Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

$\begin{array}{rl}& {\sin }^{-1}\left(\frac{\sqrt{3}}{2}x+\frac{1}{2}\sqrt{1-x^2}\right),\frac{-1}{2}<x<\frac{1}{\sqrt{2}}\\ & \Rightarrow \text{ Let }{\sin }^{-1}(x)=\theta \frac{-\pi }{6}<\theta <\frac{\pi }{4}\\ & \Rightarrow x=\sin \theta ,\text{ then }\\ & \Rightarrow {\sin }^{-1}\left(\frac{\sqrt{3}}{2}\sin \theta +\frac{1}{2}\cos \theta \right)\\ & \Rightarrow {\sin }^{-1}\left(\sin \left(\theta +\frac{\pi }{6}\right)\right)=\theta +\frac{\pi }{6}\\ & \Rightarrow {\sin }^{-1}(x)+\frac{\pi }{6}\end{array}$