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

Answer 1

Answer: (D)

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