If cosh x = operatorname{cosec} theta,then co | vedclass

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(D) Given $\cosh x = \operatorname{cosec} 	heta$. Using the identity $\cosh x = \frac{1 + 	anh^2(x/2)}{1 - 	anh^2(x/2)}$,we have: $\frac{1 + 	anh^

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$\cot^2 \left( \frac{\pi}{4} - \frac{	heta}{2} ight)$

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(D) Given $\cosh x = \operatorname{cosec} 	heta$. Using the identity $\cosh x = \frac{1 + 	anh^2(x/2)}{1 - 	anh^2(x/2)}$,we have: $\frac{1 + 	anh^2(x/2)}{1 - 	anh^2(x/2)} = \operatorname{cosec} 	heta$. Applying componendo and dividendo: $\frac{2}{2 	anh^2(x/2)} = \frac{\operatorname{cosec} 	heta + 1}{\operatorname{cosec} 	heta - 1}$. $\coth^2 \frac{x}{2} = \frac{1 + \sin 	heta}{1 - \sin 	heta} = \frac{(\cos(	heta/2) + \sin(	heta/2))^2}{(\cos(	heta/2) - \sin(	heta/2))^2}$. Dividing numerator and denominator by $\cos^2(	heta/2)$: $\coth^2 \frac{x}{2} = \left( \frac{1 + 	an(	heta/2)}{1 - 	an(	heta/2)} ight)^2 = 	an^2 \left( \frac{\pi}{4} + \frac{	heta}{2} ight)$. Note: The expression $	an^2(\pi/4 + 	heta/2)$ is equivalent to $\cot^2(\pi/4 - 	heta/2)$.

If $\cosh x = \operatorname{cosec} \theta$,then $\coth^2 \frac{x}{2} = $

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