Question

If $x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$, then the value of $ta{n}^{-1}\left(\frac{tanx}{4}\right)+ta{n}^{-1}\left(\frac{3sin2x}{5+3cos2x}\right)$ is

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

Answer 1

Answer: (D)

$ta{n}^{-1}\left(\frac{tanx}{4}\right)+ta{n}^{-1}\left(\frac{3sin2x}{5+3cos2x}\right)$
$=ta{n}^{-1}\left(\frac{tanx}{4}\right)+ta{n}^{-1}\left(\frac{\frac{6tanx}{1+ta{n}^{2}x}}{5+3\left(\frac{1-ta{n}^{2}x}{1+ta{n}^{2}x}\right)}\right)$
$=ta{n}^{-1}\left(\frac{tanx}{4}\right)+ta{n}^{-1}\left(\frac{3tanx}{4+ta{n}^{2}x}\right)$
$=ta{n}^{-1}\left(\frac{\frac{tanx}{4}+\frac{3tanx}{4+ta{n}^{2}x}}{1-\frac{3tanx}{4\left(4+ta{n}^{2}x\right)}}\right)$
$=ta{n}^{-1}\left(\frac{16tanx+ta{n}^{3}x}{16+ta{n}^{2}x}\right)$
$=ta{n}^{-1}\left(tanx\right)=x$