The value of `cosec 18^@` is the root of which of the following quadratic equation

Options

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

Answer

To find the quadratic equation for which \( \csc 18^\circ \) is a root, we can follow these steps:

### Step 1: Find the value of \( \sin 18^\circ \)
We know that:
\[ \sin 18^\circ = \frac{\sqrt{5} - 1}{4} \]

### Step 2: Calculate \( \csc 18^\circ \)
Since \( \csc \theta = \frac{1}{\sin \theta} \), we can find \( \csc 18^\circ \) as follows:
\[ \csc 18^\circ = \frac{1}{\sin 18^\circ} = \frac{1}{\frac{\sqrt{5} - 1}{4}} = \frac{4}{\sqrt{5} - 1} \]

### Step 3: Rationalize the denominator
To rationalize the denominator, we multiply the numerator and denominator by \( \sqrt{5} + 1 \):
\[ \csc 18^\circ = \frac{4(\sqrt{5} + 1)}{(\sqrt{5} - 1)(\sqrt{5} + 1)} = \frac{4(\sqrt{5} + 1)}{5 - 1} = \frac{4(\sqrt{5} + 1)}{4} = \sqrt{5} + 1 \]

### Step 4: Set up the equation
Let \( x = \csc 18^\circ \). Therefore, we have:
\[ x = \sqrt{5} + 1 \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ x - 1 = \sqrt{5} \]

### Step 6: Square both sides
Squaring both sides results in:
\[ (x - 1)^2 = (\sqrt{5})^2 \]
Expanding the left side:
\[ x^2 - 2x + 1 = 5 \]

### Step 7: Form the quadratic equation
Rearranging gives:
\[ x^2 - 2x + 1 - 5 = 0 \implies x^2 - 2x - 4 = 0 \]

### Final Answer
Thus, the quadratic equation for which \( \csc 18^\circ \) is a root is:
\[ x^2 - 2x - 4 = 0 \]