Question

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

• A. $\frac{k^2+1}{2k}$
• B. $\frac{2k}{k^2+1}$
• C. $\frac{k}{k^2+1}$
• D. $\frac{k}{k^2-1}$

Answer 1

The correct option is B

$\frac{2k}{k^2+1}$

Explanation for the correct option:

Step 1: Given information

$\sec \left(\theta \right)+\tan \left(\theta \right)=k....\left(1\right)$

Step 2: Simplification and calculate the value of $\cos \left(\theta \right)$

As we know that the trigonometrical ratio formula

${\sec }^2\theta -{\tan }^2\theta =1$

$\left[\sec \left(\theta \right)+\tan \left(\theta \right)\right]\left[\sec \left(\theta \right)-\tan \left(\theta \right)\right]=1\left[\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right]$

From equation $\left(1\right)$, put the value of $\sec \left(\theta \right)+\tan \left(\theta \right)=k$

$\begin{array}{rcl}\Rightarrow k\left[\sec \left(\theta \right)-\tan \left(\theta \right)\right]& =& 1\\ \Rightarrow \sec \left(\theta \right)-\tan \left(\theta \right)& =& \frac{1}{k}...\left(2\right)\end{array}$

Step 3: Add equation $\left(1\right)$ and equation $\left(2\right)$,

$\begin{array}{l}\Rightarrow 2\sec \left(\theta \right)=k+\frac{1}{k}\\ \Rightarrow \frac{2}{\cos \left(\theta \right)}=\frac{\left(k^2+1\right)}{k}\\ \begin{array}{rcl}& \Rightarrow & 2k=\frac{k^2+1}{k}\end{array}\\ \Rightarrow \begin{array}{rcl}& & \cos \end{array}\begin{array}{rcl}& & \left(\theta \right)\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \frac{2k}{k^2+1}\end{array}\end{array}$

Hence , option $\left(\mathrm{B}\right)$ is correct.