Question

Evaluate $\tan \left[{\sec }^{-1}\left(\sqrt{1+x^2}\right)\right]$

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

Answer 1

The correct option is B

$x$

Explanation for correct option

Given, $\tan \left[{\sec }^{-1}\left(\sqrt{1+x^2}\right)\right]$

Putting $\mathrm{x}=\tan \left(\mathrm{\theta }\right)$ in the above expression we get,

$\Rightarrow \tan \left[{\sec }^{-1}\left(\sqrt{1+{\tan }^2\left(\theta \right)}\right)\right]$

$\Rightarrow \tan \left[{\sec }^{-1}\sec \right(\mathrm{\theta }\left)\right]\left[\because 1+{\tan }^2\left(\mathrm{\theta }\right)={\sec }^2\left(\mathrm{\theta }\right)\right]$

$$\begin{gathered} \Rightarrow \tan \left(\mathrm{\theta }\right) \\ \Rightarrow \mathrm{x} \end{gathered}$$

Therefore the value of $\tan \left[{\sec }^{-1}\left(\sqrt{1+x^2}\right)\right]$ is $\mathrm{x}$.

Hence, the correct answer is Option (B).