Question : If $0°<\theta<90°$ and $2\sec\theta =3 \operatorname{cosec}^2 \theta$, then $\theta$ is:
Option 1: $\frac{\pi}{6}$
Option 2:
$\frac{\pi}{4}$
Option 3:
$\frac{\pi}{3}$
Option 4:
$\frac{\pi}{5}$
Correct Answer:
Solution : Given: $0°<\theta<90°$ and $2\sec\theta =3 \operatorname{cosec}^2 \theta$ $2\sec\theta =3 \operatorname{cosec}^2 \theta$ $⇒\frac{2}{\cos\theta}=\frac{3}{\sin^2\theta}$ Squaring both sides, we get, $\frac{4}{\cos^2\theta}=\frac{9}{\sin^4\theta}$ $⇒4\sin^4\theta+9\sin^2\theta-9=0$ $⇒(\sin^2\theta+3)(4\sin^2\theta-3)=0$ $\therefore \sin\theta= \sqrt{-3} , \frac{\sqrt{3}}{2}$ The value of $\sin\theta$ can't be $\sqrt{-3}$. $⇒\sin\theta= \frac{\sqrt{3}}{2}=\sin60°$ $\therefore \theta=60°=\frac{\pi}{3}$ Hence, the correct answer is $\frac{\pi}{3}$.
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