Question

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

Answer 1

Given expression:$\frac{1-\sin A\cos A}{\cos A(\sec A-\cos ecA)}\times \frac{{\sin }^{2}A-{\cos }^{2}A}{{\sin }^{3}A+{\cos }^{3}A}$

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

$=\frac{1-\sin A\cos A}{\cos A\left(\frac{\sin A-\cos A}{\sin A\cos A}\right)}\times \frac{(\sin A-\cos A)}{1-\sin A\cos A}$

$=\sin A$