Question

Prove that: $\frac{{\tan }^{2}2x-{\tan }^{2}x}{1-{\tan }^{2}2x{\tan }^{2}x}=\tan 3x\tan x$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\frac{{\tan }^{2}2x-{\tan }^{2}x}{1-{\tan }^{2}2x{\tan }^{2}x} \\ =\frac{(\tan 2x+\tan x)(\tan 2x-\tan x)}{1-{\tan }^{2}2x{\tan }^{2}x}\left\{\mathrm{Using}{A}^2-B^2=\left(A+B\right)\left(A-B\right)\right\} \\ \tan 3x=\tan (2x+x)\mathrm{and}\tan x=\tan (2x-x). \\ =\frac{\tan 3x\left(1-\tan 2x\tan x\right)\times \tan x\left(1+\tan 2x\tan x\right)}{1-{\tan }^{2}2x{\tan }^{2}x}\left[\because \tan 2x+\tan x=\tan 3x\left(1-\tan 2x\tan x\right)\&\tan 2x-\tan x=\tan x\left(1+\tan 2x\tan x\right)\right] \\ =\frac{\tan 3x\tan x(1-{\tan }^{2}2x{\tan }^{2}x)}{1-{\tan }^{2}2x{\tan }^{2}x} \\ =\tan 3x\tan x \end{gathered}$$

$$\begin{gathered} =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$