operatorname{Tanh}^{-1}(sin theta) = | vedclass

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(D) Let $x = \operatorname{Tanh}^{-1}(\sin 	heta)$.Then $	anh x = \sin 	heta$.We know that $	anh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}$.Using the

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$\operatorname{Cosh}^{-1}(\sec 	heta)$

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(D) Let $x = \operatorname{Tanh}^{-1}(\sin 	heta)$.Then $	anh x = \sin 	heta$.We know that $	anh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}$.Using the identity $\cosh^2 x - \sinh^2 x = 1$,we have $\operatorname{sech}^2 x = 1 - 	anh^2 x = 1 - \sin^2 	heta = \cos^2 	heta$.Thus,$\operatorname{sech} x = \cos 	heta$,which implies $\cosh x = \sec 	heta$.Using the identity $\cosh^2 x - \sinh^2 x = 1$,we get $\sinh^2 x = \cosh^2 x - 1 = \sec^2 	heta - 1 = 	an^2 	heta$.Therefore,$\sinh x = 	an 	heta$.This implies $x = \operatorname{Sinh}^{-1}(	an 	heta)$.However,checking the options,we use the identity $\operatorname{Tanh}^{-1}(y) = \operatorname{Sinh}^{-1}\left(\frac{y}{\sqrt{1-y^2}}ight)$.Substituting $y = \sin 	heta$,we get $\operatorname{Sinh}^{-1}\left(\frac{\sin 	heta}{\sqrt{1-\sin^2 	heta}}ight) = \operatorname{Sinh}^{-1}\left(\frac{\sin 	heta}{\cos 	heta}ight) = \operatorname{Sinh}^{-1}(	an 	heta)$.Wait,checking the options again,let us re-evaluate $\operatorname{Cosh}^{-1}(\sec 	heta)$.Since $\cosh x = \sec 	heta$,then $x = \operatorname{Cosh}^{-1}(\sec 	heta)$.Thus,$\operatorname{Tanh}^{-1}(\sin 	heta) = \operatorname{Cosh}^{-1}(\sec 	heta)$.

$\operatorname{Tanh}^{-1}(\sin \theta) =$

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