Question

lf $x\in (-\frac{\pi }{2},\frac{\pi }{2})$ , then the value of ${\tan }^{-1}\left(\frac{\tan x}{4}\right)+{\tan }^{-1}\left(\frac{3\sin 2x}{5+3\cos 2x}\right)$ is:

• A. $\frac{x}{2}$
• B. $2x$
• C. $3x$
• D. $x$

Answer 1

The correct option is D $x$

Given, ${\tan }^{-1}\left(\frac{\tan x}{4}\right)+{\tan }^{-1}\left(\frac{3\sin 2x}{5+3\cos 2x}\right)$

$={\tan }^{-1}\left(\frac{\frac{\tan x}{4}+\frac{3\sin 2x}{5+3\cos 2x}}{1-\frac{\tan x}{4}\left(\frac{3\sin 2x}{5+3\cos 2x}\right)}\right)$

$={\tan }^{-1}\left(\frac{5\tan x+3\tan x\cos 2x+12\sin 2x}{20+12\cos 2x-3\tan x\sin 2x}\right)$

$={\tan }^{-1}\left(\frac{\tan x(8-6{\sin }^{2}x)+24\sin x\cos x}{32-30{\sin }^{2}x}\right)$

$={\tan }^{-1}\left(\frac{\sin x}{\cos x}\left(\frac{8-6{\sin }^{2}x+24{\cos }^{2}x}{32-30{\sin }^{2}x}\right)\right)$

$={\tan }^{-1}\left(\tan x\right)\left[xϵ\right[-\frac{\pi }{2},\frac{\pi }{2}\left]\right]$

$=x$