Question

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

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

Answer 1

The correct option is C $x^2-2x-4=0$

We know that
$\text{cosec}{18}^{\circ }=\frac{4}{\sqrt{5}-1}\Rightarrow \text{cosec}{18}^{\circ }=\frac{4}{\sqrt{5}-1}\times \frac{\sqrt{5}+1}{\sqrt{5}+1}\Rightarrow \text{cosec}{18}^{\circ }=1+\sqrt{5}$
In all the given options, the coefficients of the quadratic are rational, so the roots will occur in conjugate pair,
Other root is $=1-\sqrt{5}$
Sum of root $=2$
Product of roots $=(1+\sqrt{5})(1-\sqrt{5})=1-5=-4$
$\therefore$ Required equation is
​​​​​​​$x^2-2x-4=0$