Question:
$\operatorname{cosec} 18^{\circ}$ is a root of the equation:
Correct Option: , 4
Solution:
$\operatorname{cosec} 18^{\circ}=\frac{1}{\sin 18^{\circ}}=\frac{4}{\sqrt{5}-1}=\sqrt{5}+1$
Let $\operatorname{cosec} 18^{\circ}=x=\sqrt{5}+1$
$\Rightarrow x-1=\sqrt{5}$
Squaring both sides, we get
$x^{2}-2 x+1=5$
$\Rightarrow x^{2}-2 x-4=0$
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