Question

The number of solution(s) of ${\sec }^2\theta {\text{cosec}}^2\theta +2{\text{cosec}}^2\theta =8$ for $\theta \in [0,2\pi ]$ is

• A. $0$
• B. $2$
• C. $4$
• D. $8$

Answer 1

The correct option is D $8$

${\sec }^2\theta {\text{cosec}}^2\theta +2{\text{cosec}}^2\theta =8$
$\Rightarrow \frac{1}{{\sin }^2\theta {\cos }^2\theta }+\frac{2}{{\sin }^2\theta }=8\Rightarrow 1+2{\cos }^2\theta =8{\sin }^2\theta {\cos }^2\theta (\because \sin \theta \ne 0,\cos \theta \ne 0)\Rightarrow 1+2{\cos }^2\theta =8{\cos }^2\theta (1-{\cos }^2\theta )$
$\Rightarrow 8{\cos }^4\theta -6{\cos }^2\theta +1=0$
$\Rightarrow (4{\cos }^2\theta -1)(2{\cos }^2\theta -1)=0$
$\Rightarrow {\cos }^2\theta =\frac{1}{4}$ or ${\cos }^2\theta =\frac{1}{2}$
$\Rightarrow \cos \theta =\pm \frac{1}{2},\pm \frac{1}{\sqrt{2}}$

Hence, the number of solutions is $8$.