If u = log tan left(frac{pi}{4} + frac{theta} | vedclass

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(A) Given $u = \log 	an \left(\frac{\pi}{4} + \frac{	heta}{2}ight)$.By definition of the hyperbolic cosine function,$\cosh u = \frac{e^u + e^{-u}}

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$\sec 	heta$

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(A) Given $u = \log 	an \left(\frac{\pi}{4} + \frac{	heta}{2}ight)$.By definition of the hyperbolic cosine function,$\cosh u = \frac{e^u + e^{-u}}{2}$.From the given equation,$e^u = 	an \left(\frac{\pi}{4} + \frac{	heta}{2}ight)$.Then $e^{-u} = \frac{1}{	an \left(\frac{\pi}{4} + \frac{	heta}{2}ight)} = \cot \left(\frac{\pi}{4} + \frac{	heta}{2}ight) = 	an \left(\frac{\pi}{2} - (\frac{\pi}{4} + \frac{	heta}{2})ight) = 	an \left(\frac{\pi}{4} - \frac{	heta}{2}ight)$.Now,$\cosh u = \frac{1}{2} \left[ 	an \left(\frac{\pi}{4} + \frac{	heta}{2}ight) + 	an \left(\frac{\pi}{4} - \frac{	heta}{2}ight) ight]$.Using the identity $	an(A+B) + 	an(A-B) = \frac{2 \sin(2A)}{\cos(2A) + \cos(2B)}$,we get:$\cosh u = \frac{1}{2} \left[ \frac{2 \sin(\pi/2)}{\cos(\pi/2) + \cos(	heta)} ight] = \frac{1}{0 + \cos 	heta} = \frac{1}{\cos 	heta} = \sec 	heta$.

If $u = \log \tan \left(\frac{\pi}{4} + \frac{\theta}{2}\right)$,then $\cosh u =$

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