Question

If $A=$ ${30}^{\circ }$; then, $\frac{{\cos }^{3}A-\cos 3A}{\cos A}+\frac{{\sin }^{3}A+\sin 3A}{\sin A}=$

Answer 1

Given: $A={30}^{\circ }$
To Prove: $\frac{{\cos }^{3}A-\cos 3A}{\cos A}+\frac{{\sin }^{3}A+\sin 3A}{\sin A}=3$
Now, L.H.S:$\frac{{\cos }^{3}A-\cos 3A}{\cos A}+\frac{{\sin }^{3}A+\sin 3A}{\sin A}$
= $\frac{{\cos }^3{30}^0-\cos {90}^0}{\cos {30}^0}+\frac{{\sin }^3{30}^0+\sin {90}^0}{\sin {30}^0}$
= $\frac{{\left(\frac{\sqrt{3}}{2}\right)}^3}{\frac{\sqrt{3}}{2}}+\frac{1+{\left(\frac{1}{2}\right)}^3}{\frac{1}{2}}$
= $\frac{3}{4}+\frac{9}{4}$ = $\frac{12}{4}$ = $3$