Answer
$$\frac{\tan\theta-\cot\theta}{\tan\theta+\cot\theta}=1-2\cos^2\theta$$
The proof is explained in the Work step by step.
Work Step by Step
$$\frac{\tan\theta-\cot\theta}{\tan\theta+\cot\theta}=1-2\cos^2\theta$$
The left side is more complicated, which is the reason why we need to deal with it first.
$$X=\frac{\tan\theta-\cot\theta}{\tan\theta+\cot\theta}$$
- The following identities would be applied to rewrite $\tan\theta$ and $\cot\theta$.
$$\tan\theta=\frac{\sin\theta}{\cos\theta}\hspace{2cm}\cot\theta=\frac{\cos\theta}{\sin\theta}$$
$$X=\frac{\frac{\sin\theta}{\cos\theta}-\frac{\cos\theta}{\sin\theta}}{\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}}$$
$$X=\frac{\frac{\sin^2\theta-\cos^2\theta}{\cos\theta\sin\theta}}{\frac{\sin^2\theta+\cos^2\theta}{\cos\theta\sin\theta}}$$
$$X=\frac{\sin^2\theta-\cos^2\theta}{\sin^2\theta+\cos^2\theta}$$
- Pythagorean identity: $\sin^2\theta+\cos^2\theta=1$
$$X=\frac{\sin^2\theta-\cos^2\theta}{1}$$
$$X=\sin^2\theta-\cos^2\theta$$
- Again, Pythagorean identity: $\sin^2\theta=1-\cos^2\theta$
$$X=1-\cos^2\theta-\cos^2\theta$$
$$X=1-2\cos^2\theta$$
That means $$\frac{\tan\theta-\cot\theta}{\tan\theta+\cot\theta}=1-2\cos^2\theta$$
The equation is an identity.
