Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Let $\alpha =\frac{1}{2}{\sin }^{-1}\left(\frac{6\tan \theta }{9+{\tan }^2\theta }\right)$ $\begin{array}{rl}& \theta ={\tan }^{-1}(2\tan \theta )-\alpha \\ & \theta +\alpha ={\tan }^{-1}(2\tan \theta )\\ & \tan (\theta +\alpha )=2\tan \theta \\ & \frac{\tan \theta +\tan \alpha }{1-\tan \alpha \tan \theta }=2\tan \theta ....(1)\\ & \sin 2\alpha =\frac{6\tan \theta }{9+{\tan }^2\theta }=\frac{2\tan \alpha }{1+{\tan }^2\alpha }\\ & 3\tan \theta +3\tan \theta {\tan }^2\alpha =9\tan \alpha +\tan \alpha {\tan }^2\theta \\ & 3(\tan \theta -3\tan \alpha )=\tan \alpha \tan \theta (\tan \theta -3\tan \alpha )\\ & \tan \theta =\frac{3}{\tan \alpha }\text{ or }\tan \theta =3\tan \alpha \end{array}$ $\begin{array}{rl}& \text{ Case-I : }\tan \theta =3\tan \alpha \\ & \frac{\tan \theta +\frac{\tan \theta }{3}}{1-\frac{{\tan }^2\theta }{3}}=2\tan \theta \end{array}$ $\tan \theta =0,\frac{2}{3}=1-\frac{{\tan }^2\theta }{3}\Rightarrow \tan \theta =1,-1$ $\tan \theta=0,-1,1$\begin{array}{rl}& \text{ Case-II : }\tan \theta =\frac{3}{\tan \alpha }\\ & \frac{\tan \theta +\frac{3}{\tan \theta }}{-2}=2\tan \theta \end{array}$