Q. If $y = \sec^{- 1}{\frac{x + 1}{x - 1} + \sin^{- 1}{\frac{x - 1}{x + 1},\ then\frac{dy}{dx\ }}}$ is
A) 1
B) 0
C) $\frac{x - 1}{x + 1}$
D) $\frac{x + 1}{x - 1}$
Explanation
(b)
$\because\ \ \ \ \ y = \sec^{- 1}{\left( \frac{x + 1}{x - 1} \right) + \sin^{- 1}\left( \frac{x - 1}{x + 1} \right)}$
$= \cos^{- 1}{\left( \frac{x - 1}{x + 1} \right) + \sin^{- 1}\left( \frac{x - 1}{x + 1} \right)\ }$
$\Rightarrow$ $y$ = $\frac{\pi}{2}\ \ \ \ \ \Rightarrow \ \ \frac{dy}{dx}$ =0
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.