Question

If $\tan h^2\left(x\right)={\tan }^2\left(\theta \right)$ then $\cos h\left(2x\right)=?$

• A. $-\sin \left(2\theta \right)$
• B. $\sec \left(2\theta \right)$
• C. $\cos \left(3\theta \right)$
• D. $\cos \left(2\theta \right)$

Answer 1

The correct option is B

$\sec \left(2\theta \right)$

Explanation for the correct option

Given that $\tan h^2\left(x\right)={\tan }^2\left(\theta \right)$

We know that $\cos h\left(2x\right)=\frac{1+\tan h^2\left(x\right)}{1-\tan h^2\left(x\right)}$

$\begin{array}{cccll}\therefore & \cos h\left(2x\right)& =& \frac{1+{\tan }^2\left(\theta \right)}{1-{\tan }^2\left(\theta \right)}& \left[\because \tan h^2\left(x\right)={\tan }^2\left(\theta \right)\right]\\ & & =& \frac{1}{{\displaystyle \frac{1-{\tan }^2\left(\theta \right)}{1+{\tan }^2\left(\theta \right)}}}& \\ & & =& \frac{1}{\cos \left(2\theta \right)}& \left[\because \cos \left(2\theta \right)=\frac{1-{\tan }^2\left(\theta \right)}{1+{\tan }^2\left(\theta \right)}\right]\\ \Rightarrow & \cos h\left(2x\right)& =& \sec \left(2\theta \right)& \left[\because \sec \left(\theta \right)=\frac{1}{\cos \left(\theta \right)}\right]\end{array}$

Hence the correct option is option(B) i.e. $\sec \left(2\theta \right)$