Question

If $si{n}^2\theta +3cos\theta =2$, then $co{s}^3\theta +se{c}^3\theta$ is

• A. 1
• B. 4
• C. 9
• D. 18

Answer 1

Answer: (D)

Given ${\sin }^2\theta +3\cos \theta =2$
$\Rightarrow 1-{\cos }^2\theta +3\cos \theta =2$
$\Rightarrow {\cos }^2\theta -3\cos \theta +1=0$
$\Rightarrow \cos \theta =\frac{3\pm \sqrt{9-4}}{2}$ (by quadratic formula)
$\Rightarrow \cos \theta =\frac{3\pm \sqrt{5}}{2}$
Here, $\cos \theta =\frac{3-\sqrt{5}}{2}$
$\left(\because \cos \theta \ne \frac{3+\sqrt{5}}{2}-1\le \cos \theta \le 1\right)$
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, ${\cos }^3\theta +{\sec }^3\theta =(\cos \theta +\sec \theta )$
$\left({\cos }^2\theta +{\sec }^2\theta -\cos \theta \cdot \sec \theta \right)$
$=\left(\frac{3-\sqrt{5}+3+\sqrt{5}}{2}\right)$
$\left\{(\cos \theta +\sec \theta {)}^2-3\cos \theta \cdot \sec \theta \right\}$
$=3\cdot \left\{(3{)}^2-3\right\}=3(9-3)=3\times 6=18$