Question

prove that sin cube theta + cos cube theta / sin theta + cos theta + sin theta cos theta =1

Answer 1

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