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(D) Given $	heta = \sec^{-1}(\cosh u)$,we have $\sec 	heta = \cosh u$. Using the definition of the inverse hyperbolic cosine function,$u = \cosh^{-1

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given $	heta = \sec^{-1}(\cosh u)$,we have $\sec 	heta = \cosh u$. Using the definition of the inverse hyperbolic cosine function,$u = \cosh^{-1}(\sec 	heta) = \log_e(\sec 	heta + \sqrt{\sec^2 	heta - 1})$. Since $\sqrt{\sec^2 	heta - 1} = 	an 	heta$,we get $u = \log_e(\sec 	heta + 	an 	heta)$. Rewriting in terms of sine and cosine: $u = \log_e\left(\frac{1}{\cos 	heta} + \frac{\sin 	heta}{\cos 	heta}ight) = \log_e\left(\frac{1 + \sin 	heta}{\cos 	heta}ight)$. Using trigonometric identities $\sin 	heta = \cos(\frac{\pi}{2} - 	heta)$ and $\cos 	heta = \sin(\frac{\pi}{2} - 	heta)$ is not ideal here; instead,use $1 + \sin 	heta = 1 + \cos(\frac{\pi}{2} - 	heta) = 2\cos^2(\frac{\pi}{4} - \frac{	heta}{2})$ and $\cos 	heta = \sin(\frac{\pi}{2} - 	heta) = 2\sin(\frac{\pi}{4} - \frac{	heta}{2})\cos(\frac{\pi}{4} - \frac{	heta}{2})$. Alternatively,using $1 + \sin 	heta = (\cos \frac{	heta}{2} + \sin \frac{	heta}{2})^2$ and $\cos 	heta = \cos^2 \frac{	heta}{2} - \sin^2 \frac{	heta}{2} = (\cos \frac{	heta}{2} - \sin \frac{	heta}{2})(\cos \frac{	heta}{2} + \sin \frac{	heta}{2})$,we get $\frac{1 + \sin 	heta}{\cos 	heta} = \frac{\cos \frac{	heta}{2} + \sin \frac{	heta}{2}}{\cos \frac{	heta}{2} - \sin \frac{	heta}{2}} = 	an(\frac{\pi}{4} + \frac{	heta}{2})$. Thus,$u = \log_e(	an(\frac{\pi}{4} + \frac{	heta}{2}))$.

If $\theta = \sec^{-1}(\cosh u)$,then $u =$

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