Question

If $sec\left(\theta \right)=x+\frac{1}{4x}$ then prove that, $sec\left(\theta \right)+\tan \left(\theta \right)=2x$ or $\frac{1}{2x}$.

Answer 1

Step 1: Find the possible values of $\tan \theta$ using trigonometric identities

Given that: $sec\left(\theta \right)=x+\frac{1}{4x}$

$$\begin{gathered} \because 1+{\tan }^2\left(\theta \right)=se{c}^2\left(\theta \right) \\ \Rightarrow {\tan }^2\left(\theta \right)=se{c}^2\left(\theta \right)-1 \end{gathered}$$

On expanding, we get

$\begin{array}{rcl}{\tan }^2\left(\theta \right)& =& {\left(x+\frac{1}{4x}\right)}^2-1\\ & \Rightarrow & {\tan }^2\left(\theta \right)=\left(x^2+\frac{1}{16x^2}+\frac{1}{2}-1\right)\\ & \Rightarrow & {\tan }^2\left(\theta \right)=\left(x^2+\frac{1}{16x^2}-\frac{1}{2}\right)\\ & \Rightarrow & {\tan }^2\left(\theta \right)={\left(x-\frac{1}{4x}\right)}^2\\ & \Rightarrow & {\tan }^2\left(\theta \right)=\pm \left(x-\frac{1}{4x}\right)\end{array}$

Step 2: Prove the required result

When $\tan \left(\theta \right)=\left(x-\frac{1}{4x}\right)$ we get,

$$\begin{gathered} \Rightarrow sec\left(\theta \right)+\tan \left(\theta \right)=x+\frac{1}{4x}+x-\frac{1}{4x} \\ \Rightarrow sec\left(\theta \right)+\tan \left(\theta \right)=2x \end{gathered}$$

When $\tan \left(\theta \right)=-\left(x-\frac{1}{4x}\right)$ we get,

$$\begin{gathered} \Rightarrow sec\left(\theta \right)+\tan \left(\theta \right)=\left(x+\frac{1}{4x}\right)-\left(x-\frac{1}{4x}\right) \\ \Rightarrow sec\left(\theta \right)+\tan \left(\theta \right)=\frac{1}{2x} \end{gathered}$$

Hence proved that $sec\left(\theta \right)+\tan \left(\theta \right)=\frac{1}{2x}$ and, $sec\left(\theta \right)+\tan \left(\theta \right)=2x$.