Prove that

Solution:

To Prove: cot x – 2cot 2x = tan x

Taking LHS,

$=\cot x-2 \cot 2 x \ldots$ (i)

We know that

$\cot x=\frac{\cos x}{\sin x}$

Replacing x by 2x, we get

$\cot 2 x=\frac{\cos 2 x}{\sin 2 x}$

So, eq. (i) becomes

$=\frac{\cos x}{\sin x}-2\left(\frac{\cos 2 x}{\sin 2 x}\right)$

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

$=\frac{\cos x}{\sin x}-\left(\frac{\cos 2 x}{\sin x \cos x}\right)$

$=\frac{\cos x(\cos x)-\cos 2 x}{\sin x \cos x}$

$=\frac{\cos ^{2} x-\cos 2 x}{\sin x \cos x}$

$=\frac{\cos ^{2} x-\left[2 \cos ^{2} x-1\right]}{\sin x \cos x}\left[\because 1+\cos 2 x=2 \cos ^{2} x\right]$

$=\frac{\cos ^{2} x-2 \cos ^{2} x+1}{\sin x \cos x}$

$=\frac{-\cos ^{2} x+1}{\sin x \cos x}$

$=\frac{1-\cos ^{2} x}{\sin x \cos x}$

$=\frac{\cos ^{2} x+\sin ^{2} x-\cos ^{2} x}{\sin x \cos x}\left[\because \cos ^{2} \theta+\sin ^{2} \theta=1\right]$

$=\frac{\sin ^{2} x}{\sin x \cos x}$

$=\frac{\sin x}{\cos x}$

$=\tan x\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$

= RHS

∴ LHS = RHS

Hence Proved