Answer
$$\cot\theta-\tan\theta=\frac{2\cos^2\theta-1}{\sin\theta\cos\theta}$$
The proof of identity verification is shown below.
Work Step by Step
$$\cot\theta-\tan\theta=\frac{2\cos^2\theta-1}{\sin\theta\cos\theta}$$
From the left side:
$$X=\cot\theta-\tan\theta$$
$$X=\frac{\cos\theta}{\sin\theta}-\frac{\sin\theta}{\cos\theta}$$
(according to quotient identities)
$$X=\frac{\cos^2\theta-\sin^2\theta}{\sin\theta\cos\theta}$$
Now $\sin^2\theta=1-\cos^2\theta$ (Pythagorean identity) can be replaced into $X$
$$X=\frac{\cos^2\theta-(1-\cos^2\theta)}{\sin\theta\cos\theta}$$
$$X=\frac{\cos^2\theta-1+\cos^2\theta}{\sin\theta\cos\theta}$$
$$X=\frac{2\cos^2\theta-1}{\sin\theta\cos\theta}$$
Therefore,
$$\cot\theta-\tan\theta=\frac{2\cos^2\theta-1}{\sin\theta\cos\theta}$$
The equation is an identity as a result.
