Question 689775
<pre>
tan(sin<sup>-1</sup>(x))

First let's find sin<sup>-1</sup>(x).

sin<sup>-1</sup>(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 {{{SINE}}}{{{""=""}}}{{{OPPOSITE/HYPOTENUSE}}}

Second, we observe that x = {{{x/1}}}.  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 &#8730;<span style="text-decoration: overline">1-x²</span>

Here's a right triangle which contains an angle whose sine is {{{x/1}}} or x.

{{{drawing(300,150,-.1,1.1,-.3,.9, triangle(0,0,1,0,1,.7),
locate(.5,.5,1), arc(0,0,1,-1,0,36), locate(.4,-.05,sqrt(1-x^2)),
locate(.2,.18,"sin"^(-1) ), locate(.4,.13,(x)),locate(1.02,.35,x)
)}}}

Let's go back to the original problem:

tan(sin<sup>-1</sup>(x))

We have sin<sup>-1</sup>x as an angle in the right triangle above,
so all we need is the TANGENT of the angle marked sin<sup>-1</sup>x.
Since {{{TANGENT}}}{{{""=""}}}{{{OPPOSITE/ADJACENT}}}, 

tan(sin<sup>-1</sup>(x)) = {{{x/sqrt(1-x^2)}}}

We have now verified that.

Edwin</pre>