<s>[INST] 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?
 [/INST] $$1+\cot^2\theta=1+\frac{\cos^2\theta}{\sin^2\theta}=\frac1{\sin^2\theta}=\csc^2\theta$$ or
$$1+\cot^2\theta=\frac1{\sin^2\theta}$$
Now take reciprocal of both side:
$$\frac{1}{1+\cot^2\theta}=\sin^2\theta$$
If $y=b\cot\theta$, then:
$$\frac{y}{y^2+b^2}=\frac{b\cot\theta}{b^2\cot^2\theta+b^2}=\frac{\cot\theta}{b\left(1+\cot^2\theta\right)}=\frac{1}{b}\cot\theta\sin^2\theta=\frac{1}{b}\frac{\cos\theta}{\sin\theta}\sin^2\theta=\frac{1}{2b}2\cos\theta\sin\theta=\frac{1}{2b}\sin{2\theta}$$
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