Question

Evaluate $\frac{\cot A+\tan B}{\cot B+\tan A}-\cot A\tan B.$

Answer 1

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

$\left(\frac{\cos A.\cos B+\sin B.\sin A}{\cos B.\cos A+\sin A.\sin B}\right)\times \frac{\sin B.\cos A}{\sin A.\cos B}-\cot A.\tan B$

$=\frac{\cos (A-B)}{\cos (A-B)}\times \cot A.\tan B-\cot A.\tan B$

$=0$

$\therefore$ The result is zero.