Q126 / 200 · WBJEE 2013 · Mathematics

If $\sin ^{2} \theta+3 \cos \theta=2,$ then $\cos ^{3} \theta+\sec ^{3} \theta$ is equal out to

  • A
    1
  • B
    4
  • C
    9
  • Correct answer
    D
    18

Solution

Given, $\sin ^{2} \theta+3 \cos \theta=2$ $\begin{array}{lr}\Rightarrow & 1-\cos ^{2} \theta+3 \cos \theta=2 \\ \Rightarrow & \cos ^{2} \theta-3 \cos \theta+1=0\end{array}$ $\Rightarrow \quad \cos \theta=\frac{3 \pm \sqrt{9-4}}{2}$ (by quadratic formula) $\Rightarrow \quad \cos \theta=\frac{3 \pm \sqrt{5}}{2}$ Here, $\cos \theta=\frac{3-\sqrt{5}}{2}$ $\left(\because \frac{3+\sqrt{5}}{2} > 1 \text{ and } -1 \leq \cos \theta \leq 1\right)$ and $\sec \theta=\frac{2}{3-\sqrt{5}} \times \frac{3+\sqrt{5}}{3-\sqrt{5}}$ $=\frac{2(3+\sqrt{5})}{4}=\frac{3+\sqrt{5}}{2}$ Now, $\cos ^{3} \theta+\sec ^{3} \theta=(\cos \theta+\sec \theta)$ $\quad\left(\cos ^{2} \theta+\sec ^{2} \theta-\cos \theta \cdot \sec \theta\right)$ $\begin{array}{l} =\left(\frac{3-\sqrt{5}+3+\sqrt{5}}{2}\right) \\ \quad \quad\left\{(\cos \theta+\sec \theta)^{2}-3 \cos \theta \cdot \sec \theta\right\} \\ =3 \cdot\left\{(3)^{2}-3\right\}=3(9-3)=3 \times 6=18 \end{array}$

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

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