NCERT Solutions for Exercise 3.3 Class 11 Maths Chapter 3 - Trigonometric Functions

Question 1: Prove that $\small \sin ^{2} \left ( \frac{\pi }{6} \right ) + \cos ^{2}\left ( \frac{\pi }{3} \right ) - \tan ^{2}\left ( \frac{\pi }{4} \right ) = -\frac{1}{2}$

Answer:

We know the values of sin (30 degree), cos (60 degree) and tan (45 degree). That is:

$\sin \left ( \frac{\pi}{6} \right ) = \left ( \frac{1}{2} \right )\\ \\ \cos \left ( \frac{\pi}{3} \right ) = \left ( \frac{1}{2} \right )\\ \\ \tan \left ( \frac{\pi}{4} \right ) = 1$
$\sin^{2}\frac{\pi}{6}+\cos^{2}\frac{\pi}{3}-\tan^{2}\frac{\pi}{4}=$ $\left ( \frac{1}{2} \right )^{2}+ \left ( \frac {1}{2} \right )^{2}-1^{2}$

$= \frac{1}{4}+\frac{1}{4}-1= -\frac{1}{2}$
= R.H.S.

Question 2: Prove that $\small 2\sin ^{2}\left ( \frac{\pi }{6} \right ) + cosec ^{2}\left ( \frac{7\pi }{6} \right )\cos ^{2}\frac{\pi }{3} = \frac{3}{2}$

Answer:

The solutions for the given problem is done as follows.

$\sin\frac{\pi}{6} = \frac {1}{2}\\ \\ cosec\frac{7\pi}{6} = cosec\left ( \pi + \frac{\pi}{6} \right ) = -cosec \frac{\pi}{6}=-2\\ \\ \cos \frac{\pi}{3} = \frac{1}{2}$
$2\sin^{2}\frac{\pi}{6} +cosec^{2}\frac{7\pi}{6}\cos^{2}\frac{\pi}{3} = 2\left ( \frac{1}{2} \right )^{2}+\left ( -2 \right )^{2}\left ( \frac{1}{2} \right )^{2}\\$ $\\ \Rightarrow 2\times\frac{1}{4} + 4\times\frac{1}{4} = \frac {1}{2} + 1= \frac{3}{2}$
R.H.S.

Question 3: Prove that $\small \cot ^{2}\left ( \frac{\pi }{6} \right ) + \csc \left ( \frac{5\pi }{6} \right ) + 3\tan ^{2}\left ( \frac{\pi }{6} \right ) = 6$

Answer:

We know the values of cot(30 degree), tan (30 degree) and cosec (30 degree)

$\cot \frac{\pi}{6} = \sqrt{3}\\ \\ cosec\frac{5\pi}{6} = cosec\left ( \pi - \frac{\pi}{6} \right )=cosec\frac{\pi}{6} = 2\\ \\ \tan\frac{\pi}{6}= \frac{1}{\sqrt{3}}$

$\cot^{2}\frac{\pi}{6} + cosec\frac{5\pi}{6} +3\tan^{2}\frac{\pi}{6} = \left ( \sqrt(3) \right )^{2} + 2 + 3\times\left ( \frac{1}{\sqrt{3}} \right )^{2}\\$ $\\ \Rightarrow 3+2+1 = 6$
R.H.S.

Question 4: Prove that $\small 2\sin ^{2}\left ( \frac{3\pi }{4} \right ) + 2\cos ^{2}\left ( \frac{\pi }{4} \right ) + 2\sec ^{2}\left ( \frac{\pi }{3} \right ) = 10$

Answer:

$\sin \frac{3\pi}{4} = \sin\left ( \pi-\frac{\pi}{4} \right ) = \sin \frac{\pi}{4}= \frac{1}{\sqrt{2}}\\ \\ \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}\\ \\ \sec\frac{\pi}{3}= 2$
Using the above values

$2\sin^{2}\frac{3\pi}{4} +2\cos^{2}\frac{\pi}{4}+2\sec^{2}\frac{\pi}{3} = 2\times\left ( \frac{1}{\sqrt{2}} \right )^{2}+2\times\left ( \frac{1}{\sqrt{2}} \right )^{2}+2\left ( 2 \right )^{2}\\$ $\\ \Rightarrow 1+1+8=10$
R.H.S.

Question 5:(i) Find the value of $\small (i) \sin 75^\circ$

Answer:

$\sin 75^\circ = \sin(45^\circ + 30^\circ)$
We know that
(sin(x+y)=sinxcosy + cosxsiny)
Using this idendity

$\sin 75^\circ = \sin(45^\circ + 30^\circ) = \sin45^\circ\cos30^\circ + \cos45^\circ\sin30^\circ\\$ $\\ \Rightarrow \frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\times\frac{1}{2}\\$

$ \\ \Rightarrow \frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}} = \frac{\sqrt{3}+1}{2\sqrt{2}}$

Question:5(ii) Find the value of
$\small (ii) \tan 15^\circ$

Answer:

$\tan 15^\circ = \tan (45^\circ - 30^\circ)$
We know that,

$\left [ \tan(x-y)= \frac{\tan x - \tan y}{1+\tan x\tan y} \right ]$
By using this we can write

$\tan (45^\circ - 30^\circ)= \frac{\tan 45^\circ - tan30^\circ}{1+\tan45^\circ\tan30^\circ}\\$

$\\ \Rightarrow \frac{1-\frac{1}{\sqrt{3}}}{1+1\left ( \frac{1}{\sqrt{3}} \right )} = \frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\left ( \sqrt{3}-1 \right )^{2}}{\left ( \sqrt{3}+1 \right )\left ( \sqrt{3} -1\right )}=\frac{3+1-2\sqrt{3}}{\left ( \sqrt{3} \right )^{2}-\left ( 1 \right )^{2}}\\ $

$\\ \Rightarrow \frac {4-2\sqrt{3}}{3-1}=\frac{2\left ( 2-\sqrt{3} \right )}{2}= 2-\sqrt{3}$

Question 6: Prove the following: $\small \cos \left ( \frac{\pi }{4}-x \right )\cos \left ( \frac{\pi }{4}-y \right ) - \sin \left ( \frac{\pi }{4} -x\right )\sin \left ( \frac{\pi }{4}-y \right ) =\sin (x+y)$

Answer:

$\cos\left ( \frac{\pi}{4}-x \right )\cos\left ( \frac{\pi}{4}-y \right ) - \sin\left ( \frac{\pi}{4}-x \right )\sin\left ( \frac{\pi}{4}-y \right )$

Multiply and divide by 2 both cos and sin functions
We get,

$\frac{1}{2}\left [2 \cos\left ( \frac{\pi}{4}-x \right )\cos\left ( \frac{\pi}{4}-y \right ) \right ] + \frac{1}{2}\left [- 2\sin\left ( \frac{\pi}{4}-x \right )\sin\left ( \frac{\pi}{4}-y \right ) \right ]$

Now, we know that

2cosAcosB = cos(A+B) + cos(A-B) -(i)
-2sinAsinB = cos(A+B) - cos(A-B) -(ii)
We use these two identities

In our question A = $\left (\frac{\pi}{4}-x \right )$

B = $\left (\frac{\pi}{4}-y \right )$
So,

$\frac{1}{2}\left [ \cos \left \{ \left ( \frac{\pi}{4}-x \right) +\left ( \frac{\pi}{4}-y \right ) \right \} + \cos \left \{ \left ( \frac{\pi}{4}-x \right) -\left ( \frac{\pi}{4}-y \right ) \right \} \right ] +\\$ $\\ \frac{1}{2}\left [ \cos \left \{ \left ( \frac{\pi}{4}-x \right) +\left ( \frac{\pi}{4}-y \right ) \right \} - \cos \left \{ \left ( \frac{\pi}{4}-x \right) +\left ( \frac{\pi}{4}-y \right ) \right \} \right ]$

$\Rightarrow 2 \times \frac{1}{2} \left [ \cos \left \{ \left ( \frac{\pi}{4}-x \right )+\left ( \frac{\pi}{4}-y \right ) \right \} \right ]$

$= \cos \left [ \frac{\pi}{2}-(x+y) \right ]$

As we know that

$(\cos \left ( \frac{\pi}{2} - A \right ) = \sin A)$
By using this

$= \cos \left [ \frac{\pi}{2}-(x+y) \right ]$ $=\sin(x+y)$

R.H.S

Question 7: Prove the following $\small \frac{\tan \left ( \frac{\pi }{4}+x \right )}{\tan \left ( \frac{\pi }{4} -x\right )} = \left ( \frac{1+\tan x}{1-\tan x} \right )^{2}$

Answer:

As we know that

$(\tan (A +B ) = \frac {\tan A + \tan B}{1- \tan A\tan B})$ and $\tan (A-B) = \frac {\tan A - \tan B }{1+ \tan A \tan B}$

So, by using these identities

$\frac{\tan \left ( \frac{\pi}{4}+x \right )}{\tan \left ( \frac{\pi}{4}-x \right )} = \frac{\frac{\tan \frac {\pi}{4} + \tan x}{1- \tan \frac{\pi}{4}\tan x}} {\frac{\tan \frac {\pi}{4} - \tan x}{1+ \tan \frac{\pi}{4}\tan x}} =\frac{ \frac {1+\tan x }{1- \tan x}} { \frac {1-\tan x }{1+ \tan x}} = \left ( \frac{1 + \tan x}{1 - \tan x} \right )^{2}$
R.H.S

Question 8: Prove the following $\small \frac{\cos (\pi +x)\cos (-x)}{\sin (\pi -x)\cos \left ( \frac{\pi }{2}+x \right )} = \cot ^{2} x$

Answer:

As we know that,
$\cos(\pi+x) = -\cos x$ , $\sin (\pi - x ) = \sin x$ , $\cos \left ( \frac{\pi}{2} + x\right ) = - \sin x$
and
$\cos (-x) = \cos x$

By using these our equation simplify to

$\frac{\cos x \times -\cos x}{sin x \times - \sin x} = \frac{- \cos^{2}x}{-\sin^{2}x} = \cot ^ {2}x$ $(\because \cot x = \frac {\cos x}{\sin x})$
R.H.S.

Question 9: Prove the following $\small \cos \left ( \frac{3\pi }{2} +x\right )\cos (2\pi +x)\left [ \cot \left ( \frac{3\pi }{2}-x \right ) +\cot (2\pi +x)\right ] = 1$

Answer:

We know that

$\cos \left ( \frac{3\pi}{2}+x \right ) = \sin x\\ \\ \cos (2\pi +x)= \cos x\\ \\ \cot\left ( \frac{3\pi}{2} -x\right ) = \tan x\\ \\ \cot (2\pi + x) = \cot x$

So, by using these our equation simplifies to

$\cos \left ( \frac{3\pi }{2} +x\right )\cos (2\pi +x)\left [ \cot \left ( \frac{3\pi }{2}-x \right ) +\cot (2\pi +x)\right ] \\=\sin x\cos x [\tan x + \cot x] = \sin x\cos x [\frac {\sin x}{\cos x} + \frac{\cos x}{\sin x}]\\$ $\\ \Rightarrow \sin x\cos x\left [ \frac{\sin^{2}x+\cos^{2}x}{\sin x\cos x } \right ] =\sin^{2}x+\cos^{2}x = 1$ R.H.S.

Question 10: Prove the following $\small \sin (n+1)x\sin(n+2)x + \cos(n+1)x\cos(n+2)x =\cos x$

Answer:

Multiply and divide by 2

$= \frac {2\sin(n+1)x \sin(n+2)x + 2\cos (n+1)x\cos(n+2)x}{2}$

Now by using identities

-2sinAsinB = cos(A+B) - cos(A-B)
2cosAcosB = cos(A+B) + cos(A-B)

$\frac{\left \{ -\left (\cos(2n+3)x - \cos (-x) \right ) + \left ( \cos(2n+3) +\cos(-x) \right )\right \}}{2}\\$ $\\ \left ( \because \cos(-x) = \cos x \right )\\ \\ = \frac{2\cos x}{2} = \cos x$

R.H.S.

Question 11: Prove the following $\small \cos \left ( \frac{3\pi }{4}+x \right ) - \cos\left ( \frac{3\pi }{4} -x\right ) = -\sqrt{2} \sin x$

Answer:

We know that

[ cos(A+B) - cos (A-B) = -2sinAsinB ]

By using this identity

$\cos \left ( \frac {3\pi}{4}+x \right ) - \cos \left ( \frac {3\pi}{4}-x \right ) = -2\sin\frac{3\pi}{4}\sin x = -2\times \frac{1}{\sqrt{2}}\sin x\\$ $\\ = -\sqrt{2}\sin x$ R.H.S.

Question 12: Prove the following $\small \sin^{2}6x - \sin^{2}4x = \sin2x\sin10x$

Answer:

We know that
$a^{2} - b^{2} = (a+b)(a-b)$

So,
$\sin^{2}6x - \sin^{2}4x =(\sin6x + \sin4x)(\sin6x - \sin4x)$

Now, we know that

$\sin A + \sin B = 2\sin \left ( \frac{A+B}{2} \right )\cos\left ( \frac{A-B}{2} \right )\\$ $\\ \sin A - \sin B = 2\cos \left ( \frac{A+B}{2} \right )\sin\left ( \frac{A-B}{2} \right )$
By using these identities
sin6x + sin4x = 2sin5x cosx
sin6x - sin4x = 2cos5x sinx

$\Rightarrow \sin^{2}6x - \sin^{2}4x = (2\cos5x\sin5x)(2\sin x\cos x)$

Now,

2sinAcosB = sin(A+B) + sin(A-B)
2cosAsinB = sin(A+B) - sin(A-B)

by using these identities

2cos5x sin5x = sin10x - 0
2sinx cosx = sin2x + 0

hence
$\sin^{2}6x-\sin^{2}4x = \sin2x\sin10x$

Question 13: Prove the following $\small \cos^{2}2x - \cos^{2}6x = \sin4x\sin8x$

Answer:

As we know that

$a^{2}-b^{2} =(a-b)(a+b)$

$\therefore \cos^{2}2x -\cos^{2}6x = (\cos2x-\cos6x)(\cos2x+\cos6x)$
Now
$\cos A - \cos B = -2\sin\left ( \frac{A+B}{2} \right )\sin\left ( \frac{A-B}{2} \right )\\$ $\\ \cos A + \cos B = 2\cos\left ( \frac{A+B}{2} \right )\cos\left ( \frac{A-B}{2} \right )$
By using these identities

cos2x - cos6x = -2sin(4x)sin(-2x) = 2sin4xsin2x ( $\because$ sin(-x) = -sin x
cos(-x) = cosx)
cos2x + cos 6x = 2cos4xcos(-2x) = 2cos4xcos2x

So our equation becomes

R.H.S.

Question 14: Prove the following $\small \sin2x +2\sin4x + \sin6x = 4\cos^{2}x\sin4x$

Answer:

We know that
$\sin A+ \sin B = 2\sin \left ( \frac{A+B}{2} \right )\cos\left ( \frac{A-B}{2} \right )$
We are using this identity
sin2x + 2sin4x + sin6x = (sin2x + sin6x) + 2sin4x

sin2x + sin6x = 2sin4xcos(-2x) = 2sin4xcos(2x) ( $\because$ cos(-x) = cos x)

So, our equation becomes
sin2x + 2sin4x + sin6x = 2sin4xcos(2x) + 2sin4x
Now, take the 2sin4x common
sin2x + 2sin4x + sin6x = 2sin4x(cos2x +1) ( $\because \cos2x = 2\cos^{2}x - 1$ )
=2sin4x( $2\cos^{2}x - 1$ +1 )
=2sin4x( $2\cos^{2}x$ )
= $4\sin4x\cos^{2}x$
R.H.S.

Question 15: Prove the following $\small \cot4x(\sin5x + \sin3x) = \cot x(\sin5x - \sin3x)$

Answer:

We know that
$\sin x + \sin y = 2\sin\left ( \frac{x+y}{2} \right )\cos\left (\frac{x-y}{2} \right )$
By using this , we get

sin5x + sin3x = 2sin4xcosx

$\frac{\cos4x}{\sin4x}\left ( 2\sin4x\cos x \right ) = 2\cos4x\cos x\\ \\$

now multiply and divide by sin x

$\\\ \\ \frac{2\cos4x\cos x \sin x}{\sin x } \ \ \ \ \ \ \ \ \ \ \\$ $\\ =\cot x (2\cos4x\sin x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left ( \because \frac{\cos x}{\ sin x} = \cot x \right )\\ \\$

Now we know that

$\\ 2\cos x\sin y = \sin(x+y) - \sin(x-y)\\ \\$

By using this our equation becomes

$\\ \\=\cot x (\sin5x - sin3x)\\$
R.H.S.

Question 16: Prove the following $\small \frac{\cos 9x - \cos 5x}{\sin17x - \sin3x} = -\frac{\sin2x}{\cos10x}$

Answer:

As we know that

$\\ \cos x - \cos y = -2\sin\frac{x+y}{2}\sin\frac{x-y}{2 }\\ \\ \cos 9x - \cos 5x = -2\sin 7x \sin2x \\$ $\\ \sin x - \sin y = 2\cos\frac{x+y}{2}\sin\frac{x-y}{2 }\\ \\ \sin 17x - \sin 3x = 2\cos10x \sin7x\\$ $\\ \frac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x} =\frac{-2\sin 7x \sin2x}{2\cos10x \sin7x} = -\frac{\sin 2x}{\cos10x}$
R.H.S.

Question 17: Prove the following $\small \frac{\sin5x + \sin3x}{\cos5x + \cos3x} = \tan4x$

Answer:

We know that

$\\ \sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}\\and\\$ $\\ \cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \\$

We use these identities

$\\ \sin5x + \sin3x = 2\sin4x\cos x\\ \cos5 x + \cos 3x = 2\cos4x\cos x \\$ $\\ \frac{\sin5x + \sin3x}{\cos5 x + \cos 3x} = \frac{ 2\sin4x\cos x}{2\cos4x\cos x} = \frac{\sin4x}{\cos 4x} = \tan 4x$
R.H.S.

Question 18: Prove the following $\small \frac{\sin x - \sin y}{\cos x+\cos y} = \tan \frac{(x-y)}{2}$

Answer:

We know that
$\sin x - \sin y = 2\cos\frac{x+y}{2 }\sin\frac{x-y}{2}\\and \\$ $\\ \cos x +\cos y = 2\cos\frac{x+y}{2 }\cos\frac{x-y}{2}\\$

We use these identities

$\\ We \ use \ these \ identities\\$ $\\ \frac{\sin x - \sin y}{\cos x +\cos y} =\frac{2\cos\frac{x+y}{2 }\sin\frac{x-y}{2}}{ 2\cos\frac{x+y}{2 }\cos\frac{x-y}{2}} = \frac{\sin\frac{x-y}{2}}{\cos\frac{x-y}{2}} = \tan \frac{x-y}{2}$

R.H.S.

Question 19: Prove the following $\small \frac{\sin x + \sin 3x}{\cos x + \cos3x} = \tan2x$

Answer:

We know that

$\\ \sin x + \sin y = 2\sin\frac{x+y}{2}\cos\frac{x-y}{2}\\and\\ \\ \cos x + \cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2}\\$ $\\ We \ use \ these \ equations \\ \\ \sin x + \sin3x = 2\sin2x\cos(-x) = 2\sin2x\cos x \ \ \ \ \ (\because \cos(-x) = \cos x)\\$ $\\ \cos x + \cos3x = 2\cos2x\cos(-x) =2\cos2x\cos x \ \ \ \ \ (\because \cos(-x) = \cos x)\\$ $\\ \frac{\sin x + \sin3x}{\cos x + \cos3x} = \frac {2\sin2x\cos x}{2\cos2x\cos x}= \frac{\sin2x}{\cos2x} = \tan2x$ R.H.S.

Question 20: Prove the following $\small \frac{\sin x - \sin 3x}{\sin^{2}x-\cos^{2}x} = 2\sin x$

Answer:
We know that

$\sin3x = 3\sin x - 4\sin^{3}x \ \ \ , \ \ \cos^{2}-\sin^{2}x = \cos2x \\and \\ \cos2x = 1 - 2\sin^{2}x \\$

We use these identities

$\sin x-\sin 3 x=\sin x-\left(3 \sin x-4 \sin ^3 x\right)=4 \sin ^3 x-2 \sin x$

$=2 \sin x\left(2 \sin ^2 x-1\right) \cos ^2 x-\sin ^2=\cos 2 x \cos 2 x=1-2 \sin ^2 x$

$\sin x-\sin 3 x=\sin x-\left(3 \sin x-4 \sin ^3 x\right)=4 \sin ^2 x-2 \sin x$

$=2 \sin x\left(2 \sin ^2 x-1\right) \sin ^2-\cos ^2 x=-\cos 2 x$

$\left.\left(\cos 2 x=1-2 \sin ^2 x\right) \sin ^2-\cos ^2 x=-\left(1-2 \sin ^2 x\right)=2 \sin ^4 2\right) x-1 \frac{\sin x-\sin ^2 x}{\sin ^2-\cos ^2 x}=\frac{2 \sin x\left(2 \sin ^3 x-1\right)}{\left.2 \sin ^4 2\right) \varepsilon-1}=2 \sin x$

$\sin x-\sin 3 x=\sin x-\left(3 \sin x-4 \sin ^3 x\right)-4 \sin ^3 x-2 \sin x$.

$=2 \sin \pm\left(2 \sin ^2 x-1\right) \sin ^2-\cos ^2 x--\cos 2 \pi$

$\left.\left(\because \cos 2 x-1-2 \sin ^2 x\right) \sin ^2-\cos ^2 x-\left(1-2 \sin ^2 x\right)=2 \sin ^{\prime} 2\right) x-1 \frac{\sin x-\sin 3 x}{\sin ^2-\cos ^2 x}-\frac{2 \sin x\left(2 \operatorname{nin}^2 \sigma-1\right)}{2 \operatorname{cin}^2 x-1}=2 \sin x$
R.H.S.

Question 21: Prove the following $\small \frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x$

Answer:

We know that

$\cos x + \cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2}\\$ and $\\ \sin x + \sin y = 2\sin\frac{x+y}{2}\cos\frac{x-y}{2}$
We use these identities

$\frac{(\cos4x + \cos2x) + \cos3x}{(\sin4x+\sin2x)+\sin3x} = \frac{2\cos3x\cos x + \cos3x}{2\sin3x\cos x+\sin3x} = \frac{2\cos3x(1+\cos x)}{2\sin3x(1+\cos x)}\\$ $\\ \ \ \ \ \ \ \ = cot 3x$

=RHS

Question 22: prove the following $\small \cot x \cot2x - \cot2x\cot3x - \cot3x\cot x =1$

Answer:

cot x cot2x - cot3x(cot2x - cotx)
Now we can write cot3x = cot(2x + x)

and we know that

$cot(a+b) = \frac{\cot a \cot b - 1}{\cot a + \cot b}$
So,
$cotx\ cot2x-\frac{\cot 2x \cot x - 1}{\cot 2x + \cot x}(cot2x+cotx)$
= cotx cot2x - (cot2xcotx -1)
= cotx cot2x - cot2xcotx +1
= 1 = R.H.S.

Question 23: Prove that $\small \tan4x = \frac{4\tan x(1-\tan^{2}x)}{1-6 \tan^{2}x+\tan^{4}x}$

Answer:

We know that

$tan2A=\frac{2\tan A}{1 - \tan^{2}A}$

and we can write tan 4x = tan 2(2x)
So, $tan4x=\frac{2\tan 2x}{1 - \tan^{2}2x}$ = $\frac{2( \frac{2\tan x}{1 - \tan^{2}x})}{1 - (\frac{2\tan x}{1 - \tan^{2}x})^{2}}$


= $\frac{2 (2\tan x)(1 - \tan^{2}x)}{(1-\tan x)^{2} - (4\tan^{2} x)}$

= $\frac{(4\tan x)(1 - \tan^{2}x)}{(1)^{2}+(\tan^{2} x)^{2} - 2 \tan^{2} x - (4\tan^{2} x)}$

= $\frac{(4\tan x)(1 - \tan^{2}x)}{1^{2}+\tan^{4} x - 6 \tan^{2} x }$ = R.H.S.

Question 24: Prove the following $\small \cos4x = 1 - 8\sin^{2}x\cos^{2}x$

Answer:

We know that
$cos2x=1-2\sin^{2}x$
We use this in our problem
cos 4x = cos 2(2x)
= $1-2\sin^{2}2x$
= $1-2(2\sin x \cos x)^{2}$ $(\because \sin2x = 2\sin x \cos x)$
= $1-8\sin^{2}x\cos^{2}x$ = R.H.S.

Question 25: Prove the following $\small \cos6x = 32\cos^{6}x -48\cos^{4}x + 18\cos^{2}x-1$

Answer:

We know that
cos 3x = 4 $\cos^{3}x$ - 3cos x
we use this in our problem
we can write cos 6x as cos 3(2x)
cos 3(2x) = 4 $\cos^{3}2x$ - 3 cos 2x
= $4(2\cos^{2}x - 1)^{3}$ - $3(2\cos^{2}x - 1)$ $(\because \cos 2x = 2\cos^{2}x - 1)$
= $4[(2cos^{2}x)^{3} -(1)^{3} -3(2cos^{2}x)^{2}(1) + 3(2cos^{2}x)(1)^{2}]$ $-6\cos^{2}x + 3$ $(\because (a-b)^{3} = a^{3} - b^{3} - 3a^{2}b+ 3ab^{2})$
= 32 $cos^{6}x$ - 4 - 48 $cos^{4}x$ + 24 $cos^{2}x$ - $6\cos^{2}x + 3$
= 32 $cos^{6}x$ - 48 $cos^{4}x$ + 18 $cos^{2}x$ - 1 = R.H.S.