Question

If $tan{h}^{2}x=cos\theta$, then $cosh2x=$

• A. ${\tan }^2\left(\theta /2\right)$
• B. ${\cot }^2\left(\theta /2\right)$
• C. ${\sec }^2\left(\theta /2\right)$
• D. none of these

Answer 1

The correct option is A ${\tan }^2\left(\theta /2\right)$

Given ${\tanh }^{2}x=\cos \theta$

$⟹\cosh 2x=\frac{1+{\tanh }^{2}x}{1-{\tanh }^{2}x}$

$=\frac{1+\cos \theta }{1-\cos \theta }$

$=\frac{2{\cos }^2\frac{\theta }{2}}{2{\sin }^2\frac{\theta }{2}}$

$={\cot }^2\frac{\theta }{2}$