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}}}} $