SOLUTION: please help me varify tan(sin^-1(x)) = {{{x/sqrt(1-x^2)}}}

tan(sin-1(x))

First let's find sin-1(x).

sin-1(x) means "The ANGLE whose SINE is x".

So let's draw a right triangle with an ANGLE whose SINE is x.

First we observe that 

Second, we observe that x = .  Se we draw a right triangle
with x for the OPPOSITE side and 1 for the HYPOTENUSE, the Pythagorean
theorem tells us that the ADJACENT side in √1-x²

Here's a right triangle which contains an angle whose sine is  or x.



Let's go back to the original problem:

tan(sin-1(x))

We have sin-1x as an angle in the right triangle above,
so all we need is the TANGENT of the angle marked sin-1x.
Since , 

tan(sin-1(x)) = 

We have now verified that.

Edwin