If x = log left[ cot left( frac{pi}{4} + thet | vedclass

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(B) Given that, $x = \log \left[ \cot \left( \frac{\pi}{4} + 	heta ight) ight]$.This implies $e^x = \cot \left( \frac{\pi}{4} + 	heta ight)$ a

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$-	an 2	heta$

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(B) Given that, $x = \log \left[ \cot \left( \frac{\pi}{4} + 	heta ight) ight]$.This implies $e^x = \cot \left( \frac{\pi}{4} + 	heta ight)$ and $e^{-x} = 	an \left( \frac{\pi}{4} + 	heta ight)$.We know that $\sinh x = \frac{e^x - e^{-x}}{2}$.Substituting the values, we get:$\sinh x = \frac{\cot \left( \frac{\pi}{4} + 	heta ight) - 	an \left( \frac{\pi}{4} + 	heta ight)}{2}$Using the identity $\cot A - 	an A = \frac{\cos A}{\sin A} - \frac{\sin A}{\cos A} = \frac{\cos^2 A - \sin^2 A}{\sin A \cos A} = \frac{\cos 2A}{\frac{1}{2} \sin 2A} = 2 \cot 2A$, we have:$\sinh x = \frac{1}{2} \left[ 2 \cot \left( 2 \left( \frac{\pi}{4} + 	heta ight) ight) ight]$$\sinh x = \cot \left( \frac{\pi}{2} + 2	heta ight)$Since $\cot \left( \frac{\pi}{2} + \alpha ight) = -	an \alpha$, we get:$\sinh x = -	an 2	heta$.

If $x = \log \left[ \cot \left( \frac{\pi}{4} + \theta \right) \right]$, then the value of $\sinh x$ is

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