Question

If ${\sin }^{-1}\left(\frac{2a}{1+a^2}\right)-{\cos }^1\left(\frac{1-b^2}{1+b^2}\right)={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$, then what is the value of $x$?

• A. $\frac{a}{b}$
• B. $ab$
• C. $\frac{b}{a}$
• D. $\frac{a-b}{1+ab}$

Answer 1

The correct option is D $\frac{a-b}{1+ab}$

Given, ${\sin }^{-1}\left(\frac{2a}{1+a^2}\right)-{\cos }^1\left(\frac{1-b^2}{1+b^2}\right)={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$
$\Rightarrow 2{\tan }^{-1}a-2{\tan }^{-1}b=2{\tan }^{-1}x$
$\Rightarrow {\tan }^{-1}a-{\tan }^{-1}b={\tan }^{-1}x$
$\Rightarrow {\tan }^{-1}\left(\frac{a-b}{1+ab}\right)={\tan }^{-1}x$
$\Rightarrow x=\frac{a-b}{1+ab}$