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 ha

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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 $\cosh y = \frac{1}{\sin 	heta} = \operatorname{cosec} 	heta$.Using the logarithmic form of the inverse hyperbolic cosine function,$\cosh^{-1}(x) = \log(x + \sqrt{x^2 - 1})$.Substituting $x = \operatorname{cosec} 	heta$,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 simplify:$\operatorname{cosec} 	heta + \cot 	heta = \frac{1 + \cos 	heta}{\sin 	heta} = \frac{2 \cos^2(	heta/2)}{2 \sin(	heta/2) \cos(	heta/2)} = \cot(	heta/2)$.Thus,$y = \log(\cot(	heta/2))$.

$(a)$ Placenta	$(i)$ Androgens
$(b)$ Zona pellucida	$(ii)$ Human Chorionic Gonadotropin $(hCG)$
$(c)$ Bulbourethral glands	$(iii)$ Layer of the ovum
$(d)$ Leydig cells	$(iv)$ Lubrication of the penis

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

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$(b)$ Zona pellucida $(ii)$ Human Chorionic Gonadotropin $(hCG)$
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$(d)$ Leydig cells $(iv)$ Lubrication of the penis

$(a)\quad (b)\quad (c)\quad (d)$

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