The number of real roots of the equation `cosec theta + sec theta-sqrt(15)=0` lying in `[0, pi]` is

Options

A. 6
B. 8
C. 4
D. 0

Explanation

To find the number of real roots of the equation \( \csc \theta + \sec \theta - \sqrt{15} = 0 \) in the interval \( [0, \pi] \), we will follow these steps:

### Step 1: Rewrite the Equation
We start by rewriting the equation in terms of sine and cosine:
\[ \csc \theta = \frac{1}{\sin \theta} \quad \text{and} \quad \sec \theta = \frac{1}{\cos \theta} \]
Thus, the equation becomes:
\[ \frac{1}{\sin \theta} + \frac{1}{\cos \theta} - \sqrt{15} = 0 \]

### Step 2: Combine the Fractions
We can combine the fractions on the left-hand side:
\[ \frac{\cos \theta + \sin \theta}{\sin \theta \cos \theta} = \sqrt{15} \]
This leads to:
\[ \cos \theta + \sin \theta = \sqrt{15} \sin \theta \cos \theta \]

### Step 3: Square Both Sides
Next, we square both sides to eliminate the square root:
\[ (\cos \theta + \sin \theta)^2 = 15 \sin^2 \theta \cos^2 \theta \]
Expanding the left side gives:
\[ \cos^2 \theta + 2\sin \theta \cos \theta + \sin^2 \theta = 15 \sin^2 \theta \cos^2 \theta \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ 1 + 2\sin \theta \cos \theta = 15 \sin^2 \theta \cos^2 \theta \]

### Step 4: Substitute for \( \sin 2\theta \)
Recall that \( \sin 2\theta = 2\sin \theta \cos \theta \):
\[ 1 + \sin 2\theta = \frac{15}{4} \sin^2 2\theta \]
Rearranging gives:
\[ \frac{15}{4} \sin^2 2\theta - \sin 2\theta - 1 = 0 \]

### Step 5: Solve the Quadratic Equation
Let \( x = \sin 2\theta \). The equation becomes:
\[ \frac{15}{4}x^2 - x - 1 = 0 \]
Multiplying through by 4 to eliminate the fraction:
\[ 15x^2 - 4x - 4 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 15 \cdot (-4)}}{2 \cdot 15} \]
Calculating the discriminant:
\[ = \frac{4 \pm \sqrt{16 + 240}}{30} = \frac{4 \pm \sqrt{256}}{30} = \frac{4 \pm 16}{30} \]
This gives us two solutions:
\[ x_1 = \frac{20}{30} = \frac{2}{3}, \quad x_2 = \frac{-12}{30} = -\frac{2}{5} \]

### Step 6: Determine Valid Solutions for \( \sin 2\theta \)
The valid solution for \( \sin 2\theta \) is \( \frac{2}{3} \) since \( \sin 2\theta \) must be between -1 and 1. The solution \( -\frac{2}{5} \) is also valid but will not contribute to the count of roots in the interval.

### Step 7: Find \( \theta \) Values
For \( \sin 2\theta = \frac{2}{3} \), we find \( 2\theta \):
\[ 2\theta = \arcsin\left(\frac{2}{3}\right) \quad \text{and} \quad 2\theta = \pi - \arcsin\left(\frac{2}{3}\right) \]
Thus,
\[ \theta = \frac{1}{2} \arcsin\left(\frac{2}{3}\right) \quad \text{and} \quad \theta = \frac{\pi}{2} - \frac{1}{2} \arcsin\left(\frac{2}{3}\right) \]
Both of these solutions are valid in the interval \( [0, \pi] \).

### Step 8: Count the Roots
Since \( \sin 2\theta = -\frac{2}{5} \) also gives us two solutions in the interval \( [0, 2\pi] \), we can conclude that:
- For \( \sin 2\theta = \frac{2}{3} \), we have 2 solutions.
- For \( \sin 2\theta = -\frac{2}{5} \), we also have 2 solutions.

Thus, the total number of real roots in the interval \( [0, \pi] \) is:
\[ \text{Total Roots} = 4 \]

### Final Answer
The number of real roots of the equation \( \csc \theta + \sec \theta - \sqrt{15} = 0 \) lying in \( [0, \pi] \) is **4**.