Question

Prove the following identities (1-17)

$\frac{1-\sin A\cos A}{\cos A\left(secA-\mathrm{cosec}A\right)}·\frac{{\sin }^{2}A-{\cos }^{2}A}{{\sin }^{3}A+{\cos }^{3}A}=\sin A$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\frac{1-\sin A\cos A}{\cos A\left(secA-\mathrm{cosec}A\right)}\times \frac{{\sin }^{2}A-{\cos }^{2}A}{{\sin }^{3}A+{\cos }^{3}A} \\ =\frac{1-\sin A\cos A}{\cos A\left({\displaystyle \frac{1}{\cos A}}-{\displaystyle \frac{1}{\sin A}}\right)}\times \frac{{\left(\sin A\right)}^2-{\left(\cos A\right)}^2}{{\left(\sin A\right)}^3+{\left(\cos A\right)}^3} \\ =\frac{1-\sin A\cos A}{\cos A\left({\displaystyle \frac{\sin A-\cos A}{\cos A\sin A}}\right)}\times \frac{\left(\sin A+\cos A\right)\left(\sin A-\cos A\right)}{\left(\sin A+\cos A\right)\left[{\left(\sin A\right)}^2+{\left(\cos A\right)}^2-\sin A\cos A\right]} \\ =\frac{\sin A\left(1-\sin A\cos A\right)}{\left({\displaystyle \sin A-\cos A}\right)}\times \frac{\left(\sin A+\cos A\right)\left(\sin A-\cos A\right)}{\left(\sin A+\cos A\right)\left[{\left(\sin A\right)}^2+{\left(\cos A\right)}^2-\sin A\cos A\right]} \\ =\frac{\sin A\left(1-\sin A\cos A\right)}{1}\times \frac{1\times 1}{\left[{\sin }^{2}A+{\cos }^{2}A-\sin A\cos A\right]} \\ =\sin A\left(1-\sin A\cos A\right)\times \frac{1}{\left(1-\sin A\cos A\right)} \\ =\sin A \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}\mathit{.} \end{gathered}$$