Question

\(\csc 18^\circ\) 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. \(4x^2 + 2x - 1 = 0\)

Hint:

Memorizing common trigonometric values like \(\sin 18^\circ = \frac{\sqrt{5}-1}{4}\) and \(\cos 36^\circ = \frac{\sqrt{5}+1}{4}\) is very helpful in competitive exams.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We use the known value of \(\sin 18^\circ\) to determine the numerical value of \(\csc 18^\circ\) and then find which quadratic equation it satisfies.
Step 2: Detailed Explanation:
We know that \(\sin 18^\circ = \frac{\sqrt{5} - 1}{4}\).
Therefore, \(\csc 18^\circ = \frac{1}{\sin 18^\circ} = \frac{4}{\sqrt{5} - 1}\).
Rationalizing the denominator:
\[ x = \frac{4(\sqrt{5} + 1)}{(\sqrt{5} - 1)(\sqrt{5} + 1)} = \frac{4(\sqrt{5} + 1)}{5 - 1} = \frac{4(\sqrt{5} + 1)}{4} = \sqrt{5} + 1 \]
Now, isolate the square root and square:
\[ x - 1 = \sqrt{5} \implies (x - 1)^2 = 5 \]
\[ x^2 - 2x + 1 = 5 \implies x^2 - 2x - 4 = 0 \]
Step 3: Final Answer:
\(\csc 18^\circ\) is a root of \(x^2 - 2x - 4 = 0\).