Question
Prove that: $\frac{\tan^22\text{x}-\tan^2\text{x}}{1-\tan^22\text{x}\tan^2\text{x}}=\tan3\text{x}\tan\text{x}$

Answer

We have,
$\text{R.H.S}=\tan3\text{x} \tan\text{x}$
$=\tan(2\text{x}+{\text{x})}\times\tan(2\text{x}-\text{x)}$
$=\Big[\frac{\tan2\text{x}+\tan\text{x}}{1-\tan2\text{x}\tan\text{x}}\Big]\times\Big[\frac{\tan2\text{x}-\tan\text{x}}{1+\tan2\text{x}\tan\text{x}}\Big]$
$=\frac{(\tan2\text{x}+\tan\text{x)}(\tan2\text{x}-\tan\text{x)}}{(1-\tan2\text{x}\tan \text{x)}(1+\tan2\text{x}\tan\text{x)}}$
$=\frac{\tan^22\text{x}-\tan^2\text{x}}{1-\tan^22\text{x}\tan\text{x}}$
$=\text{L.H.S}$
$\therefore\text{L.H.S}=\text{R.H.S}$
hence proved.

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