Write the given expression as an algebraic expression in $x$: $\tan(2 \cos^{−1} x)$. OK I first assigned $\theta = \cos^{−1} x$
and so $\cos(\theta) = x$
drew a triangle where adj = $x$  hyp = $1$ and opp = $\sqrt{1-x^2}$ by pyth theorem.
Then $\tan(2\cos^{−1} x) = \tan(2\theta)$
and I used the double angle formula for tangent so
$\tan(2\theta) = \frac{2\tan(\theta)}{1-\tan^2(\theta)}$
Then used the triangle: $\tan(\theta) = \frac{\sqrt{1-x^2}}{x}$
but my answer is 
The computer states I am wrong. Am I setting the problem up wrong? 

You set the problem up fine:
Let me retrace your work for you.
So you are correct with all your steps.
$$tan(\theta) = \frac{\sqrt{1-x^2}}{x} $$
Now, 
$$\tan{2\theta} = \frac{2*\frac{\sqrt{1-x^2}}{x}}{1 - (\frac{\sqrt{1-x^2}}{x})^2}$$
$$ = \frac{\frac{2\sqrt{1-x^2}}{x}}{1 - \frac{1-x^2}{x^2}}$$
$$= \frac{\frac{2\sqrt{1-x^2}}{x}}{\frac{2x^2 - 1}{x^2}}$$
$$= \frac{2\sqrt{1-x^2}}{x}*\frac{x^2}{2x^2 -1}$$
$$= \frac{2x\sqrt{1-x^2}}{2x^2 -1}$$
