5=\overline{\mathrm{HI}}
\end{aligned}
\)
Remember that this is not the full height of the larger building. To find that, we need to add \(55.5+40=95.5\) feet. Therefore, the taller building is \(95.5\) feet tall.

Hint

We know that \(\tan (\pi / 6)=1 /(\sqrt{3})\)
Since, \(\tan (\pi-\pi / 6)=-\tan (\pi / 6)=-1 /(\sqrt{ } 3)\)
Further, \(\tan (2 \pi-\pi / 6)=-\tan (\pi / 6)=-1 /(\sqrt{3})\)
Hence, the principal solutions are \(\tan (\pi-\pi / 6)=\tan (5 \pi / 6)\) and \(\tan (2 \pi-\pi / 6)=\tan (11 \pi / 6)\)

Hint

\(
\begin{aligned}
& \sin (11 \pi / 12) \text { can be written as } \sin (2 \pi / 3+\pi / 4) \\
& \text { using formula, } \sin (x+y)=\sin x \cos y+\cos x \sin y \\
& \sin (11 \pi / 12)=\sin (2 \pi / 3+\pi / 4)=\sin (2 \pi / 3) \cos \pi / 4+\cos (2 \pi / 3) \sin \pi / 4 \\
& =(\sqrt{ } 3) / 2 \times \sqrt{ } 2 / 2+(-1 / 2) \times \sqrt{ } 2 / 2 \\
& =\sqrt{6} / 4-(\sqrt{ } 2) / 4 \\
& =(\sqrt{ } 6-\sqrt{ } 2) / 4
\end{aligned}
\)

Hint

We know, \(\operatorname{cosec} x=\operatorname{cosec} \pi / 6=2\) or \(\sin x=\sin \pi / 6=1 / 2\).
Or, \(x=n \pi+(-1)^n \pi / 6\)

Hint

\(5 \cos ^2 y+2 \sin y=0\)
or \(5\left(1-\sin ^2 y\right)+2 \sin y=0\)
or \(5 \sin ^2 y-2 \sin y-5=0\)
i.e., \(\sin y=1.2\) or \(\sin y=-0.8\).
Since sin y can not be greater than 1 ,
\(\sin y=-0.8=\sin (\pi+\pi / 3)\)
or \(\sin y=\sin 4 \pi / 3\), and hence, the solution is given by \(y=n \pi+(-1)^n 4 \pi / 3\)

Hint

We know that, \(\sin \pi / 3=(\sqrt{3}) / 2\) and \(\sin 2 \pi / 3=\sin (\pi-\pi / 3)=\sin \pi / 3=(\sqrt{3}) / 2\)
Therefore, the principal solutions are \(x=\pi / 3\) and \(2 \pi / 3\).

Hint

I can use a trig identity to get a quadratic in cosine:
\(
\begin{aligned}
& \cos ^2(\alpha)+\cos (\alpha)=\sin ^2(\alpha) \\
& \cos ^2(\alpha)+\cos (\alpha)=1-\cos ^2(\alpha) \\
& 2 \cos ^2(\alpha)+\cos (\alpha)-1=0 \\
& (2 \cos (\alpha)-1)(\cos (\alpha)+1)=0 \\
& \cos (\alpha)=\frac{1}{2} \text { or } \cos (\alpha)=-1
\end{aligned}
\)
The first trig equation, \(\cos (\alpha)=\frac{1}{2}\), gives me \(\alpha=60^{\circ}\) and \(\alpha=300^{\circ}\). The second equation gives me \(\alpha=\) \(180^{\circ}\). So my complete solution is:
\(
\alpha=60^{\circ}, 180^{\circ}, 300^{\circ}
\)

Hint

First, I’ll get everything over to one side of the “equals” sign:
\(
\sin ^2(\theta)-\sin (\theta)-2=0
\)
This equation is “a quadratic in sine”; that is, the form of the equation is the quadratic-equation format:
\(
a \mathrm{X}^2+b \mathrm{X}+c=0
\)
In the case of the equation they’re wanting me to solve, \(\mathrm{X}=\sin (\theta), a=1, b=-1\), and \(c=-2\).
Since this is quadratic in form, I can apply some quadratic-equation methods. In the case of this equation, I can factor the quadratic:
\(
\begin{aligned}
& \sin ^2(\theta)-\sin (\theta)-2=0 \\
& (\sin (\theta)-2)(\sin (\theta)+1)=0
\end{aligned}
\)
The first factor gives me the related trig equation:
\(
\sin (\theta)=2
\)
But the sine is never more than 1 , so this equation is not solvable; it has no solution.
The other factor gives me the second related trig equation:
\(
\begin{aligned}
& \sin (\theta)+1=0 \\
& \sin (\theta)=-1 \\
& \theta=\frac{3}{2} \pi
\end{aligned}
\)

Hint

The natural log (well, any log) is zero when the argument is 1 , so this gives me:
\(
\begin{aligned}
& 2-\sin ^2(x)=1 \\
& 1-\sin ^2(x)=0 \\
& (1-\sin (x))(1+\sin (x))=0 \\
& 1=\sin (x) \text { or } 1=-\sin (x)
\end{aligned}
\)
From what I know of the sine wave, my solution is:
\(
x=90^{\circ}, 270^{\circ}
\)

Hint

By nature of logarithms, the equivalent exponential equation is:
\(
\begin{aligned}
& 2 \sin (x)=3^{1 / 2} \\
& 2 \sin (x)=\sqrt{3} \\
& \sin (x)=\frac{\sqrt{3}}{2}
\end{aligned}
\)
The sine takes on this value at \(x=\frac{\pi}{3}\) and also at \(x=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}\). Then my solution is:
\(
x=\frac{\pi}{3}, \frac{2 \pi}{3}
\)

Hint

\(
\begin{aligned}
& \sin x \sqrt{3}-2 \sin x \cos x=0 \\
& (\sin x)(\sqrt{3}-2 \cos x)=0 \\
& \sin x=0, x=0, \pi \\
&
\end{aligned}
\)
\(
\begin{aligned}
& \sqrt{3}-2 \cos x=0 \\
& \cos x=\frac{\sqrt{3}}{2} \\
& x=\frac{\pi}{6}, \frac{11 \pi}{6}
\end{aligned}
\)
\(
x=0, \frac{\pi}{6}, \pi, \frac{11 \pi}{6}
\)

Hint

\(
\begin{aligned}
& \sin ^2 x+\sin x-2=0 \\
& (\sin x-1)(\sin x+2)=0 \\
& \sin x-1=0 \\
& \sin x=1 \\
& x=\frac{\pi}{2}
\end{aligned}
\)
\(
\begin{aligned}
& \sin x+2=0 \\
& \sin x=-2
\end{aligned}
\)
No solution. (Since the minimum value of \(\sin x\) is \(-1\), it cannot equal \(-2\).)

Hint

\(
\begin{aligned}
& \text { (c) } \because k=\frac{2 \tan x}{1+\tan ^2 x}=\sin 2 x \\
& \Rightarrow \quad \sin 2 C=\sin 2 B \\
& \text { But } \quad \angle C>\angle B \\
& \Rightarrow \quad 2 C=\pi-2 B \Rightarrow B+C=\frac{\pi}{2} \\
& \therefore \quad \angle A=\frac{\pi}{2} \\
&
\end{aligned}
\)

Hint

(c) Given, \(f(x)=\frac{\tan ^2 x+4 \tan x+9}{1+\tan ^2 x}\) \(=\frac{2(2 \tan x)}{1+\tan ^2 x}+4\left(\frac{1-\tan ^2 A}{1+\tan ^2 A}\right)+5\)
\(
\begin{aligned}
& =2 \sin 2 x+4 \cos 2 x+5 \\
\therefore \quad R_f & =[{5}-\sqrt{20}, 5+\sqrt{20}]
\end{aligned}
\)
Hence, \((M+m)=10\).

Hence, the maximum and minimum values of trigonometrical functions of the form \(a \sin x+b \cos x\) are \(\sqrt{a^2+b^2}\) and \(-\sqrt{a^2+b^2}\), respectively.

Hint

(d) We have, \(4 \cos 18^{\circ}-\frac{3}{\cos 18^{\circ}}-2 \tan 18^{\circ}\)
\(
\begin{aligned}
& =\frac{4 \cos ^2 18^{\circ}-3-2 \sin 18^{\circ}}{\cos 18^{\circ}} \\
& =\frac{2\left(1+\cos 36^{\circ}\right)-2 \sin 18^{\circ}-3}{\cos 18^{\circ}} \\
& =\frac{2\left(1+\cos 36^{\circ}-\sin 18^{\circ}\right)-3}{\cos 18^{\circ}} \\
& =\frac{2\left(1+\frac{1}{2}\right)-3}{\cos 18^{\circ}}=0
\end{aligned}
\)

Hint

(d)
\(
\begin{aligned}
& \text { As, }\left(\frac{1}{1+2 \cos ^2 A}+\frac{2}{\sec ^2 A+2}\right) \\
& =\left(\frac{1}{1+2 \cos ^2 A}+\frac{2 \cos ^2 A}{1+2 \cos ^2 A}\right) \\
& =\frac{\left(1+2 \cos ^2 A\right)}{\left(1+2 \cos ^2 A\right)}=1 \\
&
\end{aligned}
\)
Hence, \(\log _{\frac{1}{2}}(1)=0\).

Hint

\(
\begin{aligned}
T_1=\frac{1}{\sin 1^{\circ}} & {\left[\frac{\sin \left(46^{\circ}-45^{\circ}\right)}{\sin 45^{\circ} \sin 46^{\circ}}\right]=\frac{1}{\sin 1^{\circ}}\left[\cot 45^{\circ}-\cot 46^{\circ}\right] } \\
T_2 & =\frac{1}{\sin 1^{\circ}}\left[\frac{\sin \left(48^{\circ}-47^{\circ}\right)}{\sin 48^{\circ} \sin 47^{\circ}}\right] \\
& =\frac{1}{\sin 1^{\circ}}\left[\cot 47^{\circ}-\cot 48^{\circ}\right] \\
T_l & =\frac{1}{\sin 1^{\circ}}\left[\frac{\sin \left(133^{\circ}-134^{\circ}\right)}{\sin 133^{\circ} \sin 134^{\circ}}\right] \\
& =\frac{1}{\sin 1^{\circ}}\left[\cot 133^{\circ}-\cot 134^{\circ}\right]
\end{aligned}
\)
On adding
\(
\sum_{r=1}^l T_r=\frac{1}{\sin 1^{\circ}}\left[\left\{\cot 45^{\circ}+\cot 47^{\circ}\right.\right.\left.+\cot 49^{\circ}+\ldots+\cot 133^{\circ}\right\}\left.-\left\{\cot 46^{\circ}+\cot 48^{\circ}+\cot 50^{\circ}+\ldots+\cot 134^{\circ}\right\}\right]
\)
\(
=\operatorname{cosec} 1^{\circ}
\)
[all terms cancelled except \(\cot 45^{\circ}\) remains]

Hint

\(
\begin{aligned}
& \text { (a) } f(\theta)=\frac{1-\sin 2 \theta+\cos 2 \theta}{2 \cos 2 \theta} \\
& =\frac{(\cos \theta-\sin \theta)^2+\left(\cos ^2 \theta-\sin ^2 \theta\right)}{2(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)} \\
& =\frac{(\cos \theta-\sin \theta)+(\cos \theta+\sin \theta)}{2(\cos \theta+\sin \theta)} \\
& =\frac{2 \cos \theta}{2(\cos \theta+\sin \theta)}=\frac{1}{1+\tan \theta} \\
& f\left(11^{\circ}\right) \cdot f\left(34^{\circ}\right)=\frac{1}{\left(1+\tan 11^{\circ}\right)} \cdot \frac{1}{\left(1+\tan 34^{\circ}\right)} \\
& =\frac{1}{\left(1+\tan 11^{\circ}\right)} \cdot \frac{1}{\left(1+\tan \left(45^{\circ}-11^{\circ}\right)\right)} \\
& =\frac{1}{\left(1+\tan 11^{\circ}\right)} \cdot \frac{1}{1+\frac{1-\tan 11^{\circ}}{1+\tan 11^{\circ}}} \\
& =\frac{1}{\left(1+\tan 11^{\circ}\right)} \cdot \frac{1+\tan 11^{\circ}}{2}=\frac{1}{2} \\
&
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (d) }|\sin x \cos x|+|\tan x+\cot x|=\sqrt{3} \\
& \Rightarrow \quad|\sin x \cos x|+\frac{1}{|\sin x \cos x|}=\sqrt{3} \\
& \text { but } \quad|\sin x \cos x|+\frac{1}{|\sin x \cos x|} \geq 2
\end{aligned}
\)
Hence, no solution.

Hint

(c) We have, \(4 \cos A \cos B+4 \sin A \sin B \sin C=4\)
\(
\begin{array}{ll}
\Rightarrow & \sin C=\frac{1-\cos A \cos B}{\sin A \cos B} \leq 1 \\
\Rightarrow & 1 \leq \sin A \sin B+\cos A \cos B \\
\Rightarrow & \cos (A-B) \geq 1 \\
\Rightarrow & A=B \text { and } \sin C=\frac{1-\cos ^2 A}{\sin ^2 A}=1 \\
\therefore & C=90^{\circ} \\
\text { and } & A=B=\frac{\pi}{4} \text { (each). }
\end{array}
\)

Hint

\(
\begin{aligned}
& \text { (d) }(\Sigma \cos A)^2+(\Sigma \sin A)^2=9 \\
& \Sigma\left(\cos ^2 A+\sin ^2 A\right)+2(\Sigma \cos A \cos B+\sin A \sin B) \\
& 3+2 \Sigma \cos (A-B) \leq 3+2(3)=9 .
\end{aligned}
\)
Equality holds if \(A=B=C\)
\(\Rightarrow \triangle A B C\) is equilateral \(\Rightarrow\) Infinite many equilateral
[Note We can vary side length of equilateral triangle]

Hint

\(
\text { (a) } T_1=\operatorname{cosec} \theta=\frac{\sin \left(\theta-\frac{\theta}{2}\right)}{\sin \frac{\theta}{2} \sin \theta} ; \theta=\frac{\pi}{32}
\)
\(
\left.\begin{array}{l}
T_1=\cot \frac{\theta}{2}-\cot \theta \\
T_2=\cot \theta-\cot 2 \theta \\
T_3=\cot 2 \theta-\cos 2^2 \theta \\
T_4=\cot 2^2 \theta-\cos 2^3 \theta \\
T_5=1
\end{array}\right] \text { sum }=1+\cot \frac{\theta}{2}-\cot 8 \theta
\)
\(
\begin{aligned}
\text { Sum } & =1+\cot \frac{\theta}{2}-\cot 8 \theta \\
& =1+\cot \frac{\pi}{64}-\cot \frac{\pi}{4}=\cot \frac{\pi}{64}=\cot \frac{\pi}{k} \therefore k=64
\end{aligned}
\)

Hint

(d) Put \(\theta=\frac{\pi}{2}\) in the given inequality, we get \(d=0\)
Put \(\theta=0\) in the given inequality, we get
\(
a+b+c+d=1 \dots(i)
\)
So, (d) is correct and (c) is not correct.
Now differentiate both sides with respect to \(\theta\), we get
\(
\begin{array}{r}
-5 \sin \theta=-a \sin \theta-3 b \cos ^2 \theta \sin \theta \\
-5 c \cos ^4 \theta \sin \theta \dots(ii)
\end{array}
\)

Put
\(
\theta=\frac{\pi}{2} \text {, then } a=5 \dots(iii)
\)

Again putting \(\theta=\frac{\pi}{4}\) in the given expression or in (ii), we get
\(
4 a+2 b+c=-4 \dots(iv)
\)

From (i), (iii) and (iv) we have \(b=-20\) and \(c=16\)

Hint

(b) From the given relations, we have
\(
\sin 2 B=\left(\frac{3}{2}\right) \sin 2 A \text { and } 3 \sin ^2 A=1-2 \sin ^2 B=\cos 2 B
\)
so that
\(
\begin{aligned}
\cos (A+2 B) & =\cos A \cos 2 B-\sin A \sin 2 B \\
& =\cos A \cdot 3 \sin ^2 A-\left(\frac{3}{2}\right) \sin A \sin 2 A \\
& =3 \cos A \sin ^2 A-3 \sin ^2 A \cos A=0 \\
A+2 B & =\frac{\pi}{2}
\end{aligned}
\)

Hint

(b) We can write
\(
\begin{aligned}
& k_1=\tan 27 \theta-\tan 9 \theta+\tan 9 \theta-\tan 3 \theta+\tan 3 \theta-\tan \theta \\
& \text { But } \tan 3 \theta-\tan \theta=\frac{\sin 3 \theta \cos \theta-\cos 3 \theta \sin \theta}{\cos 3 \theta \cos \theta} \\
&=\frac{\sin 2 \theta}{\cos 3 \theta \cos \theta} \\
&=\frac{2 \sin \theta}{\cos 3 \theta} \\
& \therefore \quad k_1=2\left[\frac{\sin 9 \theta}{\cos 27 \theta}+\frac{\sin 3 \theta}{\cos 9 \theta}+\frac{\sin \theta}{\cos 3 \theta}\right]=2 k_2
\end{aligned}
\)

Hint

(a)
\(
\begin{aligned}
674=a^2 & -2 a \frac{1-\tan ^2\left(\frac{x}{2}\right)}{1+\tan ^2\left(\frac{x}{2}\right)}+1 \\
& =a^2-2 a \times \frac{1-49}{1+49}+1 \\
& =a^2+2 a \times \frac{48}{50}+1
\end{aligned}
\)
\(
\begin{aligned}
& 25 a^2+48 a-673 \times 25=0 \\
&(a-25)(25 a+673)=0 \\
& a=25 \quad \text { (taking the integral value of } a \text { ). }
\end{aligned}
\)

Hint

(a) Let \(\frac{\sin ^3 x}{1+\cos x}+\frac{\cos ^3 x}{1-\sin x}=A\), then
\(
\begin{aligned}
& A=\frac{\left(\sin ^3 x+\cos ^3 x\right)+\left(\cos ^4 x-\sin ^4 x\right)}{(1+\cos x)(1-\sin x)} \\
& \left\{\left(\sin ^3 x+\cos ^3 x\right)\right\} \\
& \frac{+\left\{\begin{array}{l}
(\cos x+\sin x)(\cos x-\sin x) \\
\left(\cos ^2 x+\sin ^2 x\right)
\end{array}\right\}}{(1+\cos x)(1-\sin x)} \\
& (\sin x+\cos x)\{(1-\sin x \cos x) \\
&
\end{aligned}
\)
\(
\begin{aligned}
& A=\sin x+\cos x \\
& A=\sqrt{2}\left[\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x\right] \dots(i)
\end{aligned}
\)
\(
\begin{aligned}
A & =\sqrt{2}\left[\cos \frac{\pi}{4} \sin x+\sin \frac{\pi}{4} \cos x\right] \\
& =\sqrt{2} \sin \left[\frac{\pi}{4}+x\right]
\end{aligned}
\)
Again, by Eq. (i)
\(
\begin{aligned}
A & =\sqrt{2}\left[\sin \frac{\pi}{4} \sin x+\cos \frac{\pi}{4} \cos x\right] \\
& =\sqrt{2} \cos \left[\frac{\pi}{4}-x\right]
\end{aligned}
\)

Hint

\(
\text { (c) } \begin{aligned}
x^2+y^2 & =X^2+Y^2 \\
x y & =\left(X^2-Y^2\right) \sin \theta \cdot \cos \theta-X Y\left(\cos ^2 \theta-\sin ^2 \theta\right) \\
x^2 & +4 x y+y^2=X^2+Y^2+2\left(X^2-Y^2\right)\sin 2 \theta-2 X Y \cos 2 \theta
\end{aligned}
\)
\(
=(1+2 \sin 2 \theta) X^2+(1-2 \sin 2 \theta) Y^2-2 \cos 2 \theta \cdot X Y
\)
From the question,
\(
\begin{aligned}
a & =1+2 \sin 2 \theta, b=1-2 \sin 2 \theta, \cos 2 \theta=0 \\
\cos 2 \theta & =0 \Rightarrow \theta=\frac{\pi}{4}, \text { then } \\
a & =1+2 \sin \frac{\pi}{2}, b=1-2 \sin \frac{\pi}{2} \\
\therefore \quad a & =3, b=-1
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (b) } a=\sin \frac{\pi}{18} \sin \frac{5 \pi}{18} \sin \frac{7 \pi}{18} \\
& =\sin 10^{\circ} \sin 50^{\circ} \sin 70^{\circ} \\
& =\frac{1}{2}\left[2 \sin 70^{\circ} \sin 10^{\circ}\right] \sin 50^{\circ} \\
& =\frac{1}{2}\left[\cos 60^{\circ}-\cos 80^{\circ}\right] \sin 50^{\circ} \\
& =\frac{1}{4} \sin 50^{\circ}-\frac{1}{4}\left(2 \cos 80^{\circ} \sin 50^{\circ}\right) \\
& =\frac{1}{4} \sin 50^{\circ}-\frac{1}{4}\left(\sin 130^{\circ}-\sin 30^{\circ}\right) \\
& =\frac{1}{4} \sin 50^{\circ}-\frac{1}{4} \sin 50^{\circ}+\frac{1}{4} \cdot \frac{1}{2}=\frac{1}{8} \\
& y=2[x]+2 \text { and } y=3[x-2] \\
& \Rightarrow 2[x]+2=3[x-2] \\
& =3[x]+3[-2] \Rightarrow[x]=8 \\
& \therefore \quad a=\frac{1}{[x]} \\
&
\end{aligned}
\)

Hint

\(
\begin{aligned}
\cos \left(\frac{2 \pi}{7}\right) & +\cos \left(\frac{4 \pi}{7}\right)+\cos \left(\frac{6 \pi}{7}\right) \\
& =\operatorname{Re}\left\{e^{\frac{2 \pi i}{7}}+e^{\frac{4 \pi i}{7}}+e^{\frac{6 \pi i}{7}}\right\}
\end{aligned}
\)
\(
\begin{aligned}
& =\frac{e^{\frac{2 \pi i}{7}}+e^{\frac{4 \pi i}{7}}+e^{\frac{6 \pi i}{7}}+e^{\frac{-4 \pi i}{7}}+e^{\frac{-4 \pi i}{7}}+e^{\frac{-6 \pi i}{7}}}{2} \\
& =\frac{-1+\left(1+e^{\frac{2 \pi i}{7}}+e^{\frac{4 \pi i}{7}}+e^{\frac{6 \pi i}{7}}+e^{\frac{-2 \pi i}{7}}+e^{\frac{-4 \pi i}{7}}+e^{\frac{-6 \pi i}{7}}\right)}{2} \\
& =\frac{-1+\text { (Sum of seven roots of unity) }}{2} \\
& =\frac{-1+0}{2}=-\frac{1}{2}
\end{aligned}
\)

Hint

(b) From the third relation we get
\(
\begin{aligned}
& \cos \theta \cos \phi+\sin \theta \sin \phi=\sin \beta \sin \gamma \\
& \Rightarrow \quad \sin ^2 \theta \sin ^2 \phi=(\cos \theta \cos \phi-\sin \beta \sin \gamma)^2 \\
& \Rightarrow \quad\left(1-\frac{\sin ^2 \beta}{\sin ^2 \alpha}\right)\left(1-\frac{\sin ^2 \gamma}{\sin ^2 \alpha}\right) \\
& =\left(\frac{\sin \beta \sin \gamma}{\sin ^2 \alpha}-\sin \beta \sin \gamma\right)^2 \\
& \Rightarrow \quad\left(\sin ^2 \alpha-\sin ^2 \beta\right)\left(\sin ^2 \alpha-\sin ^2 \gamma\right) \\
& =\sin ^2 \beta \sin ^2 \gamma\left(1-\sin ^2 \alpha\right)^2 \\
& \Rightarrow \quad \sin ^4 \alpha\left(1-\sin ^2 \beta \sin ^2 \gamma\right) \\
& -\sin ^2 \alpha\left(\sin ^2 \beta+\sin ^2 \gamma-2 \sin ^2 \beta \sin ^2 \gamma\right)=0 \\
&
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \quad \sin ^2 \alpha=\frac{\sin ^2 \beta+\sin ^2 \gamma-2 \sin ^2 \beta \sin ^2 \gamma}{1-\sin ^2 \beta \sin ^2 \gamma} \\
& \text { and } \quad \cos ^2 \alpha=\frac{1-\sin ^2 \beta-\sin ^2 \gamma+\sin ^2 \beta \sin ^2 \gamma}{1-\sin ^2 \beta \sin ^2 \gamma} \\
& \Rightarrow \quad \tan ^2 \alpha=\frac{\sin ^2 \beta-\sin ^2 \beta \sin ^2 \gamma+\sin ^2 \gamma-\sin ^2 \beta \sin ^2 \gamma}{\cos ^2 \beta-\sin ^2 \gamma\left(1-\sin ^2 \beta\right)} \\
& =\frac{\sin ^2 \beta \cos ^2 \gamma+\cos ^2 \beta \sin ^2 \gamma}{\cos ^2 \beta \cos ^2 \gamma} \\
& =\tan ^2 \beta+\tan ^2 \gamma \\
& \Rightarrow \quad \tan ^2 \alpha-\tan ^2 \beta-\tan ^2 \gamma=0 \\
&
\end{aligned}
\)

Hint

(c) \(\sqrt{2} \cos A=\cos B+\cos ^3 B \dots(i)\)
and \(\sqrt{2} \sin A=\sin B-\sin ^3 B \dots(ii)\)
\(
\begin{aligned}
\Rightarrow \sqrt{2} & \sin A \cos B-\sqrt{2} \cos A \sin B \\
& =\left(\sin B-\sin ^3 B\right) \cos B-\left(\cos B+\cos ^3 B\right) \sin B \\
& =-\sin B \cos B
\end{aligned}
\)
\(
\Rightarrow \sin (A-B)=\frac{-1}{2 \sqrt{2}} \sin 2 B
\)
Now squaring and adding Eqs. (i) and (ii), we get
\(
2=\cos ^2 B+\sin ^2 B+\cos ^6 B+\sin ^6 B+2\left(\cos ^4 B-\sin ^4 B\right)
\)
\(
\begin{aligned}
& \Rightarrow \quad 1=\left(\cos ^2 A+\sin ^2 A\right)^3-3 \cos ^2 A \sin ^2 \\
& A\left(\cos ^2 A+\sin ^2 A\right)+2 \cos 2 B
\end{aligned}
\)
\(
\begin{array}{lc}
\Rightarrow & 1=1-\left(\frac{3}{4}\right) \sin ^2 2 B+2 \cos 2 B \\
\Rightarrow & -3 \sin ^2 2 B+8 \cos 2 B=0 \\
\Rightarrow & 3 \cos ^2 2 B+8 \cos 2 B-3=0 \\
\Rightarrow & \cos 2 B=\frac{1}{3} \\
\Rightarrow & \sin 2 B= \pm \frac{2 \sqrt{2}}{3} \\
\therefore & \sin (A-B)= \pm \frac{1}{3}
\end{array}
\)

Hint

\(
\begin{aligned}
& =\cos 55^{\circ}+\cos 65^{\circ}+\cos 175^{\circ} \\
& =2 \cos \frac{55^{\circ}+65^{\circ}}{2} \cos \frac{55^{\circ}-65^{\circ}}{2}+\cos 17 \\
& =2 \cos 60^{\circ} \cos \left(-5^{\circ}\right)+\cos 175^{\circ} \\
& =2 \times \frac{1}{2} \cos 5^{\circ}+\cos \left(180^{\circ}-5^{\circ}\right) \\
& =\cos 5^{\circ}-\cos 5^{\circ}=0
\end{aligned}
\)

Hint

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

Hint

\(
\begin{aligned}
& \sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ}=\frac{1}{2}\left(2 \sin 80^{\circ} \sin 40^{\circ}\right) \sin 20^{\circ} \\
& =\frac{1}{2}\left[\cos \left(80^{\circ}-40^{\circ}\right)-\cos \left(80^{\circ}+40^{\circ}\right) \sin 20^{\circ}\right. \\
& =\frac{1}{2}\left(\cos 40^{\circ}-\cos 120^{\circ}\right) \sin 20^{\circ} \\
& =\frac{1}{4}\left(2 \cos 40^{\circ} \sin 20^{\circ}-2 \cos 120^{\circ} \sin 20^{\circ}\right) \\
& =\frac{1}{4}\left[\sin \left(40^{\circ}+20^{\circ}\right)-\sin \left(40^{\circ}-20^{\circ}\right)-2\left(-\frac{1}{2}\right) \sin 20^{\circ}\right] \\
& =\frac{1}{4}\left[\sin 60^{\circ}-\sin 20^{\circ}+\sin 20^{\circ}\right]=\frac{1}{4} \sin 60^{\circ} \\
& =\frac{1}{4} \cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{8}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { We have, }\sqrt{2+\sqrt{2+\sqrt{2(1+\cos 8 \theta)}}} \\
& \Rightarrow \sqrt{2+\sqrt{2+\sqrt{2\left(2 \cos ^2 4 \theta\right)}}} \left[\because 1+\cos 8 \theta=2 \cos ^2 \frac{8 \theta}{2}\right]
\end{aligned}
\)
\(
\begin{array}{ll}
\Rightarrow & \sqrt{2+\sqrt{2+\sqrt{\left(4 \cos ^2 4 \theta\right)}}} \\
\Rightarrow & \sqrt{2+\sqrt{2+2 \cos 4 \theta}} \\
\Rightarrow & \sqrt{2+\sqrt{2(1+\cos 4 \theta)}} \\
\Rightarrow & \sqrt{2+\sqrt{2\left(2 \cos ^2 2 \theta\right)}} \quad\left[\because 1+\cos 4 \theta=2 \cos ^2 2 \theta\right] \\
\Rightarrow & \sqrt{2+2 \cos 2 \theta}=\sqrt{2(1+\cos 2 \theta)}
\end{array}
\)
\(
\begin{aligned}
& =\sqrt{2+2 \cos 2 \theta}=\sqrt{2(1+\cos 2 \theta)} \\
& =\sqrt{2\left(2 \cos ^2 \theta\right)}=2 \cos \theta
\end{aligned}
\)

Hint

We have \(\cos \frac{7 \pi}{8}=\cos \left(\pi-\frac{\pi}{8}\right)=-\cos \frac{\pi}{8}\)
\(
\begin{aligned}
& \cos \frac{5 \pi}{8}=\cos \left(\pi-\frac{3 \pi}{8}\right)=-\cos \frac{3 \pi}{8} \\
& \left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)=\frac{1}{b} \\
& =\left(1-\cos ^2 \frac{\pi}{8}\right)\left(1-\cos ^2 \frac{3 \pi}{8}\right) \\
& =\sin ^2 \frac{\pi}{8} \sin ^2 \frac{3 \pi}{8} \\
& =\frac{1}{4}\left(2 \sin ^2 \frac{\pi}{8}\right)\left(2 \sin ^2 \frac{3 \pi}{8}\right) \\
& =\frac{1}{4}\left[\left(1-\cos \frac{\pi}{4}\right)\left(1-\cos \frac{3 \pi}{4}\right)\right] \because\left[1-\cos \theta=2 \sin ^2 \frac{\theta}{2}\right] \\
& =\frac{1}{4}\left[\left(1-\frac{1}{\sqrt{2}}\right)\left(1+\frac{1}{\sqrt{2}}\right)\right]=\frac{1}{4}\left(1-\frac{1}{2}\right)=\frac{1}{8}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { We have, } \tan \left(x+100^{\circ}\right)=\tan \left(x+50^{\circ}\right) \tan x \tan \left(x-50^{\circ}\right) \\
& \Rightarrow \quad \frac{\tan \left(x+100^{\circ}\right)}{\tan \left(x+50^{\circ}\right)}=\tan \left(x+50^{\circ}\right) \tan x^{\circ} \\
& \Rightarrow \quad \frac{\sin \left(x+100^{\circ}\right) \cos \left(x-50^{\circ}\right)}{\cos \left(x+100^{\circ}\right) \sin \left(x-50^{\circ}\right)}=\frac{\sin \left(x+50^{\circ}\right) \sin x}{\cos \left(x+50^{\circ}\right) \cos x} \\
& \Rightarrow \frac{\sin \left(x+100^{\circ}\right) \cos \left(x-50^{\circ}\right)+\cos \left(x+100^{\circ}\right) \sin \left(x-50^{\circ}\right)}{\sin \left(x+100^{\circ}\right) \cos \left(x-50^{\circ}\right)-\cos \left(x+100^{\circ}\right) \sin \left(x-50^{\circ}\right)} \\
& \quad=\frac{\sin \left(x+50^{\circ}\right) \sin x+\cos \left(x+50^{\circ}\right) \cos x}{\sin \left(x+50^{\circ}\right) \sin x-\cos \left(x+50^{\circ}\right) \cos x} \\
& \Rightarrow \quad \frac{\sin \left(x+100^{\circ}+x-50^{\circ}\right)}{\sin \left(x+100^{\circ}-x+50^{\circ}\right)}=\frac{\cos \left(x+50^{\circ}-x\right)}{-\cos \left(x+50^{\circ}+x\right)} \\
& \Rightarrow \quad \frac{\sin \left(2 x+50^{\circ}\right)}{\sin 150^{\circ}}=\frac{\cos 50^{\circ}}{-\cos \left(2 x+50^{\circ}\right)}
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \quad \sin \left(2 x+50^{\circ}\right) \cos \left(2 x+50^{\circ}\right)=-\sin 150^{\circ} \cos 50^{\circ} \\
& \Rightarrow \quad 2 \sin \left(2 x+50^{\circ}\right) \cos \left(2 x+50^{\circ}\right)=-2 \cos 60^{\circ} \cos 50^{\circ} \left[\because \sin 150^{\circ}=\cos 60^{\circ}\right]
\end{aligned}
\)
\(
\begin{array}{ll}
\Rightarrow & \sin \left(4 x+100^{\circ}\right)=\sin \left(270-50^{\circ}\right) \\
\Rightarrow & \sin \left(4 x+100^{\circ}\right)=\sin 220^{\circ} \\
\Rightarrow & 4 x+100^{\circ}=220^{\circ} \Rightarrow x=30^{\circ}
\end{array}
\)

Hint

Given, \(\alpha=112^{\circ} 30^{\prime}\)
\(
\begin{aligned}
& \therefore \quad 2 \alpha=225^{\circ} \\
& \text { or } \quad \cos 2 \alpha=\cos 225^{\circ}=\cos \left(180^{\circ}+45^{\circ}\right) \\
& =-\cos 45^{\circ}=-\frac{1}{\sqrt{2}} \\
&
\end{aligned}
\)
Now, \(\sin ^2 \alpha=\frac{1-\cos 2 \alpha}{2}\)
Since \(\alpha\) lies in the 2 nd quadrant \(\therefore \sin \alpha\) is positive
\(
\begin{aligned}
\therefore \quad \sin \alpha & =\sqrt{\frac{1-\cos 2 \alpha}{2}}=\sqrt{\frac{1-\left(-\frac{1}{\sqrt{2}}\right)}{2}} \\
& =\sqrt{\frac{\sqrt{2}+1}{2 \sqrt{2}}}=\frac{\sqrt{2+2 \sqrt{2}}}{2} \\
\text { Hence, } \sin \alpha & =\frac{\sqrt{2+2 \sqrt{2}}}{2}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \cos 6^{\circ} \cos 42^{\circ} \cos 66^{\circ} \cos 78^{\circ} \\
& =\frac{1}{4}\left(2 \cos 66^{\circ} \cos 6^{\circ}\right)\left(2 \cos 78^{\circ} \cos 42^{\circ}\right) \\
& =\frac{1}{4}\left[\cos \left(66^{\circ}+6^{\circ}\right)+\cos \left(66^{\circ}-6^{\circ}\right)\right] \\
& \quad \times\left[\cos \left(78^{\circ}+42^{\circ}\right)+\cos \left(78^{\circ}-42^{\circ}\right)\right] \\
& =\frac{1}{4}\left(\cos 72^{\circ}+\cos 60^{\circ}\right)\left(\cos 120^{\circ}+\cos 36^{\circ}\right)
\end{aligned}
\)
\(
\begin{aligned}
& =\frac{1}{4}\left(\sin 18^{\circ}+\frac{1}{2}\right)\left(-\frac{1}{2}+\cos 36^{\circ}\right) \\
& =\frac{1}{4}\left[\frac{(\sqrt{5}-1)}{4}+\frac{1}{2}\right]\left[-\frac{1}{2}+\frac{(\sqrt{5}-1)}{4}\right] \\
& {\left[\because \sin 18^{\circ}=\frac{(\sqrt{5}-1)}{4} \text { and } \cos 36^{\circ}=\frac{(\sqrt{5}+1)}{4}\right]} \\
& =\frac{1}{4} \cdot \frac{(\sqrt{5}+1)}{4} \cdot \frac{(\sqrt{5}-1)}{4}=\frac{(5-1)}{64} \\
& =\frac{4}{64}=\frac{1}{16}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \sin \frac{\pi}{5} \sin \frac{2 \pi}{5} \sin \frac{3 \pi}{5} \sin \frac{4 \pi}{5} \\
& =\sin \frac{\pi}{5} \sin \frac{2 \pi}{5} \sin \left(\pi-\frac{2 \pi}{5}\right) \sin \left(\pi-\frac{\pi}{5}\right) \\
& =\sin ^2 \frac{\pi}{5} \sin ^2 \frac{2 \pi}{5} \\
& =\left(\sin 36^{\circ}\right)^2 \times\left(\sin 72^{\circ}\right)^2 \\
& =\left(\sin 36^{\circ}\right)^2 \times\left(\cos 18^{\circ}\right)^2 \\
& {\left[\because \sin 72^{\circ}=\sin \left(90^{\circ}-18^{\circ}\right)=\cos 18^{\circ}\right]}
\end{aligned}
\)
\(
\begin{aligned}
& =\frac{(10-2 \sqrt{5})}{16} \times \frac{(10+2 \sqrt{5})}{16}=\frac{(100-20)}{(16 \times 16)} \\
& {\left[\begin{array}{l}
\because \sin 36^{\circ}=\frac{\sqrt{10-2 \sqrt{5}}}{4} \\
\text { and } \cos 18^{\circ}=\frac{\sqrt{10+2 \sqrt{5}}}{4}
\end{array}\right]} \\
& =\frac{80}{(16 \times 16)}=\frac{5}{16} \\
&
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \frac{\sqrt{3}-1}{2 \sqrt{2} \sin x}+\frac{\sqrt{3}+1}{2 \sqrt{2} \cos x}=2 \\
& \sin \frac{\pi}{12} \cos x+\cos \frac{\pi}{12} \sin x=\sin 2 x \\
& \sin 2 x=\sin \left(x+\frac{\pi}{12}\right)
\end{aligned}
\)
\(
\begin{aligned}
2 x & =x+\frac{\pi}{12} \\
\text { or } 2 x & =\pi-x-\frac{\pi}{12} \\
x & =\frac{\pi}{12} \\
\text { or } 3 x & =\frac{11 \pi}{12} \\
x & =\frac{\pi}{12} \text { or } \frac{11 \pi}{36}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (a,c) } \tan A+\tan B+\tan C=\tan A \tan B \tan C \\
& \Rightarrow \quad \tan A+5=3 \tan B \\
& \Rightarrow \quad 5+\tan A=3(5-\tan C) \\
& \Rightarrow \quad 5+\tan A=15-\frac{9}{\tan A} \\
& \Rightarrow \quad \tan ^2 A-10 \tan A+9=0 \\
& \Rightarrow \quad \tan A=1 \text { or } \tan A=9 \\
&
\end{aligned}
\)
\(\Rightarrow \tan B\) and \(\tan C\) are 2,3 or \(\frac{14}{3}, \frac{1}{3}\), respectively
\(\Rightarrow \triangle A B C\) is always on acute angled triangle and sum of all possible values of \(\tan A\) is 10 .

Hint

(b, c) \(y=\cos ^4 x-\cos ^2 x+1\)
\(
=\left(\cos ^2 x-\frac{1}{2}\right)^2+\frac{3}{4}
\)
\(\therefore y_{\min }=\frac{3}{4}\) and \(y\) is maximum when \(\left(\cos ^2 x-\frac{1}{2}\right)^2\) is then maximum
\(
\therefore \quad y_{\max }=\frac{1}{4}+\frac{3}{4}=1
\)

Hint

\(
\begin{aligned}
& (b, c) \tan A+\tan B+\tan C=6 \dots(i) \\
& \Rightarrow \tan A \tan B \tan C=6 \\
& 2 \tan C=6 \\
& \therefore \quad \tan C=3 \\
& \therefore \quad \sin ^2 C=\frac{\tan ^2 C}{1+\tan ^2 C}=\frac{9}{1+9}=\frac{9}{10} \\
&
\end{aligned}
\)
From Eq. (i), \(\tan A+\tan B=3\) and \(\tan A \tan B=2\)
\(\tan A-\tan B\)
\(
\begin{aligned}
& = \pm \sqrt{\left.\left\{(\tan A+\tan B)^2-4 \tan A \tan B\right)\right\}} \\
& = \pm 1
\end{aligned}
\)
we get, \(\tan A=2,1\) and \(\tan B=1,2\)
\(
\begin{array}{ll}
\therefore & \sin ^2 A=\frac{4}{1+4}, \frac{1}{1+1} \text { and } \sin ^2 B=\frac{1}{1+1}, \frac{4}{1+4} \\
\Rightarrow & \sin ^2 A=\frac{8}{10}, \frac{5}{10} \text { and } \sin ^2 B=\frac{5}{10}, \frac{8}{10} \\
\therefore & \sin ^2 A: \sin ^2 B: \sin ^2 C=8: 5: 9 \text { or } 5: 8: 9
\end{array}
\)

Hint

(a) We have, \(x \sin ^3 \theta+y \cos ^3 \theta=\sin \theta \cos \theta\) and \(x \sin \theta-y \cos \theta=0 \dots(i)\)
From Eq. (ii), \(\tan \theta=\frac{y}{x} \dots(ii)\)
\(
\therefore \quad \sin \theta=\frac{y}{\sqrt{\left(x^2+y^2\right)}} \text { and } \cos \theta=\frac{x}{\sqrt{\left(x^2+y^2\right)}}
\)

\(
\begin{aligned}
& \text { From Eq., (i) } x \times \frac{y^3}{\left(x^2+y^2\right)^{\frac{3}{2}}}+y \times \frac{x^3}{\left(x^2+y^2\right)^{\frac{3}{2}}} \\
& =\frac{x y}{\left(x^2+y^2\right)} \\
& \text { or } \quad \frac{\left(x^2+y^2\right)}{\left(x^2+y^2\right)^{\frac{3}{2}}}=\frac{1}{\left(x^2+y^2\right)} \Rightarrow\left(x^2+y^2\right)^{\frac{1}{2}}=1 \\
& \text { or } \quad x^2+y^2=1 \text { which is a circle }
\end{aligned}
\)

Hint

For the expression \(a \cos ^2 \theta+b \sec ^2 \theta\) if \(b>a\), then minimum value attains at \(\cos ^2 \theta=\sec ^2 \theta=1\)
\(\Rightarrow \max\) of \(\left\{9-\left(2 \cos ^2 \theta+4 \sec ^2 \theta\right)\right\}=3\)
So, maximum of \(\left.\log _3\left(9-2 \cos ^2 \theta+4 \sec ^2 \theta\right)\right)=1\)

Hint

\(
(\sec A \sec B+\tan A \tan B)^2-(\sec A \tan B+\tan A \sec B)^2
\)
\(
\begin{aligned}
& =\left[\frac{1+\sin A \sin B}{\cos A \cos B}\right]^2-\left[\frac{\sin B+\sin A}{\cos A \cos B}\right]^2 \\
& =\frac{1+\sin ^2 A \sin ^2 B-\sin ^2 B-\sin ^2 A}{\cos ^2 A \cos ^2 B} \\
& =\frac{1-\sin ^2 B \cos ^2 A-\sin ^2 A}{\cos ^2 A \cos ^2 B} \\
& =\frac{\cos ^2 A \cos ^2 B}{\cos ^2 A \cos ^2 B}=1
\end{aligned}
\)
\(
\Rightarrow(\sec A \sec B+\tan A \tan B)^2=(91)^2+1=8282
\)

Hint

\(
\text { We know }\left(\cos 27^{\circ}+\sin 27^{\circ}\right)^2
\)
\(
\begin{aligned}
& =1+\sin 54=1+\cos 36^{\circ} \\
& \Rightarrow \quad \cos 27^{\circ}+\sin 27^{\circ}=\sqrt{\left(1+\cos 36^{\circ}\right)} \quad[\because \text { LHS }>0]
\end{aligned}
\)
Also, \(\cos 27^{\circ}-\sin 27^{\circ}=\sqrt{\left(1-\cos 36^{\circ}\right)}\)
\(
\begin{aligned}
& {\left[\because \cos 27^{\circ}>\sin 27^{\circ}\right]} \\
& \therefore \quad 2 \sin 27^{\circ}=\sqrt{\left(1+\cos 36^{\circ}\right)}-\sqrt{\left(1-\cos 36^{\circ}\right)} \\
& =\sqrt{\left(1+\left(\frac{\sqrt{5}+1}{4}\right)\right)}-\sqrt{\left(1-\left(\frac{\sqrt{5}+1}{4}\right)\right)} \\
& \therefore \quad 4 \sin 27^{\circ}=\sqrt{(5+\sqrt{5})}-\sqrt{(3-\sqrt{5})} \\
&
\end{aligned}
\)
On comparing, we get
\(
\begin{aligned}
& \alpha=5+\sqrt{5}, \beta=3-\sqrt{5} \\
& \therefore \quad \alpha+\beta=8, \alpha \beta=10-2 \sqrt{5} \\
& \alpha+\beta-\alpha \beta+2=2 \sqrt{5} \\
& \therefore(\alpha+\beta-\alpha \beta+2)^4=400
\end{aligned}
\)

Hint

\(\sin A\) and \(\cos A\) are the roots of the equation \(4 x^2-3 x+a=0\), then
\(
\sin A+\cos A=\frac{3}{4}, \sin A \cos A=\frac{a}{4}
\)
Also, \(\sin A+\cos A+\tan A+\cot A+\sec A+\operatorname{cosec} A=7\)
\(
\begin{aligned}
& \Rightarrow \quad(\sin A+\cos A)+\left(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right) \\
& +\left(\frac{1}{\cos A}+\frac{1}{\sin A}\right)=7 \\
& \Rightarrow \quad(\sin A+\cos A)+\frac{1}{\sin A \cos A}+\frac{(\sin A+\cos A)}{\sin A \cos A}=7 \\
& \Rightarrow \quad \frac{3}{4}+\frac{4}{a}+\frac{3}{a}=7 \\
& \Rightarrow \quad \frac{3}{4}+\frac{7}{a}=7 \\
& \Rightarrow \quad \frac{7}{a}=7-\frac{3}{4}=\frac{25}{4} \\
& \therefore \quad 25 a=28 \\
&
\end{aligned}
\)

Hint

Here, \(\cos 290^{\circ}=\cos \left(270^{\circ}+20^{\circ}\right)=\sin 20^{\circ}\) and \(\sin 250^{\circ}=\sin \left(270^{\circ}-20^{\circ}\right)=-\cos 20^{\circ}\)
\(
\begin{aligned}
& \therefore \text { The given expression }=\frac{1}{\sin 20^{\circ}}-\frac{1}{\sqrt{3} \cos 120^{\circ}}=\lambda \\
& \Rightarrow \quad \frac{1}{\sin 20^{\circ}}-\frac{\cos 60^{\circ}}{\sin 60^{\circ} \cos 20^{\circ}}=\lambda \\
& \Rightarrow \quad \frac{\sin 60^{\circ} \cos 20^{\circ}-\cos 60^{\circ} \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ} \sin 60^{\circ}}=\lambda \\
& \Rightarrow \quad \frac{\sin \left(60^{\circ}-20^{\circ}\right)}{\frac{\sin 40^{\circ}}{2} \times \frac{\sqrt{3}}{2}}=\lambda \\
& \therefore \quad \lambda=\frac{4}{\sqrt{3}} \\
& \Rightarrow \quad \lambda^2=\frac{16}{3} \\
&
\end{aligned}
\)
Then, \(9 \lambda^4+81 \lambda^2+97=9 \times \frac{256}{9}+81 \times \frac{16}{3}+97\)
\(
=256+432+97=785
\)

Hint

\(
\begin{gathered}
\log _{10}\left(\frac{\sin 2 x}{2}\right)=-1 \\
\frac{\sin 2 x}{2}=\frac{1}{10} \\
\sin 2 x=\frac{1}{5}
\end{gathered}
\)
\(
\text { Also } \log _{10}(\sin x+\cos x)=\frac{\log _{10}\left(\frac{n}{10}\right)}{2}
\)
\(
\begin{aligned}
& \log _{10}(\sin x+\cos x)^2=\log _{10}\left(\frac{n}{10}\right) \\
& 1+\sin 2 x=\frac{n}{10} \\
& 1+\frac{1}{5}=\frac{n}{10} \text { or } \frac{6}{5}=\frac{n}{10} \\
& \frac{n}{3}=4
\end{aligned}
\)

Hint

Given, \(5\left(\tan ^2 x-\cos ^2 x\right)=2 \cos 2 x+9\)
\(
\Rightarrow \quad 5\left(\frac{1-\cos 2 x}{1+\cos 2 x}-\frac{1+\cos 2 x}{2}\right)=2 \cos 2 x+9
\)
Put \(\cos 2 x=y\), we have
\(
5\left(\frac{1-y}{1+y}-\frac{1+y}{2}\right)=2 y+9
\)
\(
\begin{aligned}
& 5\left(2-2 y-1-y^2-2 y\right)=2(1+y)(2 y+9) \\
& 5\left(1-4 y-y^2\right)=2\left(2 y+9+2 y^2+9 y\right) \\
& 5-20 y-5 y^2=22 y+18+4 y^2 \\
& 9 y^2+42 y+13=0 \\
& 9 y^2+3 y+39 y+13=0 \\
& 3 y(3 y+1)+13(3 y+1)=0 \\
& (3 y+1)(3 y+13)=0 \\
& y=-\frac{1}{3},-\frac{13}{3} \\
& \cos 2 x=-\frac{1}{3},-\frac{13}{3} \\
& \cos 2 x=-\frac{1}{3} \quad\left[\because \cos 2 x \neq-\frac{13}{3}\right] \\
&
\end{aligned}
\)
Now,
\(
\begin{aligned}
\cos 4 x & =2 \cos ^2 2 x-1=2\left(-\frac{1}{3}\right)^2-1 \\
& =\frac{2}{9}-1=-\frac{7}{9}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}=\frac{\sin A}{\cos A} \times \frac{\sin A}{\sin A-\cos A} \\
& \quad+\frac{\cos A}{\sin A} \times \frac{\cos A}{\cos A-\sin A} \\
&= \frac{1}{\sin A-\cos A}\left\{\frac{\sin ^3 A-\cos ^3 A}{\cos A \sin A}\right\} \\
&= \frac{\sin ^2 A+\sin A \cos A+\cos ^2 A}{\sin A \cos A} \\
&= \frac{1+\sin A \cos A}{\sin A \cos A} \\
&= 1+\sec A \operatorname{cosec} A
\end{aligned}
\)

Hint

Given \(A \triangle P Q R\) such that
\(
\begin{aligned}
& 3 \sin P+4 \cos Q=6 \dots(i) \\
& 4 \sin Q+3 \cos P=1 \dots(ii)
\end{aligned}
\)
On squaring and adding Eqs. (i) and (ii) both sides we get
\(
\begin{aligned}
& (3 \sin P+4 \cos Q)^2+(4 \sin Q+3 \cos P)^2=36+1 \\
& \Rightarrow 9\left(\sin ^2 P+\cos ^2 P\right)+16\left(\sin ^2 Q+\cos ^2 Q\right) \\
& +2 \times 3 \times 4(\sin P \cos Q+\sin Q \cos P)=37 \\
& \Rightarrow \quad 24[\sin (P+Q)]=37-25 \\
& \Rightarrow \quad \sin (P+Q)=\frac{1}{2} \\
&
\end{aligned}
\)
Since, \(P\) and \(Q\) are angles of \(\triangle P Q R\), hence \(0^{\circ}<P, Q<180^{\circ}\).
\(
\begin{aligned}
& \Rightarrow \quad P+Q=30^{\circ} \text { or } 150^{\circ} \\
& \Rightarrow \quad R=150^{\circ} \text { or } 30^{\circ} \\
&
\end{aligned}
\)
Hence, two cases arise here.
\(
\begin{aligned}
& \text { Case I } \quad R=150^{\circ} \\
& R=150^{\circ} \Rightarrow P+Q=30^{\circ} \\
& \Rightarrow \quad 0<P, Q<30^{\circ} \\
& \Rightarrow \quad \sin P<\frac{1}{2}, \cos Q<1 \\
& \Rightarrow \quad 3 \sin P+4 \cos Q<\frac{3}{2}+4 \\
& \Rightarrow \quad 3 \sin P+4 \cos Q<\frac{11}{2}<6 \\
& \Rightarrow 3 \sin P+4 \cos Q \Rightarrow 6 \text { is not possible. } \\
& \text { Case II } \quad R=30^{\circ} \\
&
\end{aligned}
\)
Hence, \(R=30^{\circ}\) is the only possibility.

Hint

\(
\begin{aligned}
& A=\sin ^2 x+\cos ^4 x \\
& A=1-\cos ^2 x+\cos ^4 x
\end{aligned}
\)
\(
\begin{aligned}
& =\cos ^4 x-\cos ^2 x+\frac{1}{4}+\frac{3}{4} \\
& =\left(\cos ^2 x-\frac{1}{2}\right)^2+\frac{3}{4} \dots(i)
\end{aligned}
\)
\(
\begin{array}{ll}
\text { where, } & 0 \leq\left(\cos ^2 x-\frac{1}{2}\right)^2 \leq \frac{1}{4} \dots(ii) \\
\therefore & \frac{3}{4} \leq A \leq 1
\end{array}
\)

Hint

\(
\cos (\alpha+\beta)=\frac{4}{5} \Rightarrow \alpha+\beta \in \text { Ist quadrant }
\)
and \(\sin (\alpha-\beta)=\frac{5}{13} \Rightarrow \alpha-\beta \in\) Ist quadrant
Now, \(\quad 2 \alpha=(\alpha+\beta)+(\alpha-\beta)\)
\(
\therefore \quad \tan 2 \alpha=\frac{\tan (\alpha+\beta)+\tan (\alpha-\beta)}{1-\tan (\alpha+\beta) \tan (\alpha-\beta)}=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \cdot \frac{5}{12}}=\frac{56}{33}
\)

Hint

\(
\begin{aligned}
& \cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)=-\frac{3}{2} \\
& \Rightarrow 2[\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)]+3=0
\end{aligned}
\)
\(
\Rightarrow 2[\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)]+\sin ^2 \alpha+\cos ^2 \alpha+\sin ^2 \beta+\cos ^2 \beta+\sin ^2 \gamma+\cos ^2 \gamma=0
\)
\(
\Rightarrow(\sin \alpha+\sin \beta+\sin \gamma)^2+(\cos \alpha+\cos \beta+\cos \gamma)^2=0
\)
It is possible when,
\(
\text { and } \quad \begin{aligned}
\sin \alpha+\sin \beta+\sin \gamma & =0 \\
\cos \alpha+\cos \beta+\cos \gamma & =0
\end{aligned}
\)
Hence, both statements \(A\) and \(B\) are true.

Hint

\(
\text { Area }=\frac{1}{2} \times \text { Base } \times \text { Altitude }
\)

\(
=\frac{1}{2} \times(2 x \cos \theta) \times(x \sin \theta)=\frac{1}{2} x^2 \sin 2 \theta
\)
[since, maximum value of \(\sin 2 \theta\) is 1]
\(\therefore\) Maximum area \(=\frac{1}{2} x^2\)

Hint

Given, \(\quad \cos x+\sin x=\frac{1}{2}\)
\(
\therefore \frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}+\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}=\frac{1}{2}
\)
Let \(\tan \frac{x}{2}=t \Rightarrow \frac{1-t^2}{1+t^2}+\frac{2 t}{1+t^2}=\frac{1}{2}\)
\(\Rightarrow \quad 2\left(1-t^2+2 t\right)=1+t^2 \Rightarrow 3 t^2-4 t-1=0\)
\(\Rightarrow \quad t=\frac{2 \pm \sqrt{7}}{3}\)
As \(0<x<\pi \Rightarrow 0<\frac{x}{2}<\frac{\pi}{2}\)
So, \(\tan \frac{x}{2}\) is positive.
\(\therefore \quad t=\tan \frac{x}{2}=\frac{2+\sqrt{7}}{3}\)
Now, \(\quad \tan x=\frac{2 \tan \frac{x}{2}}{1-\tan ^2 \frac{x}{2}}=\frac{2 t}{1-t^2}\)
\(\Rightarrow \quad \tan x=\frac{2\left(\frac{2+\sqrt{7}}{3}\right)}{1-\left(\frac{2+\sqrt{7}}{3}\right)^2}\)
\(\Rightarrow \quad \tan x=\frac{-3(2+\sqrt{7})}{1+2 \sqrt{7}} \times \frac{1-2 \sqrt{7}}{1-2 \sqrt{7}}\)
\(\Rightarrow \quad \tan x=-\left(\frac{4+\sqrt{7}}{3}\right)\)

Hint

\(
\begin{aligned}
& \text { Since, } \quad \angle R=\frac{\pi}{2} \\
& \Rightarrow \quad \angle P+\angle Q=\frac{\pi}{2} \\
& \Rightarrow \quad \frac{\angle P}{2}=\frac{\pi}{4}-\frac{\angle Q}{2} \\
& \therefore \quad \tan \frac{P}{2}=\tan \left(\frac{\pi}{4}-\frac{Q}{2}\right) \\
& =\frac{\tan \frac{\pi}{4}-\tan \frac{Q}{2}}{1+\tan \frac{\pi}{4} \tan \frac{Q}{2}} \\
& \Rightarrow \quad \tan \frac{P}{2}+\tan \frac{P}{2} \tan \frac{Q}{2}=1-\tan \frac{Q}{2} \\
& \Rightarrow \quad \tan \frac{P}{2}+\tan \frac{Q}{2}=1-\tan \frac{P}{2} \tan \frac{Q}{2} \\
& \Rightarrow \quad-\frac{b}{a}=1-\frac{c}{a} \\
& \Rightarrow \quad-b=a-c \\
& \Rightarrow \quad c=a+b \\
&
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5} \\
& \Rightarrow \quad \frac{\sin ^4 x}{2}+\frac{\left(1-\sin ^2 x\right)^2}{3}=\frac{1}{5} \\
& \Rightarrow \quad \frac{\sin ^4 x}{2}+\frac{1+\sin ^4 x-2 \sin ^2 x}{3}=\frac{1}{5}
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \quad 5 \sin ^4 x-4 \sin ^2 x+2=\frac{6}{5} \\
& \Rightarrow \quad 25 \sin ^4 x-20 \sin ^2 x+4=0 \\
& \Rightarrow \quad\left(5 \sin ^2 x-2\right)^2=0 \\
& \Rightarrow \quad \sin ^2 x=\frac{2}{5} \\
& \cos ^2 x=\frac{3}{5}, \tan ^2 x=\frac{2}{3} \\
& \therefore \quad \frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125} \\
&
\end{aligned}
\)

Hint

\(
\begin{aligned}
= & \left(\sin \frac{2 \pi}{11}-\cos \frac{2 \pi}{11}\right)+\left(\sin \frac{4 \pi}{11}-\cos \frac{4 \pi}{11}\right) \\
& +\left(\sin \frac{6 \pi}{11}-\cos \frac{6 \pi}{11}\right)+\ldots+\left(\sin \frac{20 \pi}{11}-\cos \frac{20 \pi}{11}\right) \\
= & \left(\sin \frac{2 \pi}{11}+\sin \frac{4 \pi}{11}+\ldots+\sin \frac{20 \pi}{11}\right) \\
= & \frac{\sin \pi \cdot \sin \frac{10 \pi}{11}}{\sin \frac{\pi}{11}}-\frac{\cos \frac{2 \pi}{11}+\cos \frac{4 \pi}{11}+\ldots+\sin \frac{10 \pi}{11}}{\sin \frac{\pi}{11}} \\
= & 0+\frac{\sin \left(\pi-\frac{\pi}{11}\right)}{\sin \frac{\pi}{11}}=1
\end{aligned}
\)

Hint

\(
\begin{aligned}
& f(x)=9 \sin ^2 x-16 \cos ^2 x-10(3 \sin x-4 \cos x) \\
& -10(3 \sin x+4 \cos x)+100 \\
& =25 \sin ^2 x-60 \sin x+84 \\
& =(5 \sin x-6)^2+48 \\
& \therefore f(x)_{\min } \text { occurs when } \sin x=1 \\
& \text { Minimum value }=49 \\
&
\end{aligned}
\)

Hint

\(
S=\frac{1}{1+\tan ^3 0^{\circ}}+\frac{1}{1+\tan ^3 10^{\circ}}+\ldots+\frac{1}{1+\tan ^3 80^{\circ}}
\)
Now, \(\frac{1}{1+\tan ^3 \theta}+\frac{1}{1+\tan ^3(90-\theta)}\)
\(
\begin{aligned}
& =\frac{1}{1+\tan ^3 \theta}+\frac{1}{1+\cot ^3 \theta} \\
& =\frac{1}{1+\tan ^3 \theta}+\frac{\tan ^3 \theta}{1+\tan ^3 \theta} \\
& =\frac{1+\tan ^3 \theta}{1+\tan ^3 \theta}=1
\end{aligned}
\)
Hence, \(S=1+(1+1+1+1)=5\)

Hint

\(
\text { Clearly, } \sqrt{1-\sin ^2 110^{\circ}} \cdot \sec 110^{\circ}
\)
\(
\begin{aligned}
& =\left|\cos 110^{\circ}\right| \sec 110^{\circ} \\
& =-\cos 110^{\circ} \sec 110^{\circ}=-1
\end{aligned}
\)

Hint

Let \(A\) be the expression. Multiplying \(A\) by \(2^{2008}\) and using \(2 \sin \theta \cos \theta=\sin 2 \theta\),
we have
\(
2^{2008} A=\sin \frac{\pi}{2}=1 . A=\frac{1}{2^{2008}}
\)

Alternatively

\(\sin \left(\frac{\pi}{2^{2009}}\right) \cos \left(\frac{\pi}{2^{2009}}\right)=\frac{1}{2} \sin \left(\frac{\pi}{2^{2008}}\right)\)
\(
\begin{aligned}
& =\frac{1}{2^2} \cdot 2 \sin \left(\frac{\pi}{2^{2008}}\right) \cos \left(\frac{\pi}{2^{2008}}\right) \\
& =\frac{1}{2^2} \sin \left(\frac{\pi}{2^{2007}}\right)
\end{aligned}
\)
Similarly, continued product upto,
\(
\cos \left(\frac{\pi}{2^2}\right)=\frac{1}{2^{2008}} \sin \left(\frac{\pi}{2}\right)=\frac{1}{2^{2008}}
\)

Hint

\(
\begin{aligned}
&\left(\cos ^4 1^{\circ}+\cos ^4 2^{\circ}+\cos ^4 3^{\circ}+\ldots+\cos ^4 179^{\circ}\right) \\
&-\left(\sin ^4 1^{\circ}+\sin ^4 2^{\circ}+\sin ^4 3^{\circ}+\ldots+\sin ^4 179^{\circ}\right) \\
&=\cos 2^{\circ}+\cos 4^{\circ}+\cos 6^{\circ}+\ldots+\cos \left(358^{\circ}\right) \\
&=\cos \frac{\left(\frac{2^{\circ}+358^{\circ}}{2}\right) \cdot \sin \left(179 \times 1^{\circ}\right)}{\sin 1^{\circ}} \\
&= \cos \left(180^{\circ}\right)=-1 .
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \sin x+\sin y=a \dots(i)\\
& \quad \cos x+\cos y=2 a \dots(ii)
\end{aligned}
\)
On squaring and adding Eqs. (i) and (ii), we get
\(
\begin{aligned}
2+2 \cos (x-y) & =5 a^2 \\
\cos (x-y) & =\frac{5 a^2-2}{2}
\end{aligned}
\)

Hint

\(
\sum_{r=1}^{18} \cos ^2(5 r)^{\circ}=\cos ^2 5^{\circ}+\cos ^2 10^{\circ}+\cos ^2 15^{\circ}+\ldots+\cos ^2 85^{\circ}+\cos ^2 90^{\circ}
\)
\(
\begin{aligned}
= & \left(\cos ^2 5^{\circ}+\cos ^2 85^{\circ}\right)+\left(\cos ^2 10^{\circ}+\cos ^2 80^{\circ}\right) \\
& +\left(\cos ^2 15^{\circ}+\cos ^2 75^{\circ}\right)+\ldots+\left(\cos ^2 40^{\circ}+\cos ^2 50^{\circ}\right)+\cos ^2 45^{\circ} \\
= & \left(\cos ^2 5^{\circ}+\sin ^2 5^{\circ}\right)+\left(\cos ^2 10^{\circ}+\sin ^2 10^{\circ}\right) \\
& +\left(\cos ^2 15^{\circ}+\sin ^2 15^{\circ}\right)+\ldots+\left(\cos ^2 40^{\circ}+\sin ^2 40^{\circ}\right)+\cos ^2 45^{\circ}
\end{aligned}
\)
\(
=1+1+1+\ldots+1+\frac{1}{2}=8+\frac{1}{2}=\frac{17}{2}
\)

Hint

\(
\begin{aligned}
& A=\sin 44^{\circ}+\cos 44^{\circ} \\
&=\cos 46^{\circ}+\sin 46^{\circ}=C \\
& B=\sin 45^{\circ}+\cos 45^{\circ}=\sqrt{2}\left[\sin 90^{\circ}\right] \\
& A=\sqrt{2}\left[\frac{1}{\sqrt{2}} \sin 44^{\circ}+\frac{1}{\sqrt{2}} \cos 44^{\circ}\right] \\
&=\sqrt{2}\left[\sin 44^{\circ} \cdot \cos 45^{\circ}+\cos 44^{\circ} \cdot \sin 45^{\circ}\right] \\
&=\sqrt{2} \sin 89^{\circ} \\
& \Rightarrow \quad B>A
\end{aligned}
\)

Hint

\(
\begin{aligned}
\cot \left(7 \frac{1^{\circ}}{2}\right)= & \frac{1+\cos 15^{\circ}}{\sin 15^{\circ}} \\
& =\frac{1+\frac{\sqrt{3}+1}{2 \sqrt{2}}}{\frac{\sqrt{3}-1}{2 \sqrt{2}}}=\frac{2 \sqrt{2}+\sqrt{3}+1}{\sqrt{3}-1} \\
& =\frac{(2 \sqrt{2}+\sqrt{3}+1)(\sqrt{3}+1)}{2}
\end{aligned}
\)
\(
\begin{aligned}
& =\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{5}+\sqrt{6} \\
& \therefore \quad n_1=1, n_2=2, n_3=3, n_4=4, n_5=5 \text { and } n_6=6 \\
& \therefore \sum_{i=1}^6 n_i^2=n_1^2+n_2^2+n_3^2+n_4^2+n_5^3+n_6^2 \\
& =1^2+2^2+3^2+4^2+5^2+6^2 \\
& =91 \\
&
\end{aligned}
\)