Question

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

Answer 1

LHS

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

$=\frac{(1+\cot A+\tan A)(\sin A-\cos A)}{(\sec A-cosecA)({\sec }^{2}A+cose{c}^{2}A+\sec AcosecA)}$

$=\frac{(1+\cot A+\tan A)(\sin A-\cos A)}{({\sec }^{2}A+cosec2A+\sec AcosecA)}$

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

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

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

$={\sin }^{2}A{\cos }^{2}A=RHS$

Hence proved