Question

Prove the following identity :

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

Answer 1

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

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

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

= ${\displaystyle \frac{\frac{1+\tan A\tan B}{\tan A}}{\frac{1+\tan A\tan B}{\tan B}}=\frac{1+\tan A\tan B}{\tan A}.\frac{\tan B}{1+\tan A\tan B}}$

= ${\displaystyle \frac{\tan B}{\tan A}=\frac{1}{\tan A}.\tan B=\cot A\tan B}$