*How do you write `sin(2Tan^(-1)(x))` as a function of `x` ?*

(1) Let `theta=Tan^(-1)(x)` . Then there is a right triangle with acute angle `theta` such that `tan theta=x` . Then we can let the legs of the right triangle be `x` and 1.

Then the hypotenuse is `sqrt(x^2+1)`

(2) Then `sin(2Tan^(-1)(x))=sin 2 theta = 2 sin theta cos theta` .

For the triangle, `sin theta = x/(sqrt(x^2+1))` and `cos theta = 1/(sqrt(x^2+1))` .

`2sin theta cos theta = 2(x/sqrt(x^2+1))(1/sqrt(x^2+1))=(2x)/(x^2+1)`

**Thus** `sin(2Tan^(-1)(x))=(2x)/(x^2+1)` .

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