cosec 18° = is a root of the equation:


$\operatorname{cosec} 18^{\circ}$ is a root of the equation:


  1. $x^{2}+2 x-4=0$

  2. $4 x^{2}+2 x-1=0$

  3. $x^{2}-2 x+4=0$

  4. $x^{2}-2 x-4=0$

Correct Option: , 4


$\operatorname{cosec} 18^{\circ}=\frac{1}{\sin 18^{\circ}}=\frac{4}{\sqrt{5}-1}=\sqrt{5}+1$

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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