Question

If $\tan \left(\frac{x}{2}\right)\cot h\left(\frac{x}{2}\right)=1$, then the value of $\cos \left(x\right)\cos h\left(x\right)$ is

• A. $1$
• B. $-1$
• C. ${\cos }^2\left(x\right)$
• D. ${\sin }^2\left(x\right)$

Answer 1

The correct option is A

$1$

Explanation for the correct option

Given: $\tan \left(\frac{x}{2}\right)\cot h\left(\frac{x}{2}\right)=1$

$$\begin{gathered} \Rightarrow \tan \left(\frac{x}{2}\right)=\frac{1}{\cot h\left({\displaystyle \frac{x}{2}}\right)} \\ \Rightarrow \tan \left(\frac{x}{2}\right)=\tan h\left(\frac{x}{2}\right) \end{gathered}$$

squaring both sides

$\Rightarrow \frac{{\tan }^2\left(\frac{x}{2}\right)}{1}=\frac{\tan h^2\left(\frac{x}{2}\right)}{1}$

applying componendo and diviendo

$$\begin{gathered} \Rightarrow \frac{1+{\tan }^2\left({\displaystyle \frac{x}{2}}\right)}{1-{\tan }^2\left({\displaystyle \frac{x}{2}}\right)}=\frac{1+\tan h^2\left({\displaystyle \frac{x}{2}}\right)}{1-\tan h^2\left({\displaystyle \frac{x}{2}}\right)} \\ \Rightarrow \frac{1}{\cos \left(x\right)}=\cos h\left(x\right)\left[\because \frac{1-{\tan }^2\left({\displaystyle a}\right)}{1+{\tan }^2\left({\displaystyle a}\right)}=\cos \left(2a\right),\frac{1+\tan h^2\left({\displaystyle a}\right)}{1-\tan h^2\left({\displaystyle a}\right)}=\cos h\left(2a\right)\right] \\ \Rightarrow \cos \left(x\right)\cos h\left(x\right)=1 \end{gathered}$$

Hence, option A is correct.