Question

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

Answer: (D)

$tan^{-1}\left(\frac{tan\,x}{4}\right)+tan^{-1}\left(\frac{3\,sin\,2x}{5 + 3\,cos\,2x}\right)$
$= tan^{-1}\left(\frac{tan\,x}{4}\right)+tan^{-1}\left(\frac{\frac{6\,tan\,x}{1+tan^{2}\,x}}{5 + 3\left(\frac{1-tan^{2}\,x}{1+tan^{2}\,x}\right)}\right)$
$= tan^{-1}\left(\frac{tan\,x}{4}\right) + tan^{-1}\left(\frac{3\,tan\,x}{4+tan^{2}\,x}\right)$
$= tan^{-1}\left(\frac{\frac{tan\,x}{4}+\frac{3\,tan\,x}{4+tan^{2}\,x}}{1-\frac{3\,tan\,x}{4\left(4+tan^{2}\,x\right)}}\right)$
$= tan^{-1}\left(\frac{16\,tan\,x+tan^{3}\,x}{16 + tan^{2}\,x}\right)$
$= tan^{-1}\left(tanx\right) = x$