Q. If $- \frac{\pi}{2} < x < \frac{\pi}{2}$, then the value of $\log{\sec x}$ is
A) $2\coth^{- 1}\left( cosec^{2}\frac{x}{2} - 1 \right)$
B) $2\coth^{- 1}\left( cosec^{2}\frac{x}{2} + 1 \right)$
C) $2{cosech}^{- 1}\left( cot^{2}\frac{x}{2} - 1 \right)$
D) $2{cosech}^{- 1}\left( \cot^{2}\frac{x}{2} + 1 \right)$
Explanation
(a)
Let $\log{\sec{x = y}}$
$\Rightarrow \frac{1}{\cos x} = e^{y} = e^{y/2 + y/2} = e^{y/2.e^{y}/2}$
$\therefore\ \frac{1}{\cos x} = \frac{e^{y/2}}{e^{- y/2}}$
By componendo and Dividendo rule
$\frac{1 + \cos x}{1 - \cos x} = \frac{e^{y/2} + e^{- y/2}}{e^{y/2} - e^{- y/2}}$
$\Rightarrow \cot^{2}{\left( \frac{x}{2} \right) = \coth\left( \frac{y}{2} \right)}$
$\Rightarrow y = 2\coth^{- 1}\left( cosec^{2}\frac{x}{2} - 1 \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.