## Trigonometry (11th Edition) Clone

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