Prove $ \frac{\sin\theta}{1-\cos\theta} - \frac{\sin\theta}{1+\cos\theta} = 2\cot \theta$

Question

Prove $$ \frac{\sin\theta}{1-\cos\theta} - \frac{\sin\theta}{1+\cos\theta} = 2\cot \theta$$

So I started by combining the two fractions, which gave me: $$ \frac{\sin\theta(1+\cos\theta) - \sin\theta(1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)} = \frac{2\sin\theta\cos\theta}{1-\cos^2\theta} = \frac{\sin2\theta}{1-\cos^2\theta}$$ I wasn't sure where to go from here considering I'm aiming for $2\cos\theta / \sin\theta$

Answer 1

$$\frac{2\sin\theta\cos\theta}{1-\cos^2\theta} =\frac{2\sin\theta\cos\theta}{\sin^2\theta} = \frac{2\cos\theta}{\sin \theta} = 2\cot \theta$$

Answer 2

It's $$\frac{\sin2\theta}{\sin^2\theta}=\frac{2\sin\theta\cos\theta}{\sin^2\theta}=\frac{2\cos\theta}{\sin\theta}=2\cot\theta.$$

Answer 3

As $\sin^2t=1-\cos^2t=(1-?)(1+?)$

$$\dfrac{\sin t}{1\pm\cos t}=\dfrac{1\mp \cos t}{\sin t}$$