I have to show that the left equation simplifies to $\tan\theta$:
Show that:
$$\frac{1-\cos2 \theta}{\sin2 \theta} = \tan \theta$$
I do have prior knowledge that:
$$\tan \theta = \frac{\sin\theta}{\cos \theta}$$
But I'm stuck from this point, I have tried a few rules, but none have seemed to work so far.
Best Answer
Using the double angle formulae:
$$\cos(2\theta)=1-2\sin^2\theta$$ $$\sin(2\theta)=2\sin\theta\cos\theta$$
$$\frac{1-\cos(2\theta)}{\sin(2\theta)}=\frac{1-1+2\sin^2\theta}{2\sin\theta\cos\theta}=\frac{2\sin^2\theta}{2\sin\theta\cos\theta}=\frac{\sin\theta}{\cos\theta}=\tan\theta$$