Answer
$$\frac{2\cos2\theta}{\sin2\theta}=\cot\theta-\tan\theta$$
The equation is an identity. The proof is below.
Work Step by Step
$$\frac{2\cos2\theta}{\sin2\theta}=\cot\theta-\tan\theta$$
We solve from the left side, as it is more complex.
$$X=\frac{2\cos2\theta}{\sin2\theta}$$
- We replace $\cos2\theta$ and $\sin2\theta$ with the following identities from Double-Angle identities:
$$\cos2\theta=\cos^2\theta-\sin^2\theta$$
$$\sin2\theta=2\sin\theta\cos\theta$$
Therefore, $$X=\frac{2(\cos^2\theta-\sin^2\theta)}{2\sin\theta\cos\theta}$$
$$X=\frac{\cos^2\theta-\sin^2\theta}{\sin\theta\cos\theta}$$
$$X=\frac{\cos^2\theta}{\sin\theta\cos\theta}-\frac{\sin^2\theta}{\sin\theta\cos\theta}$$
$$X=\frac{\cos\theta}{\sin\theta}-\frac{\sin\theta}{\cos\theta}$$
- From Quotient Identities: $\frac{\cos\theta}{\sin\theta}=\cot\theta$ and $\frac{\sin\theta}{\cos\theta}=\tan\theta$.
Therefore, $$X=\cot\theta-\tan\theta$$
So, $$\frac{2\cos2\theta}{\sin2\theta}=\cot\theta-\tan\theta$$
We conclude that the equation is an identity.
