Simplifying $\frac{y}{y^2+b^2}$ where $y=b \cot\theta$ Could appreciate some help with this question.

I want to simplify the following trigonometric equation.
  $$\frac{y}{y^2+b^2}$$
  where $y=b \cot\theta$.

The solution I got was 
$$\frac{1}b \cos\theta$$
Can someone verify and try and guide me through the solution?

Hint:
$$b^2+b^2\cot^2(x)=b^2(1+\cot^2(x))=b^2\left(1+ \frac{\cos^2(x)}{\sin^2(x)}\right)=b^2\left(\frac{\sin^2(x)}{\sin^2(x)}+ \frac{\cos^2(x)}{\sin^2(x)}\right)$$
