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