Question : If $\alpha \sin 45^{\circ}=\beta \operatorname{cosec} 30^{\circ}$, then $\frac{\alpha^4}{ \beta^4}$ is:
Option 1: $4^4$
Option 2: $3^3$
Option 3: $2^3$
Option 4: $4^3$
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Correct Answer: $4^3$
Solution : $\sin 45^{\circ}$ = $\frac{1}{\sqrt2}$ $\operatorname{cosec} 30^{\circ}$ = 2 $\alpha \sin 45^{\circ}=\beta \operatorname{cosec} 30^{\circ}$ ⇒ $\alpha\times{\frac{1}{\sqrt2}}=\beta\times{2}$ ⇒ $\frac{\alpha}{\beta}$ = $2\times{\sqrt{2}}$ $\therefore$ $\frac{\alpha^4}{\beta^4}=(\frac{\alpha}{\beta})^4 = (2\times{\sqrt{2}})^4 = 64 = 4^3$ Hence, the correct answer is $4^3$.
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