Question

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

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is C 3

Given expression requires the values of $\sin 3A$ and $\cos 3A$

$\sin 3A=3\sin A-4{\sin }^{3}A$

$\cos 3A=4{\cos }^{3}A-3\cos A$

Substituting the values of $\sin 3A$ and $\cos 3A$, we get

$\frac{{\cos }^{3}A-(4{\cos }^{3}A-3\cos A)}{\cos A}+\frac{{\sin }^{3}A-3\sin A-4{\sin }^{3}A}{\sin A}$

$=\frac{{\cos }^{3}A-4{\cos }^{3}A+3\cos A}{\cos A}+\frac{-3{\sin }^{3}A+3\sin A}{\sin A}$

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

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

$=6-3({\sin }^{2}A+{\cos }^{2}A)$

$6-3=3$