Question

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

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

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
$x^2-2x+1=5$
$\Rightarrow x^2-2x-4=0$