Q170 / 270 · NDA 2015 (Phase 2) · Mathematics

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

  • A
    $\sin 3 \mathrm{~A}$
  • B
    $\cos 3 \mathrm{~A}$
  • C
    $\sin \mathrm{A}+\cos \mathrm{A}$
  • Correct answer
    D
    3

Solution

$\begin{aligned} & \frac{\sin ^3 \mathrm{~A}+\sin 3 \mathrm{~A}}{\sin \mathrm{~A}}+\frac{\cos ^3 \mathrm{~A}-\cos 3 \mathrm{~A}}{\cos \mathrm{~A}} \\ \Rightarrow & \frac{\sin ^3 \mathrm{~A}+3 \sin \mathrm{~A}-4 \sin ^3 \mathrm{~A}}{\sin \mathrm{~A}}+\frac{\cos ^3 \mathrm{~A}-\left[4 \cos ^3 \mathrm{~A}-3 \cos \mathrm{~A}\right]}{\cos \mathrm{A}} \\ \Rightarrow & \frac{3 \sin \mathrm{~A}-3 \sin ^3 \mathrm{~A}}{\sin \mathrm{~A}}+\frac{\left(-3 \cos ^3 \mathrm{~A}+3 \cos \mathrm{~A}\right)}{\cos \mathrm{A}} \\ = & 3-3 \sin ^2 \mathrm{~A}-3 \cos ^2 \mathrm{~A}+3 \\ = & 6-3\left(\cos ^2 \mathrm{~A}+\sin ^2 \mathrm{~A}\right) \\ = & 6-3(1) \\ = & 3\end{aligned}$

Hence, the correct answer is option: 'D'.

Relevant Questions