Question

If $\sec \left(\theta \right)+\tan \left(\theta \right)=p$,then $\tan \left(\theta \right)$ is equal to

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

Answer 1

The correct option is B

$\frac{p^2-1}{2p}$

Explanation for the correct option:

Step 1: Given information

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

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

We know that the trigonometrical ratio formula

$\begin{array}{rcl}\left[{\sec }^2\left(\theta \right)-{\tan }^2\left(\theta \right)\right]& =& 1\\ \Rightarrow \left[\sec \left(\theta \right)+\tan \left(\theta \right)\right]\left[\sec \left(\theta \right)-\tan \left(\theta \right)\right]& =& 1\left[\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right]\\ & & \end{array}$

Put the value of $\sec \left(\theta \right)+\tan \left(\theta \right)=p$

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

Step 3: Subtract the equation $\left(2\right)$ from $\left(1\right)$

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

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