If tanh ^2 x = tan ^2 theta,then cosh 2x = | vedclass

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(D) We are given $	anh ^2 x = 	an ^2 	heta$. Using the identity $\cosh 2x = \frac{1 + 	anh ^2 x}{1 - 	anh ^2 x}$,we substitute $	anh ^2 x = 	an

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$\sec 2	heta$

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(D) We are given $	anh ^2 x = 	an ^2 	heta$. Using the identity $\cosh 2x = \frac{1 + 	anh ^2 x}{1 - 	anh ^2 x}$,we substitute $	anh ^2 x = 	an ^2 	heta$: $\cosh 2x = \frac{1 + 	an ^2 	heta}{1 - 	an ^2 	heta}$. We know that $1 + 	an ^2 	heta = \sec ^2 	heta$ and $1 - 	an ^2 	heta = \frac{\cos ^2 	heta - \sin ^2 	heta}{\cos ^2 	heta} = \frac{\cos 2	heta}{\cos ^2 	heta}$. Thus,$\cosh 2x = \frac{\sec ^2 	heta}{\frac{\cos 2	heta}{\cos ^2 	heta}} = \frac{1}{\cos ^2 	heta} 	imes \frac{\cos ^2 	heta}{\cos 2	heta} = \frac{1}{\cos 2	heta} = \sec 2	heta$. Therefore,the correct option is $D$.

If $\tanh ^2 x = \tan ^2 \theta$,then $\cosh 2x =$

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