Question

If ${\sin }^2\theta +3\cos \theta -2=0$, then ${\cos }^3\theta +{\sec }^3\theta$ is equal to

• A. 18
• B. 9
• C. 4
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

Given ${\sin }^2\theta +3\cos \theta -2=0$
$\Rightarrow {\cos }^2\theta -3\cos \theta +1=0$
$\Rightarrow {\cos }^2\theta +1=3\cos \theta$
$\Rightarrow \cos \theta +\frac{1}{\cos \theta }=3,$
cubing both sides, we get
${\cos }^3\theta +\frac{1}{{\cos }^3\theta }+3\left(\cos \theta +\frac{1}{\cos \theta }\right)=27$
$\Rightarrow {\cos }^3\theta +{\sec }^3\theta =27-3\times 3=18$