Question

Prove the following identities (1-17)

$\frac{{\sin }^{3}A+{\cos }^{3}A}{\sin A+\cos A}+\frac{{\sin }^{3}A-{\cos }^{3}A}{\sin A-\cos A}=2$

Answer 1

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