Question

If $\tan \left(\theta \right)+sec\left(\theta \right)=e^x$, then $\cos \left(\theta \right)$ equals

• A. $\frac{e^x+e^{-x}}{2}$
• B. $\frac{2}{e^x+e^{-x}}$
• C. $\frac{e^x-e^{-x}}{2}$
• D. $\frac{e^x-e^{-x}}{e^x+e^{-x}}$

Answer 1

The correct option is B

$\frac{2}{e^x+e^{-x}}$

Explanation for the correct option

Given: $\tan \left(\theta \right)+sec\left(\theta \right)=e^x$ -------(1)

As we know that $se{c}^2\left(\theta \right)-{\tan }^2\left(\theta \right)=1$

$\Rightarrow (\sec \left(\theta \right)-\tan (\theta \left)\right)(\sec \left(\theta \right)+\tan (\theta \left)\right)=1$

from equation (1)

$$\begin{gathered} \Rightarrow (\sec \left(\theta \right)-\tan (\theta \left)\right)\left(e^x\right)=1 \\ \Rightarrow \sec \left(\theta \right)-\tan \left(\theta \right)=\frac{1}{e^x}-----\left(2\right) \end{gathered}$$

adding both equations

$$\begin{gathered} \Rightarrow \tan \left(\theta \right)+sec\left(\theta \right)-\tan \left(\theta \right)+sec\left(\theta \right)=e^x+\frac{1}{e^x} \\ \Rightarrow 2\sec \left(\theta \right)=e^x+e^{-x} \\ \Rightarrow \frac{1}{\cos \left(\theta \right)}=\frac{e^x+e^{-x}}{2} \\ \Rightarrow \cos \left(\theta \right)=\frac{2}{e^x+e^{-x}} \end{gathered}$$

Hence, option B is correct.