Question

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

Answer 1

Simplify the LHS of $\frac{1-\sin A\cos A}{\cos A\left(\sec A-\csc A\right)}\times \frac{{\sin }^{2}A-{\cos }^{2}A}{{\sin }^{3}A+{\cos }^{3}A}=\sin A$

$\frac{1-\sin A\cos A}{\cos A\left(\sec A-\csc A\right)}\times \frac{{\sin }^{2}A-{\cos }^{2}A}{{\sin }^{3}A+{\cos }^{3}A}$

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

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

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

$=\sin A$

$=RHS$