Question

If $x\in (-\frac{\pi }{2},\frac{\pi }{2})$, 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 $x$

${\tan }^{-1}\left(\frac{\tan x}{4}\right)+{\tan }^{-1}\left(\frac{3\sin 2x}{5+3\cos 2x}\right)$
${\tan }^{-1}\left(\frac{\tan x}{4}\right)+{\tan }^{-1}\left[\frac{3.\frac{2\tan x}{1+{\tan }^{2}x}}{5+3.\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}}\right]$
${\tan }^{-1}\left(\frac{\tan x}{4}\right)+{\tan }^{-1}\left[\frac{3\tan x}{4+{\tan }^{2}x}\right]$
${\tan }^{-1}\left[\frac{\frac{\tan x}{4}+\frac{3\tan x}{4+{\tan }^{2}x}}{1-\frac{\tan x}{4}.\frac{3\tan x}{4+{\tan }^{2}x}}\right]$
${\tan }^{-1}\left[\frac{16\tan x+{\tan }^{3}x}{16+{\tan }^{2}x}\right]$
${\tan }^{-1}\left[\frac{\tan x(16+{\tan }^{2}x)}{16+{\tan }^{2}x}\right]$
$={\tan }^{-1}\left(\tan x\right)$
$=x$