Question

If $sec\left(a\right)+\tan \left(a\right)=x$, then $sec\left(a\right)=$

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

Answer 1

The correct option is A

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

Explanation for the correct option:

Find the required value using standard trigonometric identities

Since it is given that $sec\left(a\right)+\tan \left(a\right)=x...\left[1\right]$.

Multiply and divide the left-hand side by $sec\left(a\right)-\tan \left(a\right)$.

$$\begin{gathered} \Rightarrow \left(sec\left(a\right)+\tan \left(a\right)\right)\times \frac{sec\left(a\right)-\tan \left(a\right)}{sec\left(a\right)-\tan \left(a\right)}=x \\ \Rightarrow \frac{se{c}^2\left(a\right)-{\tan }^2\left(a\right)}{sec\left(a\right)-\tan \left(a\right)}=x \\ \Rightarrow \frac{1}{sec\left(a\right)-\tan \left(a\right)}=x \\ \Rightarrow sec\left(a\right)-\tan \left(a\right)=\frac{1}{x}...\left[2\right] \end{gathered}$$ {Since, $se{c}^2\left(\theta \right)-{\tan }^2\left(\theta \right)=1$}

Add equation $1$ and equation $2$.

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

Therefore, $sec\left(a\right)=\frac{x^2+1}{2x}$.

Hence, option $A$ is the correct answer.