Can someone help me on this -
Prove that tan(sin^-1x) = x/sqrt(1-x^2) for |x| < 1 and x not equal to 0
Thanks !
$\displaystyle \tan(\sin^{-1} x) = \frac{\sin (\sin^{-1} x)}{\cos (\sin^{-1} x)} = \frac{x}{\cos (\sin^{-1} x)}$.
$\displaystyle \cos^2 (\sin^{-1} x) = 1 - \sin^2(\sin^{-1} x) = 1-x^2\implies \cos x (\sin^{-1} x) = \sqrt{1-x^2}$ (since the value is positive).
Thus, we get $\displaystyle \frac{x}{\sqrt{1-x^2}}$.