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

Question

Aicharya · Accepted Answer

Since, $\tan h^2\left(\frac{\theta}{2}\right)=\frac{\cosh (\theta)-1}{\cos h(\theta)+1}$ $\begin{aligned} & =\frac{\sec x-1}{\sec x+1} \quad(\because \cosh \theta=\sec x \text { given }) \ & =\frac{1-\cos x}{1+\cos x}=\frac{2 \sin ^2 \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}=\tan ^2 \frac{x}{2} \end{aligned}$

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

Solution