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

### Step 1: Rewrite the equation using sine and cosine
The cosecant and secant functions can be expressed in terms of sine and cosine:
\[ \csc \theta = \frac{1}{\sin \theta}, \quad \sec \theta = \frac{1}{\cos \theta} \]
Thus, the equation becomes:
\[ \frac{1}{\sin \theta} + \frac{1}{\cos \theta} - \sqrt{15} = 0 \]

### Step 2: Combine the fractions
To combine the fractions, we find a common denominator:
\[ \frac{\cos \theta + \sin \theta}{\sin \theta \cos \theta} = \sqrt{15} \]
This leads to:
\[ \cos \theta + \sin \theta = \sqrt{15} \sin \theta \cos \theta \]

### Step 3: Square both sides
To eliminate the square root, we square both sides:
\[ (\cos \theta + \sin \theta)^2 = (\sqrt{15} \sin \theta \cos \theta)^2 \]
Expanding both sides gives:
\[ \cos^2 \theta + 2 \sin \theta \cos \theta + \sin^2 \theta = 15 \sin^2 \theta \cos^2 \theta \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we simplify this to:
\[ 1 + \sin 2\theta = 15 \sin^2 \theta \cos^2 \theta \]

### Step 4: Express \( \sin^2 \theta \cos^2 \theta \) in terms of \( \sin 2\theta \)
We know that:
\[ \sin^2 \theta \cos^2 \theta = \frac{1}{4} \sin^2 2\theta \]
Substituting this into the equation gives:
\[ 1 + \sin 2\theta = \frac{15}{4} \cdot \frac{1}{4} \sin^2 2\theta \]
This simplifies to:
\[ 1 + \sin 2\theta = \frac{15}{16} \sin^2 2\theta \]

### Step 5: Rearranging the equation
Rearranging gives us a quadratic in terms of \( \sin 2\theta \):
\[ \frac{15}{16} \sin^2 2\theta - \sin 2\theta - 1 = 0 \]
Multiplying through by 16 to eliminate the fraction:
\[ 15 \sin^2 2\theta - 16 \sin 2\theta - 16 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( \sin 2\theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = 15, \quad b = -16, \quad c = -16 \]
Calculating the discriminant:
\[ D = (-16)^2 - 4 \cdot 15 \cdot (-16) = 256 + 960 = 1216 \]
Finding the roots:
\[ \sin 2\theta = \frac{16 \pm \sqrt{1216}}{30} \]
Calculating \( \sqrt{1216} \):
\[ \sqrt{1216} = 34.8 \quad \text{(approximately)} \]
Thus, we have:
\[ \sin 2\theta = \frac{16 \pm 34.8}{30} \]
Calculating the two possible values:
1. \( \sin 2\theta = \frac{50.8}{30} \approx 1.693 \) (not valid since sine cannot exceed 1)
2. \( \sin 2\theta = \frac{-18.8}{30} \approx -0.6267 \)

### Step 7: Determine the number of solutions
For \( \sin 2\theta = -0.6267 \):
- The general solutions for \( 2\theta \) are:
\[ 2\theta = \arcsin(-0.6267) + 2k\pi \quad \text{and} \quad 2\theta = \pi - \arcsin(-0.6267) + 2k\pi \]
This gives two solutions for \( 2\theta \) in the interval \([0, 2\pi]\), which corresponds to four solutions for \( \theta \) in the interval \([0, \pi]\).

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