operatorname{sech}^{-1}(sin theta) is equal t | vedclass

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(D) Let $y = \operatorname{sech}^{-1}(\sin 	heta)$.Then, $\operatorname{sech} y = \sin 	heta$.Since $\operatorname{sech} y = \frac{1}{\cosh y}$, we

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$\log \cot \frac{	heta}{2}$

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(D) Let $y = \operatorname{sech}^{-1}(\sin 	heta)$.Then, $\operatorname{sech} y = \sin 	heta$.Since $\operatorname{sech} y = \frac{1}{\cosh y}$, we have $\cosh y = \frac{1}{\sin 	heta} = \operatorname{cosec} 	heta$.Thus, $y = \cosh^{-1}(\operatorname{cosec} 	heta)$.Using the logarithmic form of the inverse hyperbolic cosine function, $\cosh^{-1}(x) = \log(x + \sqrt{x^2 - 1})$, we get:$y = \log(\operatorname{cosec} 	heta + \sqrt{\operatorname{cosec}^2 	heta - 1})$.Since $\operatorname{cosec}^2 	heta - 1 = \cot^2 	heta$, we have:$y = \log(\operatorname{cosec} 	heta + \cot 	heta)$.Using the trigonometric identities $\operatorname{cosec} 	heta = \frac{1}{\sin 	heta} = \frac{1}{2 \sin(	heta/2) \cos(	heta/2)}$ and $\cot 	heta = \frac{\cos 	heta}{\sin 	heta} = \frac{\cos^2(	heta/2) - \sin^2(	heta/2)}{2 \sin(	heta/2) \cos(	heta/2)}$, we get:$y = \log\left(\frac{1 + \cos^2(	heta/2) - \sin^2(	heta/2)}{2 \sin(	heta/2) \cos(	heta/2)}ight) = \log\left(\frac{2 \cos^2(	heta/2)}{2 \sin(	heta/2) \cos(	heta/2)}ight) = \log(\cot(	heta/2))$.Therefore, $\operatorname{sech}^{-1}(\sin 	heta) = \log \cot \frac{	heta}{2}$.

$\operatorname{sech}^{-1}(\sin \theta)$ is equal to

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