Question

How do you simplify $sec\left(ta{n}^{-1}x\right)$?

Answer 1

Simplify the given expression using the trigonometric identities.

We have to evaluate $sec\left({\tan }^{-1}x\right)$

Let's assume that

$$\begin{gathered} y=\left({\tan }^{-1}x\right) \\ \Rightarrow x=\tan y \\ \Rightarrow x=\frac{\sin y}{\cos y} \\ \Rightarrow x^2=\frac{{(\sin y)}^2}{{(\cos y)}^2} \\ \Rightarrow x^2+1=\frac{{\cos }^{2}y+{\sin }^{2}y}{{\cos }^{2}y}[{\sin }^{2}x+{\cos }^{2}x=1] \\ \Rightarrow x^2+1=se{c}^{2}y \\ \Rightarrow \sqrt{x^2+1}=se{c}^{2}y \\ \Rightarrow \sqrt{x^2+1}=sec\left({\tan }^{-1}x\right) \\ \Rightarrow sec\left({\tan }^{-1}x\right)=\sqrt{x^2+1} \end{gathered}$$

Hence, the required answer is $sec\left({\tan }^{-1}x\right)=\sqrt{x^2+1}$