The value of $\tan \left\{\frac{1}{2} \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)+\frac{1}{2} \cos…

The value of $\tan \left\{\frac{1}{2} \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)+\frac{1}{2} \cos ^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right\}$ is
  1. $\frac{x+y}{1-x y}$
  2. $\frac{x-y}{1+x y}$
  3. $\frac{x-y}{1-x y}$
  4. $\frac{x+y}{1+x y}$

Solution

Let $x=\tan \theta$ and $y=\tan \phi$ $\begin{aligned} & \Rightarrow \tan \left\{\frac{1}{2} \sin ^{-1} \sin 2 \theta+\frac{1}{2} \cos ^{-1} \cos 2 \phi\right\} \\ & =\tan (\theta+\phi) \\ & =\frac{\tan \theta+\tan \phi}{1-\tan \theta \cdot \tan \phi}=\frac{x+y}{1-x y} \end{aligned}$

Asked in: MHT CET 2022 (11 Aug Shift 1)

Practice more Inverse Trigonometric Functions questions on Aicharya