Question

On simplifying $\frac{{\sin }^{3}A+\sin 3A}{\sin A}+\frac{{\cos }^{3}A-\cos 3A}{\cos A}$ we get

• A. $\sin 3A$
• B. $\cos 3A$
• C. $\sin A+\cos A$
• D. $3$

Answer 1

Answer: (D)

$3$

Solution:

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

$\because \sin 3A=3\sin A-4{\sin }^{3}A$ and $\cos 3A=4{\cos }^{3}A-3\cos A$

Then, we get

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

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

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

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

$=6-3$

$=3$

Hence, D is the correct option.