# How to prove this trig identity turning $\tan$ into $\cot$?

$$\frac{\tan u - \tan v}{1 + \tan u \cdot \tan v} = \frac{\cot v - \cot u}{\cot u \cdot \cot v+1}$$

I've been trying to prove this for a while, no luck (I do know it's true). I've attempted to turn it all into $\tan(u - v)$, didn't work. Playing with reciprocals didn't work either. I would appreciate any pointers. Thanks.

The result follows from multiplying the top and bottom of $$\frac{\tan u - \tan v}{1 + \tan u \tan v}$$ by $\cot u\cot v$. Keep in mind $\tan u\cot u = 1 = \tan v\cot v$.
Use $$\tan u = \frac{1}{\cot u}$$ and similar for $v$.