Question

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

• A. $1$
• B. $4$
• C. $9$
• D. $18$

Answer 1

The correct option is D $18$

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