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(A) We know that the formula for the inverse hyperbolic sine function is $\sinh ^{-1}(y) = \log \left(y + \sqrt{1 + y^2}\right)$.Substituting $y = \co

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$\log \left(\cot \frac{x}{2}\right)$

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(A) We know that the formula for the inverse hyperbolic sine function is $\sinh ^{-1}(y) = \log \left(y + \sqrt{1 + y^2}\right)$.Substituting $y = \cot x$ into the formula, we get:$\sinh ^{-1}(\cot x) = \log \left(\cot x + \sqrt{1 + \cot ^2 x}\right)$Since $1 + \cot ^2 x = \operatorname{cosec}^2 x$, we have:$\sinh ^{-1}(\cot x) = \log \left(\cot x + \sqrt{\operatorname{cosec}^2 x}\right) = \log (\cot x + \operatorname{cosec} x)$Using trigonometric identities $\cot x = \frac{\cos x}{\sin x}$ and $\operatorname{cosec} x = \frac{1}{\sin x}$, we get:$\log \left(\frac{\cos x + 1}{\sin x}\right)$Using half-angle formulas $1 + \cos x = 2 \cos^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$:$\log \left(\frac{2 \cos^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}\right) = \log \left(\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}\right) = \log \left(\cot \frac{x}{2}\right)$.

For $0 < x \leq \pi$, $\sinh ^{-1}(\cot x)$ is equal to

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