Question

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

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

Answer 1

Answer: (C)

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