Let $f:R-(-\pi ,\pi )$ be a differentiable function such that $f\left(x\right)+f\left(y\right)=f\left(\frac{xy}{1-xy}\right)$. If $f\left(1\right)=\frac{\pi }{2}$ and $\underset{x\to 0}{\lim }\frac{f\left(x\right)}{x}=2$. Then, $f\left(x\right)$ is equal to.

Given:

$f\left(x\right)+f\left(y\right)=f\left(\frac{xy}{1-xy}\right)$

$f\left(1\right)=\frac{\pi }{2}$$f\left(x\right)+f\left(y\right)=f\left(\frac{x+y}{1-xy}\right)$

$\underset{x\to 0}{\lim }\frac{f\left(x\right)}{x}=2$

$\underset{x\to 0}{\lim }x$

To find the value of $f\left(x\right)$:

Evaluate,

$f\left(x\right)=a{\tan }^{-1}x,a\in$ constant

Calculate, Value of $a$ using

a=2

If $f\left(x\right)$ is a differentiable function and $f\left(x\right)+f\left(y\right)=f\left(\frac{xy}{1-xy}\right)$ then value of $f\left(x\right)=2{\tan }^{-1}x$.