Question

If $-\frac{\pi }{2}<{\sin }^{-1}x<\frac{\pi }{2},$ then $\tan ({\sin }^{-1}x)$ is equal

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

Answer 1

Answer: (C)

$\tan ({\sin }^{-1}x)=\tan \left({\tan }^{-1}\frac{x}{\sqrt{1-x^2}}\right),x\in (-1,1)$
$=\frac{x}{\sqrt{1-x^2}}$