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Question
Prove that:
\[\frac{\tan^2 2x - \tan^2 x}{1 - \tan^2 2x \tan^2 x} = \tan 3x \tan x\]
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Solution
\[\text{ LHS }= \frac{\tan^2 2x - \tan^2 x}{1 - \tan^2 2x \tan^2 x}\]
\[ = \frac{(\tan2x + \tan x)(\tan2x - \tan x)}{1 - \tan^2 2x \tan^2 x} \left\{\text{ Using }A^2 - B^2 = \left( A + B \right)\left( A - B \right) \right\}\]
\[ \tan3x=\tan(2x+x)\text{ and }\tan x=\tan(2x-x) .\]
\[ = \frac{\tan3 x\left( 1 - \tan2x \tan x \right) \times \tan x\left( 1 + \tan2x \tan x \right)}{1 - \tan^2 2x \tan^2 x} \left[ \because \tan 2x + \tan x = \tan3x\left( 1 - \tan2x \tanx \right) \tan 2x - \tanx = \tanx\left( 1 + \tan2x \tanx \right) \right]\]
\[ = \frac{\tan3x \tan x (1 - \tan^2 2x \tan^2 x)}{1 - \tan^2 2x \tan^2 x}\]
\[ = \tan3 x\tan x\]
= RHS
Hence proved .
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