Question : If $7 \sin ^2 \theta+3 \cos ^2 \theta=4,0^{\circ}<\theta<90^{\circ}$, then the value of $(\tan ^2 2 \theta+\operatorname{cosec}^2 2 \theta)$ is:
Option 1: $7$
Option 2: $\frac{15}{4}$
Option 3: $\frac{13}{3}$
Option 4: $\frac{13}{4}$
Correct Answer: $\frac{13}{3}$
Solution :
Given: The trigonometric expression is $7 \sin ^2 \theta+3 \cos ^2 \theta=4,0^{\circ}<\theta<90^{\circ}$.
Use the trigonometric identity, $\sin ^2 \theta+\cos^2\theta=1$.
$7 \sin ^2 \theta+3 \cos ^2 \theta=4$
⇒ $4 \sin ^2 \theta+3 (\cos ^2 \theta+\sin^2\theta)=4$
⇒ $4 \sin ^2 \theta+3=4$
⇒ $4 \sin ^2 \theta=1$
⇒ $\sin ^2 \theta=\frac{1}{4}$
⇒ $\sin \theta=\frac{1}{2}$
⇒ $\theta = 30^{\circ}$
The value of $(\tan ^2 2 \theta+\operatorname{cosec}^2 2 \theta)$
$=\tan ^2 60^{\circ}+\operatorname{cosec}^2 60^{\circ}$
$=3+\frac{4}{3}$
$=\frac{9+4}{3}=\frac{13}{3}$
Hence, the correct answer is $\frac{13}{3}$.
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