If x = log_e left[ cot left( frac{pi}{4} + th | vedclass

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(D) Given $x = \log_e \left[ \cot \left( \frac{\pi}{4} + 	heta ight) ight]$.By definition,$\sinh x = \frac{e^x - e^{-x}}{2}$.Here,$e^x = \cot \le

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$-	an 2	heta$

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(D) Given $x = \log_e \left[ \cot \left( \frac{\pi}{4} + 	heta ight) ight]$.By definition,$\sinh x = \frac{e^x - e^{-x}}{2}$.Here,$e^x = \cot \left( \frac{\pi}{4} + 	heta ight) = \frac{1 - 	an 	heta}{1 + 	an 	heta}$.Then $e^{-x} = \frac{1}{\cot \left( \frac{\pi}{4} + 	heta ight)} = 	an \left( \frac{\pi}{4} + 	heta ight) = \frac{1 + 	an 	heta}{1 - 	an 	heta}$.Substituting these into the formula for $\sinh x$:$\sinh x = \frac{1}{2} \left( \frac{1 - 	an 	heta}{1 + 	an 	heta} - \frac{1 + 	an 	heta}{1 - 	an 	heta} ight)$$= \frac{1}{2} \left( \frac{(1 - 	an 	heta)^2 - (1 + 	an 	heta)^2}{(1 + 	an 	heta)(1 - 	an 	heta)} ight)$$= \frac{1}{2} \left( \frac{1 - 2	an 	heta + 	an^2 	heta - (1 + 2	an 	heta + 	an^2 	heta)}{1 - 	an^2 	heta} ight)$$= \frac{1}{2} \left( \frac{-4	an 	heta}{1 - 	an^2 	heta} ight) = -2 \left( \frac{	an 	heta}{1 - 	an^2 	heta} ight)$.Since $	an 2	heta = \frac{2	an 	heta}{1 - 	an^2 	heta}$,we have $\sinh x = -	an 2	heta$.

If $x = \log_e \left[ \cot \left( \frac{\pi}{4} + \theta \right) \right]$ where $|\theta| < \frac{\pi}{4}$,then $\sinh x =$

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