Question

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

• A. $x^{2}+2 x-4=0$
• B. $4 x^{2}+2 x-1=0$
• C. $x^{2}-2 x+4=0$
• D. $x^{2}-2 x-4=0$

Answer 1

Answer: (D)

$\text{cosec} 18^{\circ}=\frac{1}{\sin 18^{\circ}}=\frac{4}{\sqrt{5}-1}=\sqrt{5}+1$
Let $\text{cosec} 18^{\circ}=x=\sqrt{5}+1$
$\Rightarrow x-1=\sqrt{5}$
Squaring both sides, we get
$x^{2}-2 x+1=5 $
$\Rightarrow x^{2}-2 x-4=0$