The value of \ (1)/(2) ^ -1 ((2 x)/(1+x^2))+(1)/(2) ^ -1 ((1-y^2)/(1+y^2))\ is

Correct answer: A. (x+y)/(1-x y) Let x= θ and y= φ aligned & \ (1)/(2) ^ -1 2 θ+(1)/(2) ^ -1 2 φ\ & = (θ+φ) & =( θ+ φ)/(1- θ · φ)=(x+y)/(1-x y) aligned

JEE Main

Exam: JEE Main
Subject: Mathematics
Chapter: Inverse Trigonometric Functions

JEE Main · Mathematics · Inverse Trigonometric Functions

The value of \ (1)/(2) ^ -1 ((2 x)/(1+x^2))+(1)/(2) ^ -1...

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\}

Options

A $\frac{x+y}{1-x y}$
B $\frac{x-y}{1+x y}$
C $\frac{x-y}{1-x y}$
D $\frac{x+y}{1+x y}$

Correct answer

A. $\frac{x+y}{1-x y}$

Step-by-step 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}

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