If x in left(-frac{pi}{2}, frac{pi}{2}right), | vedclass

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(C) Let $y = \log \sec x$. Then $\sec x = e^y$,which implies $\cos x = e^{-y}$.We know that $\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$.Also,$\

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$2 \operatorname{coth}^{-1}\left(\operatorname{cosec}^2 \frac{x}{2}-1\right)$

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(C) Let $y = \log \sec x$. Then $\sec x = e^y$,which implies $\cos x = e^{-y}$.We know that $\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$.Also,$\cos x = \frac{e^y + e^{-y} - (e^y - e^{-y})}{e^y + e^{-y} + (e^y - e^{-y})} = \frac{2e^{-y}}{2e^y} = e^{-2y}$.Alternatively,using hyperbolic functions,$\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$.We have $\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)} = \frac{\cot^2(x/2) - 1}{\cot^2(x/2) + 1}$.Using the identity $\coth(y/2) = \frac{e^{y/2} + e^{-y/2}}{e^{y/2} - e^{-y/2}}$,we relate this to the expression.Specifically,$\coth(y/2) = \frac{1 + e^{-y}}{1 - e^{-y}} = \frac{1 + \cos x}{1 - \cos x} = \frac{1 + \frac{1-	an^2(x/2)}{1+	an^2(x/2)}}{1 - \frac{1-	an^2(x/2)}{1+	an^2(x/2)}} = \frac{2}{2	an^2(x/2)} = \cot^2(x/2)$.Since $\cot^2(x/2) = \operatorname{cosec}^2(x/2) - 1$,we have $\coth(y/2) = \operatorname{cosec}^2(x/2) - 1$.Therefore,$\frac{y}{2} = \operatorname{coth}^{-1}(\operatorname{cosec}^2(x/2) - 1)$,which gives $y = 2 \operatorname{coth}^{-1}(\operatorname{cosec}^2(x/2) - 1)$.

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

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