For theta in left(0, frac{pi}{2}right), opera | vedclass

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(C) We know that the inverse hyperbolic secant function is defined as $\operatorname{sech}^{-1}(x) = \log \left( \frac{1 + \sqrt{1 - x^2}}{x} \right)$

Vedclass · Accepted Answer

$\log \left|	an \left(\frac{\pi}{4}+\frac{	heta}{2}ight)ight|$

Vedclass · Answer

(C) We know that the inverse hyperbolic secant function is defined as $\operatorname{sech}^{-1}(x) = \log \left( \frac{1 + \sqrt{1 - x^2}}{x} ight)$.Substituting $x = \cos 	heta$, we get:$\operatorname{sech}^{-1}(\cos 	heta) = \log \left( \frac{1 + \sqrt{1 - \cos^2 	heta}}{\cos 	heta} ight)$Since $	heta \in \left(0, \frac{\pi}{2}ight)$, $\sin 	heta > 0$, so $\sqrt{1 - \cos^2 	heta} = \sin 	heta$.Thus, the expression becomes $\log \left( \frac{1 + \sin 	heta}{\cos 	heta} ight)$.This can be rewritten as $\log (\sec 	heta + 	an 	heta)$.Using the identity $\sec 	heta + 	an 	heta = 	an \left( \frac{\pi}{4} + \frac{	heta}{2} ight)$, we obtain:$\operatorname{sech}^{-1}(\cos 	heta) = \log \left| 	an \left( \frac{\pi}{4} + \frac{	heta}{2} ight) ight|$.

For $\theta \in \left(0, \frac{\pi}{2}\right)$, $\operatorname{sech}^{-1}(\cos \theta)$ is equal to

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