Question

Prove that: $(1+\cot A+\tan A)\left(\sin A-\cos A\right)=\frac{\sec A}{{\csc }^{2}A}-\frac{\csc A}{\sec {}^{2}A}$

Answer 1

L.H.S. $=(\sin A-\cos A)(1+\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A})$
$=\frac{1}{\sin A\cos A}(\sin A-\cos A)({\sin }^{2}A+{\cos }^{2}A+\sin A-\cos A)$
$=\frac{{\sin }^{3}A{\cos }^{3}A}{\sin A\cos A}=\frac{{\sin }^{2}A}{\cos A}-\frac{{\cos }^{2}A}{\sin A}$
$=\frac{\sec A}{{csc}^{2}A}-\frac{\csc A}{{\sec }^{2}A}$ .