Question

If $\tan A-\tan B=x,\cot B-\cot A=y$, prove that $\cot (A-B)=\left(\frac{1}{x}\right)+\left(\frac{1}{y}\right)$

Answer 1

$\Rightarrow$$\cot B-\cot A=y$

$\therefore \frac{1}{\left(\tan B\right)}-\frac{1}{\left(\tan A\right)}=y$

$\Rightarrow$$\frac{\tan A-\tan B}{\tan A\tan B}=y$
$\Rightarrow \frac{x}{y}=\tan A\tan B$

Now $\cot (A-B)=\frac{1}{\tan (A-B)}=\frac{1+\tan A\tan B}{\tan A-\tan B}=\frac{1+\frac{x}{y}}{x}=\frac{1}{x}+\frac{1}{y}$

Hence Proved.