If `sinalpha,sinbeta` and `cosalpha` are in G.P then roots of `x^2+2xcotbeta+1=0` are always

Options

A. Real and distinct
B. both positive
C. Real and equal
D. Imaginary

Explanation

To solve the problem, we need to analyze the given condition that \( \sin \alpha, \sin \beta, \) and \( \cos \alpha \) are in geometric progression (G.P.). We will then find the nature of the roots of the quadratic equation \( x^2 + 2x \cot \beta + 1 = 0 \).

### Step-by-Step Solution:

1. **Understanding the G.P. Condition:**
Since \( \sin \alpha, \sin \beta, \) and \( \cos \alpha \) are in G.P., we can use the property of G.P. which states that the square of the middle term is equal to the product of the other two terms. Thus, we have:
\[ \sin^2 \beta = \sin \alpha \cdot \cos \alpha \]

2. **Setting Up the Quadratic Equation:**
The given quadratic equation is:
\[ x^2 + 2x \cot \beta + 1 = 0 \]
Here, we can identify \( a = 1 \), \( b = 2 \cot \beta \), and \( c = 1 \).

3. **Finding the Discriminant:**
The nature of the roots of a quadratic equation can be determined using the discriminant \( D \), which is given by:
\[ D = b^2 - 4ac \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ D = (2 \cot \beta)^2 - 4 \cdot 1 \cdot 1 \]
Simplifying this gives:
\[ D = 4 \cot^2 \beta - 4 \]
Factoring out the common term:
\[ D = 4(\cot^2 \beta - 1) \]

4. **Using the Identity for Cotangent:**
We know that:
\[ \cot^2 \beta - 1 = \frac{\cos^2 \beta - \sin^2 \beta}{\sin^2 \beta} \]
Therefore, we can rewrite \( D \) as:
\[ D = 4 \left( \frac{\cos^2 \beta - \sin^2 \beta}{\sin^2 \beta} \right) \]

5. **Analyzing the Sign of the Discriminant:**
Since \( \sin^2 \beta = \sin \alpha \cdot \cos \alpha \) (from the G.P. condition), it is always positive (as both sine and cosine are positive in the first quadrant). Thus, \( D \) can be expressed as:
\[ D = 4 \left( \frac{\cos^2 \beta - \sin^2 \beta}{\sin^2 \beta} \right) \]
Since \( \cos^2 \beta \) can be greater than or equal to \( \sin^2 \beta \), it follows that \( D \geq 0 \).

6. **Conclusion About the Roots:**
Since \( D \geq 0 \), the roots of the quadratic equation are always real. Moreover, since \( \cot^2 \beta \) can be greater than 1, \( D \) can also be greater than 0, indicating that the roots are distinct.

### Final Answer:
The roots of the equation \( x^2 + 2x \cot \beta + 1 = 0 \) are always **real and distinct**.

---