Question

Prove that $\frac{(1+\cot A+\tan A)(\sin A-\cos A)}{{\sec }^{3}A-{\csc }^{3}A}={\sin }^{2}A{\cos }^{2}A$

Answer 1

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

$=\frac{(1+\cot A+\tan A)(\sin A-\cos A)}{\frac{1}{{\cos }^{3}A}-\frac{1}{{\sin }^{3}A}}$

$=\frac{{\sin }^{3}A{\cos }^{3}A(1+\cot A+\tan A)(\sin A-\cos A)}{{\sin }^{3}A-{\cos }^{3}A}$

$=\frac{{\sin }^{3}A{\cos }^{3}A(1+\cot A+\tan A)(\sin A-\cos A)}{(\sin A-\cos A)({\sin }^{2}A+{\cos }^{2}A+\sin A\cos A)}$

$=\frac{{\sin }^{3}A{\cos }^{3}A(1+\cot A+\tan A)}{(1+\sin A\cos A)}$

$=\frac{{\sin }^{3}A{\cos }^{3}A(1+\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A})}{(1+\sin A\cos A)}$

$=\frac{{\sin }^{3}A{\cos }^{3}A\left(\frac{\sin A\cos A+{\cos }^{2}A+{\sin }^{2}A}{\sin A\cos A}\right)}{(1+\sin A\cos A)}$

$=\frac{{\sin }^{2}A{\cos }^{2}A(1+\sin A\cos A)}{(1+\sin A\cos A)}$

$={\sin }^{2}A{\cos }^{2}A=$R.H.S

Hence proved.