Simplifying $\frac{y}{y^2+b^2}$ where $y=b \cot\theta$

Question

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?

Answer 1

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

Answer 2

Observe that

$$1+\cot^2t=1+\frac{\cos^2t}{\sin^2t}=\frac1{\sin^2t}=\csc^2t\implies$$

$$\frac y{y^2+\cot^2t}=\frac{b\cot t}{b^2\cot^2t+\cot^2t}=\frac{b\cot t}{b^2(1+\cot^2t)}=\frac{b\cot t}{b^2\csc^2t}=\frac{\cos t}b=\frac1b\cos t$$

So you got it right...

Answer 3

$$1+\cot^2\theta=1+\frac{\cos^2\theta}{\sin^2\theta}=\frac1{\sin^2\theta}=\csc^2\theta$$ 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: $$\frac{y}{y^2+b^2}=\frac{b\cot\theta}{b^2\cot^2\theta+b^2}=\frac{\cot\theta}{b\left(1+\cot^2\theta\right)}=\frac{1}{b}\cot\theta\sin^2\theta=\frac{1}{b}\frac{\cos\theta}{\sin\theta}\sin^2\theta=\frac{1}{2b}2\cos\theta\sin\theta=\frac{1}{2b}\sin{2\theta}$$