Question

If $\theta ={\tan }^{-1}\left(2{\tan }^2\theta \right)-\frac{1}{2}{\sin }^{-1}\left(\frac{3\sin 2\theta }{5+4\cos 2\theta }\right)$ then find the sum of all possible values of $\tan \theta$

Answer 1

$tan\theta =\frac{2{\tan }^2\theta -\frac{3}{\tan \theta }}{1+2{\tan }^2\theta \left(\frac{3}{\tan \theta }\right)}=\frac{2{\tan }^3\theta -3}{(1+6\tan \theta )\tan \theta }$

$\therefore 4{\tan }^3\theta +{\tan }^2\theta +3=0$

$\Rightarrow \tan \theta =-1$

Or

$4{\tan }^2\theta -3\tan \theta +3=0$

$\Delta =9-48⟨0$

$\tan \theta =-1$ corresponds to $\tan \frac{\beta }{2}=\frac{3}{\tan \theta }=-3$

but $\beta E\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$

$\Rightarrow \tan \frac{\beta }{2}⟩0$

$\therefore \tan \theta$ can't be $-1$

$\therefore \tan \theta$ can be $0,1,-2$