Question

If $x \in \left( \frac{ -\pi}{2} , \frac{\pi}{2} \right)$ , then $log \,sec\, x =$

• A. $2 cosec h^{-1} \left(\cot^{2} \frac{x}{2} - 1\right) $
• B. $2 cosec h^{-1} \left(\cot^{2} \frac{x}{2} + 1\right) $
• C. $2 \cot h^{-1} \left(cosec^{2} \frac{x}{2} - 1\right) $
• D. $2 \cot h^{-1} \left(cosec^{2} \frac{x}{2} + 1\right) $

Answer 1

Answer: (C)

For $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,
$\log \sec x=y$ (let)
$\Rightarrow \sec\, x=e^{y} \Rightarrow \cos\, x=e^{-y}$
$\because \frac{\sin \,h y}{\cos \,h y}=\frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}$
$\Rightarrow \frac{\cos\, h y-\sin\,h y}{\cos\,h y+\sin\,h y}=\frac{e^{-y}}{e^{y}}=\cos ^{2} x$
$\Rightarrow \left(\frac{\cos\,h \frac{y}{2}-\sin\,h \frac{y}{2}}{\cos\, h \frac{y}{2}+\sin\, h \frac{y}{2}}\right)^{2}=\cos ^{2} x$
$\Rightarrow \frac{\cos\, h \frac{y}{2}-\sin\,h \frac{y}{2}}{\cos\, h \frac{y}{2}+\sin\,h \frac{y}{2}}=\cos\, x$
$\Rightarrow \frac{2 \,\cos \,h \frac{y}{2}}{2 \,\sin\,h \frac{y}{2}}=\frac{1+\cos \,x}{1-\cos \,x}=\cot ^{2} \frac{x}{2}$
$\Rightarrow \cot\, h \frac{y}{1} =\cot ^{2} \frac{x}{2}=cosec^{2} \frac{x}{2}-1 $
$\Rightarrow \frac{y}{2} =\cot \,h ^{-1}\left(cosec^{2} \frac{x}{2}-1\right)$
$ \Rightarrow y=2 \cot\, h ^{-1}\left(cosec^{2} \frac{x}{2}-1\right)$