Question

If $sin^2\, θ + 3cosθ = 2$, then $cos^3\, θ + sec^3\, θ$ 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 \neq \frac{3+\sqrt{5}}{2}-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) $
$\left(\cos ^{2} \theta+\sec ^{2} \theta-\cos \theta \cdot \sec \theta\right)$
$=\left(\frac{3-\sqrt{5}+3+\sqrt{5}}{2}\right)$
$\left\{(\cos \theta+\sec \theta)^{2}-3 \cos \theta \cdot \sec \theta\right\}$
$=3 \cdot\left\{(3)^{2}-3\right\}=3(9-3)=3 \times 6=18$