Question

$\frac{\tan A+\tan B}{\cot A+\cot B}=\tan A\tan B$

Answer 1

$\begin{array}{l}\\ \text{LHS=}\frac{\tan A+\tan B}{\cot A+\cot B}\\ \text{=}\frac{\tan A+\tan B}{\frac{1}{\tan A}+\frac{1}{\tan B}}\\ \text{=}\frac{\tan A+\tan B}{\frac{\tan A+\tan B}{\tan A\tan B}}\\ \text{=}\frac{\tan A\tan B(\tan A+\tan B)}{(\text{t}\mathrm{an}A+\tan B)}\\ \text{=}\tan A\tan B\\ \text{=RHS}\end{array}$

Hence, LHS = RHS