Q. If $\sin^{- 1}{\frac{2a}{1 + a^{2}} - \cos^{- 1}{\frac{1 - b^{2}}{1 + b^{2}} = \tan^{- 1}{\frac{2x}{1 - x^{2}},}}}$ then value of $x$ is
A) $a$
B) $b$
C) $\frac{a + b}{1 - ab}$
D) $\frac{a - b}{1 + ab}$
Explanation
(d)
$\sin^{- 1}{\frac{2a}{1 + a^{2}} - \cos^{- 1}\ \frac{1 - b^{2}}{1 + b^{2}} = \tan^{- 1}\ \frac{2x}{1 - x^{2}}}$
$\Rightarrow 2\tan^{- 1}{a - 2\tan^{- 1}{b = 2\tan^{- 1}x}}$
$\Rightarrow \tan^{- 1}{\frac{a - b}{1 + ab} = \tan^{- 1}x}$
$\Rightarrow x = \frac{a - b}{1 + ab}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.