Question

$\tan \left({\cos }^{-1}x\right)$is equal to?

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

Answer 1

The correct option is A

$\frac{\sqrt{1-x^2}}{x}$

Explanation for correct option

Simplifying the given expression

Given, $\tan \left({\cos }^{-1}x\right)$

Let,

$$\begin{gathered} {\cos }^{-1}\left(x\right)=𝛳 \\ \Rightarrow \cos 𝛳=x \end{gathered}$$

Solving the above expression:

$$\begin{gathered} \Rightarrow \sin 𝛳=\sqrt{1-x^2} \\ \Rightarrow \tan 𝛳=\frac{\sqrt{1-x^2}}{x} \\ \Rightarrow 𝛳={\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right) \\ \Rightarrow {\cos }^{-1}\left(x\right)={\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right) \\ \Rightarrow \tan \left({\cos }^{-1}\left(x\right)\right)=\tan \left({\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)\right) \\ =\frac{\sqrt{1-x^2}}{x} \end{gathered}$$

Hence, the correct answer is Option (A).