Question

A triangle $ABC$ is right angled at $B$; find the value of $\frac{\sec A.\sin C-\tan A.\tan C}{\sin B}$.

Answer 1

$A+90°+C=180°=>A+C=90°\frac{\sec A\sin C-\tan A\tan C}{\sin B}=(\sec A\sin C-\frac{\sin A}{cosA}\times \frac{\sin C}{cosC})÷\sin 90°=(\frac{\sin C}{\cos A}-\frac{\sin A}{cosA}\times \frac{\sin C}{cosC})=(\frac{\sin (90°-A)}{\cos A}-\frac{\sin A}{cosA}\times \frac{\sin (90°-A)}{cos(90°-A)})=\frac{\cos A}{\cos A}-\frac{\sin A}{cosA}\times \frac{\cos A}{\sin A}=1-1=0$