Question

If ${\sin }^{-1}\left(\frac{2a}{1+a^2}\right)+{\sin }^{-1}\left(\frac{2b}{1+b^2}\right)=2{\tan }^{-1}x$ , then x is equal to

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

Answer 1

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

$si{n}^{-1}\left(\frac{2a}{1+a^2}\right)+si{n}^{-1}\left(\frac{2b}{1+b^2}\right)=2ta{n}^{-1}x$
$2ta{n}^{-1}a+2ta{n}^{-1}b$
$2\left(ta{n}^{-1}\right(\frac{a+b}{1-ab}\left)\right)=2ta{n}^{-1}x$
$x=\frac{a+b}{1-ab}$