Question

Prove that $tan3x=\frac{3tanx-ta{n}^{3}x}{1-3ta{n}^{2}x}$.

Answer 1

Consider the given function:

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

$L.H.S\tan 3x$

$=\tan (2x+x)$

$=\frac{\tan \left(2x\right)+\tan \left(x\right)}{1-\tan \left(2x\right)\tan \left(x\right)}$

$=\frac{\frac{2\tan x}{1-{\tan }^{2}x}+\tan x}{1-\frac{2\tan x}{1-{\tan }^{2}x}.\tan x}$

$=\frac{\frac{2\tan x}{1-{\tan }^{2}x}+\tan x}{1-\frac{2\tan x}{1-{\tan }^{2}x}.\tan x}\ast \frac{1-{\tan }^2}{1-{\tan }^2}$

$=\frac{2\tan x+\tan x-{\tan }^{3}x}{1-{\tan }^{2}x-2{\tan }^{2}x}$

$=\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^2}=R.H.S$

Hence this is the answer.