#### Answer

$$\frac{\cot\alpha+1}{\cot\alpha-1}=\frac{1+\tan\alpha}{1-\tan\alpha}$$
The identity is proved by representing the left side in terms of $\tan\alpha$.

#### Work Step by Step

$$\frac{\cot\alpha+1}{\cot\alpha-1}=\frac{1+\tan\alpha}{1-\tan\alpha}$$
We find the left side comprising of only $\cot\alpha$, while the right side comprising of only $\tan\alpha$, so we only have to choose one side and represent it in terms of the other.
Here I would choose to represent the left side in terms of $\tan\alpha$.
$$A=\frac{\cot\alpha+1}{\cot\alpha-1}$$
We would do so by using the identity $$\cot\alpha=\frac{1}{\tan\alpha}$$
which means $$A=\frac{\frac{1}{\tan\alpha}+1}{\frac{1}{\tan\alpha}-1}$$
$$A=\frac{\frac{1+\tan\alpha}{\tan\alpha}}{\frac{1-\tan\alpha}{\tan\alpha}}$$
$$A=\frac{1+\tan\alpha}{\tan\alpha}\times\frac{\tan\alpha}{1-\tan\alpha}$$
$$A=\frac{1+\tan\alpha}{1-\tan\alpha}$$
The left side is equal with the right side. The trigonometric expression thus is an identity.