Question

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

• A. True
• B. False

Answer 1

The correct option is A True

Applying the formula ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$

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

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

We know that $\sin 2\alpha =\frac{2\tan \alpha }{1+{\tan }^2\alpha }$ and $\cos 2\alpha =\frac{1-{\tan }^2\alpha }{1+{\tan }^2\alpha }$

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

$\Rightarrow {\tan }^{-1}\left(\frac{\frac{6\tan \alpha }{5+5{\tan }^2\alpha +3-{\tan }^2\alpha }+\frac{\tan \alpha }{4}}{1-\left(\frac{6\tan \alpha }{5+5{\tan }^2\alpha +3-3{\tan }^2\alpha }\right)\times \frac{\tan \alpha }{4}}\right)=\alpha$

$\Rightarrow {\tan }^{-1}\left(\frac{\frac{6\tan \alpha }{8+4{\tan }^2\alpha }+\frac{\tan \alpha }{4}}{1-\frac{6{\tan }^2\alpha }{4(8+2{\tan }^2\alpha )}}\right)=\alpha$

$={\tan }^{-1}\left(\frac{24\tan \alpha +8\tan \alpha +4{\tan }^3\alpha }{32+8{\tan }^2\alpha -6{\tan }^2\alpha }\right)=\alpha$

$\Rightarrow \frac{4\tan \alpha ({\tan }^2\alpha +8)}{32+2{\tan }^2\alpha }=\tan \alpha$

$\Rightarrow \frac{4({\tan }^2\alpha +8)}{32+2{\tan }^2\alpha }=1$

$\Rightarrow \frac{4{\tan }^2\alpha +32}{32+2{\tan }^2\alpha }=1$

$\Rightarrow 4{\tan }^2\alpha +32=32+2{\tan }^2\alpha$

$\Rightarrow {\tan }^2\alpha =0$

$\Rightarrow \alpha =0$ where $\alpha \in (\frac{-\pi }{2},\frac{\pi }{2})$

Hence the given statement is true.