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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Question : $\left(\frac{\tan ^3 \theta}{\sec ^2 \theta}+\frac{\cot ^3 \theta}{\operatorname{cosec}^2 \theta}+2 \sin \theta \cos \theta\right) \div\left(1+\operatorname{cosec}^2 \theta+\tan ^2 \theta\right), 0^{\circ}<\theta<90^{\circ}$, is equal to:
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Question : $\frac{(1+\sec \theta \operatorname{cosec} \theta)^2(\sec \theta-\tan \theta)^2(1+\sin \theta)}{(\sin \theta+\sec \theta)^2+(\cos \theta+\operatorname{cosec} \theta)^2}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
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