Question : If $3 \tan \theta=2 \sqrt{3} \sin \theta, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{\operatorname{cosec}^2 2 \theta+\cot ^2 2 \theta}{\sin ^2 \theta+\tan ^2 2 \theta}$ is:
Option 1: $\frac{4}{13}$
Option 2: $\frac{20}{39}$
Option 3: $\frac{4}{3}$
Option 4: $\frac{20}{27}$
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Correct Answer: $\frac{20}{39}$
Solution :
$3 \tan \theta=2 \sqrt{3} \sin \theta$
⇒ $3 × \frac{\sin \theta}{\cos \theta} = 2 \sqrt{3} \sin \theta$
⇒ $\frac{3}{\cos \theta} = 2 \sqrt{3}$
⇒ $\cos \theta = \frac{3}{2 \sqrt{3}} = \frac{\sqrt3}{2}$
⇒ $\theta = 30^{\circ}$
Now, $\frac{\operatorname{cosec}^2 2 \theta+\cot ^2 2 \theta}{\sin ^2 \theta+\tan ^2 2 \theta}$
$=\frac{\operatorname{cosec}^2 (2 ×30^{\circ})+\cot ^2 (2 ×30^{\circ})}{\sin ^2 30^{\circ}+\tan ^2 (2× 30^{\circ})}$
$=\frac{\operatorname{cosec}^2 60^{\circ}+\cot ^2 60^{\circ}}{\sin ^2 30^{\circ}+\tan ^2 60^{\circ}}$
$=\frac{(\frac{2}{\sqrt3})^2+(\frac{1}{\sqrt3})^2}{(\frac{1}{2})^2+(\sqrt3)^2}$
$=\frac{(\frac{4}{3})+(\frac{1}{3})}{(\frac{1}{4})+3}$
$=\frac{(\frac{5}{3})}{(\frac{1+12}{4})}$
$= \frac{5×4}{3×13}$
$=\frac{20}{39}$
Hence, the correct answer is $\frac{20}{39}$.
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