Why $\frac{1}{\cos(\sin^{-1}(x))}=\frac{1}{\sqrt{1-x^2}}$? This is the inverse function of sin. Why is $\cos(\sin^{-1}x)=\sqrt{1-x^2}$?
Thanks a lot.

Draw a right angled triangle a marked angle $\theta $. We can scale the triangle to set the hypotenuse equal to 1. Label the side opposite $\theta $ by some length $x$, then the adjacent side has length $\sqrt{1-x^2}.$ 
$\phi = \sin^{-1} x $ denotes the angle such that $\sin \phi = x $. So in the triangle, $ \sin^{-1} (x) = \theta. $ 
Then $ \displaystyle \cos \theta = \frac{ \text{Adjacent} }{ \text{Hypotenuse} } = \frac{ \sqrt{1-x^2} }{1} = \sqrt{1-x^2} $ so $$ \cos ( \sin^{-1} (x) ) = \sqrt{1-x^2}.$$
