operatorname{Sech}^{-1}(sin alpha) = | vedclass

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(C) We know that $\operatorname{Sech}^{-1}(x) = \log \left( \frac{1 + \sqrt{1 - x^2}}{x} \right)$.Substituting $x = \sin \alpha$,we get:$\operatorname

Vedclass · Accepted Answer

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

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(C) We know that $\operatorname{Sech}^{-1}(x) = \log \left( \frac{1 + \sqrt{1 - x^2}}{x} \right)$.Substituting $x = \sin \alpha$,we get:$\operatorname{Sech}^{-1}(\sin \alpha) = \log \left( \frac{1 + \sqrt{1 - \sin^2 \alpha}}{\sin \alpha} \right)$$= \log \left( \frac{1 + \cos \alpha}{\sin \alpha} \right)$Using the half-angle formulas $1 + \cos \alpha = 2 \cos^2 \frac{\alpha}{2}$ and $\sin \alpha = 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$:$= \log \left( \frac{2 \cos^2 \frac{\alpha}{2}}{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}} \right)$$= \log \left( \frac{\cos \frac{\alpha}{2}}{\sin \frac{\alpha}{2}} \right)$$= \log \left( \cot \frac{\alpha}{2} \right)$.

$\operatorname{Sech}^{-1}(\sin \alpha) =$

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