Question

If ${\sin }^{-1}\left(\frac{2a}{1+a^2}\right)+{\cos }^{-1}\left(\frac{1-a^2}{1+a^2}\right)={\tan }^{-1}\frac{2x}{1-x^2}$, where $a,x\in ]0,1[$, then the value of $x$ is

• A. 0
• B. $\frac{a}{2}$
• C. $a$
• D. $\frac{2a}{1-a^2}$

Answer 1

Answer: (D)

Given, ${\sin }^{-1}\left(\frac{2a}{1+a^2}\right)+{\cos }^{-1}\left(\frac{1-a^2}{1+a^2}\right)={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$
$\Rightarrow 2{\tan }^{-1}a+2{\tan }^{-1}a={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$
$\left[\because {\sin }^{-1}\left(\frac{2x}{1+x^2}\right)=2{\tan }^{-1}x\text{ and }{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2{\tan }^{-1}x\right]$
$\Rightarrow 4{\tan }^{-1}a={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$
$\Rightarrow 2{\tan }^{-1}\left(\frac{2a}{1-a^2}\right)={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$
$(\because 2{\tan }^{-1}x={\tan }^{-1}\frac{2x}{1-x^2})$
$\Rightarrow 2{\tan }^{-1}\left(\frac{2a}{1-a^2}\right)=2{\tan }^{-1}x$
On comparing, we get
$x=\frac{2a}{1-a^2}$