Answer :

$\text{L.H.S.}=\sin^{2} \left(\dfrac{\pi}{6}\right)+\cos^{2} \left(\dfrac{\pi}{3}\right)-\tan^{2} \left(\dfrac{\pi}{4}\right)$

$\hspace{1.1cm} =\left(\dfrac{1}{2}\right)^{2}+\left(\dfrac{1}{2}\right)^{2}-\left(1\right)^{2}$

$\hspace{1.1cm} =\left(\dfrac{1}{4}\right)+\left(\dfrac{1}{4}\right)-1=-\left(\dfrac{1}{2}\right)$

$\hspace{1.1cm} = \text{R.H.S.}$

Answer :

$\text{L.H.S.}=2 \sin^{2} \left(\dfrac{\pi}{6}\right)+\text{cosec} ^2 \left(\dfrac{7 \pi}{6}\right) \cos^{2} \left(\dfrac{\pi}{3}\right)$

$\hspace{1cm} \begin{aligned} & =2\left(\dfrac{1}{2}\right)^{2}+\text{cosec}^{2}\left(\pi+\dfrac{\pi}{6}\right)\left(\dfrac{1}{2}\right)^{2} \\ & =2 \times \left(\dfrac{1}{4}\right)+\left(\text{- cosec} \dfrac{\pi}{6}\right)^{2}\left(\dfrac{1}{4}\right) \\ & =\left(\dfrac{1}{2}\right)+\left(-2\right)^{2}\left(\dfrac{1}{4}\right) \\ & =\left(\dfrac{1}{2}\right)+\left(\dfrac{4}{4}\right)=\left(\dfrac{1}{2}\right)+1=\left(\dfrac{3}{2}\right) \\ \\ & =\text{ R.H.S. } \end{aligned} $

Answer :

$\text{L.H.S.}=\cot^{2} \left(\dfrac{\pi}{6}\right)+\text{cosec} \left(\dfrac{5 \pi}{6}\right)+3 \tan^{2} \left(\dfrac{\pi}{6}\right)$

$\hspace{1cm}=\left(\sqrt{3}\right)^{2}+\text{cosec}\left(\pi-\dfrac{\pi}{6}\right)+3\left(\dfrac{1}{\sqrt{3}}\right)^{2}$

$\hspace{1cm}=3+\text{cosec} \left(\dfrac{\pi}{6}\right)+3 \times \left(\dfrac{1}{3}\right)$

$\hspace{1cm}=3+2+1=6$

$\hspace{1cm}= \text{R.H.S}$

Answer :

$\text{L.H.S}=2 \sin^{2} \left(\dfrac{3 \pi}{4}\right)+2 \cos^{2} \left(\dfrac{\pi}{4}\right)+2 \sec^{2} \left(\dfrac{\pi}{3}\right)$

$\hspace{1cm}=2\left \lbrace \sin \left(\pi-\dfrac{\pi}{4}\right)\right\rbrace ^{2}+2\left(\dfrac{1}{\sqrt{2}}\right)^{2}+2\left(2\right)^{2}$

$\hspace{1cm}=2\left \lbrace \sin \left(\dfrac{\pi}{4}\right)\right\rbrace ^{2}+2 \times \left(\dfrac{1}{2}\right)+8$

$\hspace{1cm}=2\left(\dfrac{1}{\sqrt{2}}\right)^{2}+1+8$

$\hspace{1cm}=1+1+8=10$

$\hspace{1cm}=\text{R.H.S}$

Answer :

(i) $\sin 75^{\circ}=\sin \left(45^{\circ}+30^{\circ}\right)$

$\hspace{1.4cm} =\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30^{\circ}\qquad\bigg[\because \ \ \sin \left(x+y\right)=\sin x \cos y+\cos x \sin y\bigg]$

$\hspace{1.4cm} =\left(\dfrac{1}{\sqrt{2}}\right)\left(\dfrac{\sqrt{3}}{2}\right)+\left(\dfrac{1}{\sqrt{2}}\right)\left(\dfrac{1}{2}\right)$

$\hspace{1.4cm} =\dfrac{\sqrt{3}}{2 \sqrt{2}}+\dfrac{1}{2 \sqrt{2}}=\dfrac{\sqrt{3}+1}{2 \sqrt{2}}$

(ii) $\tan 15^{\circ}=\tan \left(45^{\circ} - 30^{\circ}\right)$

$\qquad\qquad \begin{aligned} & =\dfrac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}} \quad\bigg[\tan \left(x-y\right)=\dfrac{\tan x-\tan y}{1+\tan x \tan y}\bigg] \\ \\ & =\dfrac{1-\dfrac{1}{\sqrt{3}}}{1+1\left(\dfrac{1}{\sqrt{3}}\right)}=\dfrac{\dfrac{\sqrt{3}-1}{\sqrt{3}}}{\dfrac{\sqrt{3}+1}{\sqrt{3}}} \\ \\ & =\dfrac{\sqrt{3}-1}{\sqrt{3}+1}=\dfrac{\left(\sqrt{3}-1\right)^{2}}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{3+1-2 \sqrt{3}}{\left(\sqrt{3}\right)^{2}-\left(1\right)^{2}} \\ \\ & =\dfrac{4-2 \sqrt{3}}{3-1}=2-\sqrt{3} \end{aligned} $

Answer :

$ \begin{aligned} & \cos \left(\dfrac{\pi}{4}-x\right) \cos \left(\dfrac{\pi}{4}-y\right)-\sin \left(\dfrac{\pi}{4}-x\right) \sin \left(\dfrac{\pi}{4}-y\right) \\ \\ &= \dfrac{1}{2}\bigg[2 \cos \left(\dfrac{\pi}{4}-x\right) \cos \left(\dfrac{\pi}{4}-y\right)\bigg]+\dfrac{1}{2}\bigg[-2 \sin \left(\dfrac{\pi}{4}-x\right) \sin \left(\dfrac{\pi}{4}-y\right)\bigg] \\ \\ &= \dfrac{1}{2}\bigg[\cos\left \lbrace \left(\dfrac{\pi}{4}-x\right)+\left(\dfrac{\pi}{4}-y\right)\right\rbrace +\cos\left \lbrace \left(\dfrac{\pi}{4}-x\right)-\left(\dfrac{\pi}{4}-y\right)\right\rbrace \bigg] \\ \\ & +\dfrac{1}{2}\bigg[\cos\left \lbrace \left(\dfrac{\pi}{4}-x\right)+\left(\dfrac{\pi}{4}-y\right)\right\rbrace -\cos\left \lbrace \left(\dfrac{\pi}{4}-x\right)-\left(\dfrac{\pi}{4}-y\right)\right\rbrace \bigg] \\ \\
& \bigg[\because \ \ 2 \cos A \cos B=\cos \left(A+B\right)+\cos \left(A-B\right), \quad -2 \sin A \sin B=\cos \left(A+B\right)-\cos \left(A-B\right)\bigg] \\ \\ &= 2 \times \dfrac{1}{2}\bigg[\cos\left \lbrace \left(\dfrac{\pi}{4}-x\right)+\left(\dfrac{\pi}{4}-y\right)\right\rbrace \bigg] \\ \\ &= \cos \bigg[\dfrac{\pi}{2}-\left(x+y\right)\bigg] = \sin \left(x+y\right) \\ \\ &= \text{ R.H.S } \end{aligned} $

Answer :

It is known that

$ \tan \left(A+B\right)=\dfrac{\tan A+\tan B}{1-\tan A \tan B} \text{ and } \tan \left(A-B\right)=\dfrac{\tan A-\tan B}{1+\tan A \tan B} $

$ \dfrac{\tan \left(\dfrac{\pi}{4}+x\right)}{\tan \left(\dfrac{\pi}{4}-x\right)}=\dfrac{\left(\dfrac{\tan \dfrac{\pi}{4}+\tan x}{1-\tan \dfrac{\pi}{4} \tan x}\right)}{\left(\dfrac{\tan \dfrac{\pi}{4}-\tan x}{1+\tan \dfrac{\pi}{4} \tan x}\right)}$

$\hspace{2.3cm}=\dfrac{\left(\dfrac{1+\tan x}{1-\tan x}\right)}{\left(\dfrac{1-\tan x}{1+\tan x}\right)}=\left(\dfrac{1+\tan x}{1-\tan x}\right)^{2}=\text{ R.H.S. } $

Answer :

$ \begin{aligned} \text{ L.H.S. } & =\dfrac{\cos \left(\pi+x\right) \cos \left(-x\right)}{\sin \left(\pi-x\right) \cos \left(\dfrac{\pi}{2}+x\right)} \\ \\ & =\dfrac{[-\cos x][\cos x]}{\left(\sin x\right)\left(-\sin x\right)} \\ \\ & =\dfrac{-\cos^{2} x}{-\sin^{2} x} \\ \\ & =\cot^{2} x \\ \\ & =\text{ R.H.S. } \end{aligned} $

Answer :

$ \begin{aligned} \text{ L.H.S. } &=\cos \left(\dfrac{3 \pi}{2}+x\right) \cos \left(2 \pi+x\right)\bigg[\cot \left(\dfrac{3 \pi}{2}-x\right)+\cot \left(2 \pi+x\right)\bigg] \\ \\ & =\sin x \cos x\bigg[\tan x+\cot x\bigg] \\ \\ & =\sin x \cos x\left(\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}\right) \\ \\ & =\left(\sin x \cos x\right)\bigg[\dfrac{\sin^{2} x+\cos^{2} x}{\sin x \cos x}\bigg] \\ \\ & =1=\text{ R.H.S. } \end{aligned} $

Answer :

$\text{L.H.S.}=\sin \left(n+1\right) x \sin \left(n+2\right) x+\cos \left(n+1\right) x \cos \left(n+2\right) x$

$\hspace{1cm}=\dfrac{1}{2}\bigg[2 \sin \left(n+1\right) x \sin \left(n+2\right) x+2 \cos \left(n+1\right) x \cos \left(n+2\right) x\bigg]$

$\hspace{1cm}=\dfrac{1}{2}\bigg [ \cos\left \lbrace \left(n+1\right) x-\left(n+2\right) x\right\rbrace -\cos\left \lbrace \left(n+1\right) x+\left(n+2\right) x\right\rbrace $

$\hspace{1.9cm} +\cos\left \lbrace \left(n+1\right) x+\left(n+2\right) x\right\rbrace +\cos\left \lbrace \left(n+1\right) x-\left(n+2\right) x\right \rbrace \bigg] $

$\qquad\qquad \begin{cases} \because-2 \sin A \sin B=\cos \left(A+B\right)-\cos \left(A-B\right) \\ 2 \cos A \cos B=\cos \left(A+B\right)+\cos \left(A-B\right) \end{cases} $

$\hspace{1cm}=\dfrac{1}{2} \times 2 \cos\left \lbrace \left(n+1\right) x-\left(n+2\right) x\right\rbrace $

$\hspace{1cm}=\cos \left(-x\right)=\cos x= \text{R.H.S.}$

Answer :

It is known that

$ \cos A-\cos B=-2 \sin \left(\dfrac{A+B}{2}\right) \cdot \sin \left(\dfrac{A-B}{2}\right) $

$\therefore \ \text{ L.H.S. }=\cos \left(\dfrac{3 \pi}{4}+x\right)-\cos \left(\dfrac{3 \pi}{4}-x\right)$

$\hspace{1.8cm}=-2 \sin\left \lbrace \dfrac{\left(\dfrac{3 \pi}{4}+x\right)+\left(\dfrac{3 \pi}{4}-x\right)}{2}\right\rbrace \cdot \sin\left \lbrace \dfrac{\left(\dfrac{3 \pi}{4}+x\right)-\left(\dfrac{3 \pi}{4}-x\right)}{2}\right\rbrace $

$\hspace{1.8cm}=-2 \sin \left(\dfrac{3 \pi}{4}\right) \sin x$

$\hspace{1.8cm}=-2 \sin \left(\pi-\dfrac{\pi}{4}\right) \sin x$

$\hspace{1.8cm}=-2 \sin \dfrac{\pi}{4} \sin x$

$\hspace{1.8cm}=-2 \times \dfrac{1}{\sqrt{2}} \times \sin x$

$\hspace{1.8cm}=-\sqrt{2} \sin x$

$\hspace{1.8cm}= \text{R.H.S.}$

Answer :

It is known that,

$\sin A+\sin B=2 \sin \left(\dfrac{A+B}{2}\right) \cos \left(\dfrac{A-B}{2}\right), \sin A-\sin B=2 \cos \left(\dfrac{A+B}{2}\right) \sin \left(\dfrac{A-B}{2}\right)$

$\therefore\quad \text{L.H.S.}=\sin^{2} 6 x -\sin^{2} 4 x$

$\hspace{1.7cm}=\left(\sin 6 x+\sin 4 x\right)\left(\sin 6 x -\sin 4 x\right)$

$\hspace{1.7cm}=\bigg[2 \sin \left(\dfrac{6 x+4 x}{2}\right) \cos \left(\dfrac{6 x-4 x}{2}\right)\bigg]\bigg[2 \cos \left(\dfrac{6 x+4 x}{2}\right) \cdot \sin \left(\dfrac{6 x-4 x}{2}\right)\bigg]$

$\hspace{1.7cm} \begin{aligned} & =\left(2 \sin 5 x \cos x\right)\left(2 \cos 5 x \sin x\right) \\ \\ & =\left(2 \sin 5 x \cos 5 x\right)\left(2 \sin x \cos x\right) \\ \\ & =\sin 10 x \sin 2 x \\ \\ & =\text{ R.H.S. } \end{aligned} $

Answer :

$\text{L.H.S.} =\sin 2 x+2 \sin 4 x+\sin 6 x$

$\hspace{1.1cm}=\bigg[\sin 2 x+\sin 6 x\bigg]+2 \sin 4 x$

$\hspace{1.1cm}=\bigg[2 \sin \left(\dfrac{2 x+6 x}{2}\right) \cos \left(\dfrac{2 x-6 x}{2}\right)\bigg]+2 \sin 4 x\quad\bigg[\because \sin A+\sin B=2 \sin \left(\dfrac{A+B}{2}\right) \cos \left(\dfrac{A-B}{2}\right)\bigg]$

$\hspace{1.1cm}=2 \sin 4 x \cos \left( 2 x\right)+2 \sin 4 x$

$\hspace{1.1cm} \begin{aligned} & =2 \sin 4 x \cos 2 x+2 \sin 4 x \\ \\ & =2 \sin 4 x\left(\cos 2 x+1\right) \\ \\ & =2 \sin 4 x\left(2 \cos^{2} x - 1+1\right) \\ \\ & =2 \sin 4 x\left(2 \cos^{2} x\right) \\ \\ & =4 \cos^{2} x \sin 4 x \\ \\ & =\text{ R.H.S. } \end{aligned} $

Answer :

$\text{L.H.S}=\cot 4 x\left(\sin 5 x+\sin 3 x\right)$

$\hspace{1cm}=\dfrac{\cos 4 x}{\sin 4 x}\bigg[2 \sin \left(\dfrac{5 x+3 x}{2}\right) \cos \left(\dfrac{5 x-3 x}{2}\right)\bigg]\qquad\bigg[\because \sin A+\sin B=2 \sin \left(\dfrac{A+B}{2}\right) \cos \left(\dfrac{A-B}{2}\right)\bigg]$

$\hspace{1cm}=\left(\dfrac{\cos 4 x}{\sin 4 x}\right)\bigg[2 \sin 4 x \cos x\bigg]$

$\hspace{1.cm}=2 \cos 4 x \cos x$

$\text{R.H.S.}=\cot x\left(\sin 5 x - \sin 3 x\right)$

$\hspace{1cm}=\dfrac{\cos x}{\sin x}\bigg[2 \cos \left(\dfrac{5 x+3 x}{2}\right) \sin \left(\dfrac{5 x-3 x}{2}\right)\bigg]\qquad \bigg[\because \sin A-\sin B=2 \cos \left(\dfrac{A+B}{2}\right) \sin \left(\dfrac{A-B}{2}\right)\bigg]$

$\hspace{1cm}=\dfrac{\cos x}{\sin x}\bigg[2 \cos 4 x \sin x\bigg]$

$\hspace{1cm}=2 \cos 4 x \cdot \cos x$

$\text{L.H.S.} = \text{R.H.S.}$

Answer :

It is known that

$\cos A-\cos B=-2 \sin \left(\dfrac{A+B}{2}\right) \sin \left(\dfrac{A-B}{2}\right), $

$\sin A-\sin B=2 \cos \left(\dfrac{A+B}{2}\right) \sin \left(\dfrac{A-B}{2}\right) $

$\therefore \text{ L.H.S } =\dfrac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x} $

$\hspace{1.5cm}=\dfrac{-2 \sin \left(\dfrac{9 x+5 x}{2}\right) \cdot \sin \left(\dfrac{9 x-5 x}{2}\right)}{2 \cos \left(\dfrac{17 x+3 x}{2}\right) \cdot \sin \left(\dfrac{17 x-3 x}{2}\right)} $

$\hspace{1.5cm}=\dfrac{-2 \sin 7 x \cdot \sin 2 x}{2 \cos 10 x \cdot \sin 7 x} =-\dfrac{\sin 2 x}{\cos 10 x} $

$\hspace{1.5cm}=\text{ R.H.S. }$

Answer :

It is known that

$ \sin A+\sin B=2 \sin \left(\dfrac{A+B}{2}\right) \cos \left(\dfrac{A-B}{2}\right), $

$ \cos A+\cos B=2 \cos \left(\dfrac{A+B}{2}\right) \cos \left(\dfrac{A-B}{2}\right) $

$\therefore \ \text{L.H.S.}=\dfrac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}$

$\hspace{1.5cm}=\dfrac{2 \sin \left(\dfrac{5 x+3 x}{2}\right) \cdot \cos \left(\dfrac{5 x-3 x}{2}\right)}{2 \cos \left(\dfrac{5 x+3 x}{2}\right) \cdot \cos \left(\dfrac{5 x-3 x}{2}\right)}$

$\hspace{1.5cm}=\dfrac{2 \sin 4 x \cdot \cos x}{2 \cos 4 x \cdot \cos x}$

$\hspace{1.5cm}=\dfrac{\sin 4 x}{\cos 4 x}$

$\hspace{1.5cm}=\tan 4 x=\text{R.H.S.}$

Answer :

It is known that

$\sin A-\sin B=2 \cos \left(\dfrac{A+B}{2}\right) \sin \left(\dfrac{A-B}{2}\right),$

$ \cos A+\cos B=2 \cos \left(\dfrac{A+B}{2}\right) \cos \left(\dfrac{A-B}{2}\right)$

$\therefore \ \text{L.H.S.}=\dfrac{\sin x-\sin y}{\cos x+\cos y}$

$\hspace{1.5cm}=\dfrac{2 \cos \left(\dfrac{x+y}{2}\right) \cdot \sin \left(\dfrac{x-y}{2}\right)}{2 \cos \left(\dfrac{x+y}{2}\right) \cdot \cos \left(\dfrac{x-y}{2}\right)}$

$\hspace{1.5cm}=\dfrac{\sin \left(\dfrac{x-y}{2}\right)}{\cos \left(\dfrac{x-y}{2}\right)}$

$\hspace{1.5cm}=\tan \left(\dfrac{x-y}{2}\right)=$ R.H.S.

Answer :

It is known that

$\sin A+\sin B=2 \sin \left(\dfrac{A+B}{2}\right) \cos \left(\dfrac{A-B}{2}\right),$

$ \cos A+\cos B=2 \cos \left(\dfrac{A+B}{2}\right) \cos \left(\dfrac{A-B}{2}\right)$

$\therefore \text{ L.H.S.}=\dfrac{\sin x+\sin 3 x}{\cos x+\cos 3 x}$

$\hspace{1.5cm} \begin{aligned} & =\dfrac{2 \sin \left(\dfrac{x+3 x}{2}\right) \cos \left(\dfrac{x-3 x}{2}\right)}{2 \cos \left(\dfrac{x+3 x}{2}\right) \cos \left(\dfrac{x-3 x}{2}\right)} \\ \\ & =\dfrac{\sin 2 x}{\cos 2 x} \\ \\ & =\tan 2 x =\text{ R.H.S } \end{aligned} $

Answer :

It is known that

$ \sin A-\sin B=2 \cos \left(\dfrac{A+B}{2}\right) \sin \left(\dfrac{A-B}{2}\right),$

$ \cos^{2} A-\sin^{2} A=\cos 2 A $

$\therefore \ \ \text{L.H.S.}=\dfrac{\sin x-\sin 3 x}{\sin^{2} x-\cos^{2} x}$

$\hspace{1.5cm}=\dfrac{2 \cos \left(\dfrac{x+3 x}{2}\right) \sin \left(\dfrac{x-3 x}{2}\right)}{-\cos 2 x}$

$\hspace{1.5cm}=\dfrac{2 \cos 2 x \sin \left(-x\right)}{-\cos 2 x}$

$\hspace{1.5cm}=-2 \times\left(-\sin x\right)$

$\hspace{1.5cm}=2 \sin x= \text{R.H.S.}$

Answer :

$\text{L.H.S.}=\dfrac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}$

$\hspace{1cm} \begin{aligned} & =\dfrac{\left(\cos 4 x+\cos 2 x\right)+\cos 3 x}{\left(\sin 4 x+\sin 2 x\right)+\sin 3 x} \\ \\ & =\dfrac{2 \cos \left(\dfrac{4 x+2 x}{2}\right) \cos \left(\dfrac{4 x-2 x}{2}\right)+\cos 3 x}{2 \sin \left(\dfrac{4 x+2 x}{2}\right) \cos \left(\dfrac{4 x-2 x}{2}\right)+\sin 3 x} \\ \\ & {\bigg[\because \cos A+\cos B=2 \cos \left(\dfrac{A+B}{2}\right) \cos \left(\dfrac{A-B}{2}\right), \sin A+\sin B=2 \sin \left(\dfrac{A+B}{2}\right) \cos \left(\dfrac{A-B}{2}\right)\bigg]} \\ \\ & =\dfrac{2 \cos 3 x \cos x+\cos 3 x}{2 \sin 3 x \cos x+\sin 3 x} \\ \\ & =\dfrac{\cos 3 x\left(2 \cos x+1\right)}{\sin 3 x\left(2 \cos x+1\right)} \\ \\ & =\cot 3 x=\text{ R.H.S. } \end{aligned} $

Answer :

$\text{L.H.S.}=\cot x \cot 2 x$ -$\cot 2 x \cot 3 x$ -$\cot 3 x \cot x$

$\hspace{1cm}=\cot x \cot 2 x$ -$\cot 3 x\left(\cot 2 x+\cot x\right)$

$\hspace{1cm}=\cot x \cot 2 x - \cot \left(2 x+x\right)\left(\cot 2 x+\cot x\right)$

$\hspace{1cm}=\cot x \cot 2 x-\bigg[\dfrac{\cot 2 x \cot x-1}{\cot x+\cot 2 x}\bigg]\left(\cot 2 x+\cot x\right)\qquad\bigg[\because \cot \left(A+B\right)=\dfrac{\cot A \cot B-1}{\cot A+\cot B}\bigg]$

$\hspace{1cm}=\cot x \cot 2 x -\left( \cot 2 x \cot x -1\right)$

$\hspace{1cm}=1=\text{R.H.S.}$

Answer :

$\text{ It is known that } \tan 2 A =\dfrac{2 \tan A}{1-\tan^{2} A} $

$ \begin{aligned} \therefore \text{ L.H.S. } & =\tan 4 x=\tan 2\left(2 x\right) =\dfrac{2 \tan 2 x}{1-\tan^{2}\left(2 x\right)} \\ \\ & =\dfrac{2\left(\dfrac{2 \tan x}{1-\tan^{2} x}\right)}{1-\left(\dfrac{2 \tan x}{1-\tan^{2} x}\right)^{2}} =\dfrac{\left(\dfrac{4 \tan x}{1-\tan^{2} x}\right)}{\bigg[1-\dfrac{4 \tan^{2} x}{\left(1-\tan^{2} x\right)^{2}}\bigg]} \\ \\ & =\dfrac{\left(\dfrac{4 \tan x}{1-\tan^{2} x}\right)}{\bigg[\dfrac{\left(1-\tan^{2} x\right)^{2}-4 \tan^{2} x}{\left(1-\tan^{2} x\right)^{2}}\bigg]} =\dfrac{4 \tan x\left(1-\tan^{2} x\right)}{\left(1-\tan^{2} x\right)^{2}-4 \tan^{2} x} \\ \\ & =\dfrac{4 \tan x\left(1-\tan^{2} x\right)}{1+\tan^{4} x-2 \tan^{2} x-4 \tan^{2} x} =\dfrac{4 \tan x\left(1-\tan^{2} x\right)}{1-6 \tan^{2} x+\tan^{4} x} \\ \\ & =\text{ R.H.S. } \end{aligned} $

Answer :

$\text{L.H.S.}=\cos 6 x =\cos 3\left(2 x\right)$

$\hspace{1cm}=4 \cos^{3} 2 x-3 \cos 2 x\qquad\quad\bigg[\because\quad\cos 3 A=4 \cos^{3} A-3 \cos A\bigg]$

$\hspace{1cm}=4\bigg[\left(2 \cos^{2} x-1\right)^{3}-3\left(2 \cos^{2} x-1\right)\bigg]\qquad \bigg[\because\quad\cos 2 x=2 \cos^{2} x-1\bigg]$

$\hspace{1cm}=4\bigg[\left(2 \cos^{2} x\right)^{3}-\left(1\right)^{3}-3\left(2 \cos^{2} x\right)^{2}+3\left(2 \cos^{2} x\right)\bigg]-6 \cos^{2} x+3$

$\hspace{1cm}=4\bigg[8 \cos^{6} x-1-12 \cos^{4} x+6 \cos^{2} x\bigg]-6 \cos^{2} x+3$

$\hspace{1cm}=32 \cos^{6} x-4-48 \cos^{4} x+24 \cos^{2} x-6 \cos^{2} x+3$

$\hspace{1cm}=32 \cos^{6} x-48 \cos^{4} x+18 \cos^{2} x-1$

$\hspace{1cm}=\text{R.H.S.}$

We know that $~~\sin ^2 2 x=4 \sin ^2 x \cos ^2 x$

$\text{R.H.S.}=1-8 \sin ^2 x \cos ^2 x$

$\hspace{1.2cm}=1-2 \sin ^2 2 x\qquad[\because\quad \cos 2A=1-2\sin^{2}A]$

$\hspace{1.2cm}=\cos 4 x$

$\text{L.H.S.}$

Using,

$ \begin{aligned} & \cos 2 x=2 \cos ^2 x-1 \\ \\ & \cos 3 x=4 \cos ^3 x-3 \cos x \end{aligned} $

$\text{L.H.S.}$

$ \begin{aligned} \cos 6 x & =2 \cos ^2 3 x-1 \\ \\ & =2\left(4 \cos ^3 x-3 \cos x\right)^2-1 \\ \\ & =2\left(16 \cos ^6 x+9 \cos ^2 x-24 \cos ^4 x\right)-1 \\ \\ & =32 \cos ^6 x-48 \cos ^4 x+18 \cos ^2 x-1 \\ \\ & =\text { RHS } \end{aligned} $