Question

If $\tan A-\tan B=x$ and $\cot B-\cot A=y$

Prove that $\cot (A-B)=\frac{1}{x}+\frac{1}{y}$ ?

Answer 1

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