Question

Prove that: $\frac{\left(1+\cot A+\tan A\right)\left(\sin A-\cos A\right)}{{\sec }^{3}A-{\mathrm{cosec}}^{3}A}={\sin }^{2}A{\cos }^{2}A$

Answer 1

$\begin{array}{c}\frac{\left(1+\cot A+\tan A\right)\left(\sin A-\cos A\right)}{{\sec }^{3}A-{\mathrm{cosec}}^{3}A}\\ =\frac{\left(1+\cot A+\tan A\right)\left(\sin A-\cos A\right)}{\frac{1}{{\cos }^{3}A}-\frac{1}{{\sin }^{3}A}}\\ =\frac{\left(1+\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}\right)\left(\sin A-\cos A\right){\sin }^{3}A{\cos }^{3}A}{\left({\sin }^{3}A-{\cos }^{3}A\right)}\\ =\frac{\left(\cos A\sin A+{\cos }^{2}A+{\sin }^{2}A\right)\left(\sin A-\cos A\right){\sin }^{3}A{\cos }^{3}A}{\left(\sin A\cos A\right)\left(\sin A-\cos A\right)\left({\cos }^{2}A+{\sin }^{2}A+\cos A\sin A\right)}\\ ={\sin }^{2}A{\cos }^{2}A\end{array}$