If sec(x) = cosh(theta),then find tanh^2left( | vedclass

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(B) We know that $	anh^2\left(\frac{	heta}{2}ight) = \frac{\cosh(	heta) - 1}{\cosh(	heta) + 1}$.Given that $\cosh(	heta) = \sec(x)$,we substitu

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$	an^2\left(\frac{x}{2}ight)$

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(B) We know that $	anh^2\left(\frac{	heta}{2}ight) = \frac{\cosh(	heta) - 1}{\cosh(	heta) + 1}$.Given that $\cosh(	heta) = \sec(x)$,we substitute this into the expression:$	anh^2\left(\frac{	heta}{2}ight) = \frac{\sec(x) - 1}{\sec(x) + 1}$.Converting $\sec(x)$ to $\frac{1}{\cos(x)}$,we get:$\frac{\frac{1}{\cos(x)} - 1}{\frac{1}{\cos(x)} + 1} = \frac{1 - \cos(x)}{1 + \cos(x)}$.Using the half-angle identities $1 - \cos(x) = 2\sin^2\left(\frac{x}{2}ight)$ and $1 + \cos(x) = 2\cos^2\left(\frac{x}{2}ight)$:$= \frac{2\sin^2\left(\frac{x}{2}ight)}{2\cos^2\left(\frac{x}{2}ight)} = 	an^2\left(\frac{x}{2}ight)$.

If $\sec(x) = \cosh(\theta)$,then find $\tanh^2\left(\frac{\theta}{2}\right)$.

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