(D) We know that $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$.Therefore,$\operatorname{cosec} 18^{\circ} = \frac{1}{\sin 18^{\circ}} = \frac{4}{\sqrt{5}-1

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(D) We know that $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$.Therefore,$\operatorname{cosec} 18^{\circ} = \frac{1}{\sin 18^{\circ}} = \frac{4}{\sqrt{5}-1}$.Rationalizing the denominator,we get $\operatorname{cosec} 18^{\circ} = \frac{4(\sqrt{5}+1)}{5-1} = \sqrt{5}+1$.Let $x = \sqrt{5}+1$.Then $x-1 = \sqrt{5}$.Squaring both sides,we get $(x-1)^{2} = 5$.$x^{2}-2x+1 = 5$.$x^{2}-2x-4 = 0$.

Class 11 Mathematics Trigonometrical Ratios, Functions and Identities Mix Examples-Trigonometrical Ratios, Functions and Identities

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

JEE MAIN 2021 All JEE MAIN

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

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Mix Examples-Trigonometrical Ratios, Functions and Identities

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