Question

If x= ${\tan }^{-1}\left[\frac{3\sin 2\alpha }{5+3\cos 2\alpha }\right]+{\tan }^{-1}\left[\frac{\tan \alpha }{4}\right]$, where $-\frac{\pi }{2}<\alpha <\frac{\pi }{2}$.
Find the value of x.

• A. $x=-\alpha$
• B. $x=+\alpha$
• C. $x=-2\alpha$
• D. $x=+2\alpha$

Answer 1

Answer: (B)

$x=+\alpha$

${\tan }^{-1}\left[\frac{3\sin 2\alpha }{5+3\cos 2\alpha }\right]+{\tan }^{-1}\left[\frac{\tan \alpha }{4}\right]={\tan }^{-1}\left(\frac{6\tan \alpha }{8+2{\tan }^2\alpha }\right)+{\tan }^{-1}\left(\frac{\tan \alpha }{4}\right)$

$={\tan }^{-1}\left(\frac{\frac{3\tan \alpha }{4+{\tan }^2\alpha }+\frac{\tan \alpha }{4}}{1-\frac{3{\tan }^2\alpha }{16+4{\tan }^2\alpha }}\right)$ $[\because \frac{3{\tan }^2\alpha }{16+4{\tan }^2\alpha }<1]$
$={\tan }^{-1}\left(\frac{12\tan \alpha +4\tan \alpha +{\tan }^3\alpha }{16+{\tan }^2\alpha }\right)$
$={\tan }^{-1}\left(\tan \alpha \right)=\alpha$