Question

The number of solutions to ${\sec }^2\theta +{\text{cosec}}^2\theta +2{\text{cosec}}^2\theta =8,0\le \theta \le \pi /2$, is

• A. 4
• B. 3
• C. 0
• D. 2

Answer 1

Answer: (D)

We have $\frac{1}{{\sin }^2\theta {\cos }^2\theta }+\frac{2}{{\sin }^2\theta }=8,\sin \theta \ne 0,\cos \theta \ne 0$
$\Rightarrow 1+2{\cos }^2\theta =8{\sin }^2\theta {\cos }^2\theta$
$\Rightarrow 1+2{\cos }^2\theta =8{\cos }^2\left(1-{\cos }^2\theta \right)$
$\Rightarrow 8{\cos }^4\theta -6{\cos }^2\theta +1=0$
$\Rightarrow \left(4{\cos }^2\theta -1\right)\left(2{\cos }^2\theta -1\right)=0$
$\Rightarrow {\cos }^2\theta =1/4={\cos }^2(\pi /3)$
or ${\cos }^2\theta =1/2={\cos }^2(\pi /4)$
$\Rightarrow \theta =n\pi \pm \pi /3$
or $\theta =n\pi \pm \pi /4,n\in Z$
Hence, for $0\le \theta \le \pi /2,\theta =\pi /3,\theta =\pi /4$