Question

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

=$\frac{\lambda \alpha }{4}$ where $-\frac{\pi }{2}<\alpha <\frac{\pi }{2}$ then $\lambda$ is

Answer 1

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

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

$={\tan }^{-1}\left(\tan \alpha \right)=\alpha$