The value of $tan(cos^{-1}x)$ is :

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

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

The correct answer is Option (2) → $\frac{\sqrt{1-x^2}}{x}$ $\tan(\cos^{-1}x)$ Using pythagoras theorem so $\tan(\cos^{-1}x)$ $=\tan θ=\frac{\sqrt{1-x^2}}{x}$