Question 905367
Let us use a letter to represent 2 acrcos(x).

How about A.

We now have tan(A).

Using the Pythagorean Theorem, we get x^2 + y^2 = 1^1.

x^2 + y^2 = 1

Solve for y.

y^2 = 1 - x^2

Taking the square root of both sides we are left with

y = sqrt{1 - x^2}.

In the right triangle, we know that y = sqrt{1 - x^2}.

What is tangent?  

Tangent is opposite over adjacent.

tan(2 arccos(x)) = 2x/(sqrt{1 - x^2})

We cannot leave a radical in the denominator of a fraction.

We rationalize the denominator and the final answer is


tan(2 arccos(x)) = 2x*sqrt{1 - x^2}/(1 - x^2)