Question : If $\cos \left(2 \theta+54^{\circ}\right)=\sin \theta, 0^{\circ}<\left(2 \theta+54^{\circ}\right)<90^{\circ}$, then what is the value of $\frac{1}{\tan 5 \theta+\operatorname{cosec} \frac{5 \theta}{2}}$?
Option 1: $3\sqrt2$
Option 2: $2-\sqrt{3}$
Option 3: $2\sqrt3$
Option 4: $2+\sqrt{3}$
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Correct Answer: $2-\sqrt{3}$
Solution :
Given that $\cos \left(2 \theta+54^{\circ}\right)=\sin \theta$ and $0^{\circ}<\left(2 \theta+54^{\circ}\right)<90^{\circ}$
$⇒\cos \left(2 \theta+54^{\circ}\right)=\cos(90^{\circ}-\theta)$
Solving for $\theta$,
$⇒ \left(2 \theta+54^{\circ}\right)= (90^{\circ}-\theta)$
$⇒3 \theta=36^{\circ}$
$⇒\theta=12^{\circ}$
$⇒\tan 5 \theta=\tan \left(60^{\circ}\right)=\sqrt{3}$
$⇒\operatorname{cosec} \frac{5 \theta}{2}=\operatorname{cosec} \left(30^{\circ}\right)=2$
Now, $\frac{1}{\tan 5 \theta+\operatorname{cosec}\frac{5 \theta}{2}}$
$=\frac{1}{\sqrt{3}+2}$
$=\frac{1}{\sqrt{3}+2}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}$
$=2-\sqrt{3}$
Hence, the correct answer is $2-\sqrt{3}$.
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