# If 2tan^-1x=sin^-1K. What will be the value of k?

##### 1 Answer
May 3, 2018

$k = \frac{2 x}{1 + {x}^{2}}$

#### Explanation:

Let ${\tan}^{- 1} x = a$ then

$\rightarrow \tan a = x$

$\rightarrow \sin 2 a = \frac{2 \tan a}{1 + {\tan}^{2} a} = \frac{2 x}{1 + {x}^{2}}$

$\rightarrow 2 a = {\sin}^{- 1} \left(\frac{2 x}{1 + {x}^{2}}\right)$

$\rightarrow 2 {\tan}^{- 1} x = {\sin}^{- 1} \left(\frac{2 x}{1 + {x}^{2}}\right)$

Given that $2 {\tan}^{- 1} x = {\sin}^{- 1} k$ By comparing, we get,

$\rightarrow k = \frac{2 x}{1 + {x}^{2}}$