Question

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

Answer 1

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

$\frac{3\sin 2a}{5+3\cos 2a}=\frac{3.2\sin \alpha \cos \alpha }{5+3(2{\cos }^2\alpha -1)}=\frac{6\sin \alpha \cos \alpha }{6{\cos }^2\alpha +2}$

divide by $2{\cos }^2\alpha$

$=>\frac{3\tan \alpha }{3+{\sec }^2\alpha }({\sec }^2\alpha =1+{\tan }^2\alpha )$

$=>\frac{3\tan \alpha }{3+1+{\tan }^2\alpha }=\frac{3\tan \alpha }{4+{\tan }^2\alpha }$

$=>\frac{\frac{3}{4}\tan \alpha }{1+\frac{1}{4}{\tan }^2\alpha }=\frac{\tan \alpha -\frac{1}{4}\tan \alpha }{1+\tan \alpha \frac{1}{4}\tan \alpha }$

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

$=>{\tan }^{-1}\left(\tan \right(\alpha -{\tan }^{-1}\frac{1}{4}\tan \alpha \left)\right)+{\tan }^{-1}\left(\frac{1}{4}\tan \alpha \right)$

$=>\alpha -{\tan }^{-1}\left(\frac{1}{4}\tan \alpha \right)+{\tan }^{-1}\frac{1}{4}\tan \alpha$

$=>\alpha$