If tan A + tan B = a and cot A + cot B = b,
Question:

If $\tan A+\tan B=a$ and $\cot A+\cot B=b$, prove that $\cot (A+B) \frac{1}{a}-\frac{1}{b}$.

Solution:

Given:

$\cot A+\cot B=b$

$\Rightarrow \frac{1}{\tan A}+\frac{1}{\tan B}=b$

$\Rightarrow \frac{\tan A+\tan B}{\tan A \tan B}=b$

NOW,

$\mathrm{RHS}=\frac{1}{a}-\frac{1}{b}$

$=\frac{1}{\tan A+\tan B}-\frac{\tan A \tan B}{\tan A+\tan B}$

$=\frac{1-\tan A \tan B}{\tan A+\tan B}$

$=\cot (A+B)$

= RHS

Hence proved.