Question

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

• A. 4
• B. 3
• C. 0
• D. 2

Answer 1

Answer: (D)

We have $\frac{1}{\sin ^{2} \theta \cos ^{2} \theta}+\frac{2}{\sin ^{2} \theta}=8, \sin \theta \neq 0, \cos \theta \neq 0$
$\Rightarrow 1+2 \cos ^{2} \theta=8 \sin ^{2} \theta \cos ^{2} \theta$
$\Rightarrow 1+2 \cos ^{2} \theta=8 \cos ^{2}\left(1-\cos ^{2} \theta\right)$
$\Rightarrow 8 \cos ^{4} \theta-6 \cos ^{2} \theta+1=0$
$\Rightarrow\left(4 \cos ^{2} \theta-1\right)\left(2 \cos ^{2} \theta-1\right)=0$
$\Rightarrow \cos ^{2} \theta=1 / 4=\cos ^{2}(\pi / 3)$
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 $0 \leq \theta \leq \pi / 2, \theta=\pi / 3, \theta=\pi / 4$