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 $x=$

• A. $\frac{a-b}{1-ab}$
• B. $\frac{a-b}{1+ab}$
• C. $\frac{ab-1}{a+b}$
• D. none of these

Answer 1

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

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

or, $2{\tan }^{-1}a-2{\tan }^{-1}b=2{\tan }^{-1}x$

or, ${\tan }^{-1}a-{\tan }^{-1}b={\tan }^{-1}x$

or, ${\tan }^{-1}\left(\frac{a-b}{1+ab}\right)={\tan }^{-1}x$

$\Rightarrow x=\frac{a-b}{1+ab}$