Question

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

Answer 1

Given: $\tan A+\tan B=a$ and $\cot A+\cot B=b,$

$\cot (A+B)=\frac{\cot A\cot B-1}{\cot A+\cot B}$

$\Rightarrow \cot (A+B)=\frac{\cot A\cot B-1}{b}\dots \left(1\right)$
We know that : $\tan A+\tan B=a$
$\Rightarrow \frac{\cot A+\cot B}{\cot A\cot B}=a$
$\Rightarrow \cot A\cot B=\frac{b}{a}$
Now, from equation $\left(1\right),$ we get
$\cot (A+B)=\frac{\frac{b}{a}-1}{b}$
$\therefore \cot (A+B)=\frac{1}{a}-\frac{1}{b}$
Hence proved.