$\begin{align}  & \text{From the left hand side} \\ & \text{Note: }\tan \theta =\frac{1}{\cot \theta } \\ & \frac{\tan u-\tan v}{1+\tan u\tan v}=\frac{\tfrac{1}{\cot u}-\tfrac{1}{\cot v}}{1+\tfrac{1}{\cot u}\cdot \tfrac{1}{\cot v}}=\frac{\tfrac{\cot v-\cot u}{\cot u\cot v}}{\tfrac{\cot u\cot v+1}{\cot u\cot v}} \\ & \frac{\tan u-\tan v}{1+\tan u\tan v}=\frac{\cot v-\cot u}{\cot u\cot v}\times \frac{\cot u\cot v}{1+\cot u\cot v} \\ & \frac{\tan u-\tan v}{1+\tan u\tan v}=\frac{\cot v-\cot u}{1+\cot u\cot v} \\\end{align}$