If `alpha ,beta, gamma` are the parameters of points A,B,C on the circle `x^2+y^2=a^2` and if the triangle ABC is equilateral ,then

Options

A. `sum cos alpha =0`
B. `sum sin alpha =0`
C. `sum tan alpha =0`
D. `sum cot alpha =0`

Answer

To solve the problem, we need to show that if points A, B, and C on the circle \( x^2 + y^2 = a^2 \) are the vertices of an equilateral triangle, then the following conditions hold:

1. \( \cos \alpha + \cos \beta + \cos \gamma = 0 \)
2. \( \sin \alpha + \sin \beta + \sin \gamma = 0 \)

### Step-by-Step Solution

1. **Identify the Circle and Points**:
The equation of the circle is \( x^2 + y^2 = a^2 \). The points A, B, and C on the circle can be represented using the parameters \( \alpha, \beta, \gamma \):
- Point A: \( (a \cos \alpha, a \sin \alpha) \)
- Point B: \( (a \cos \beta, a \sin \beta) \)
- Point C: \( (a \cos \gamma, a \sin \gamma) \)

2. **Calculate the Centroid of Triangle ABC**:
The centroid (G) of triangle ABC can be calculated using the formula:
\[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]
Substituting the coordinates of points A, B, and C:
\[ G = \left( \frac{a \cos \alpha + a \cos \beta + a \cos \gamma}{3}, \frac{a \sin \alpha + a \sin \beta + a \sin \gamma}{3} \right) \]

3. **Set the Centroid to the Origin**:
Since the triangle is equilateral and symmetric about the origin, the centroid must be at the origin \( (0, 0) \). Therefore, we set both coordinates of G to zero:
\[ \frac{a \cos \alpha + a \cos \beta + a \cos \gamma}{3} = 0 \]
\[ \frac{a \sin \alpha + a \sin \beta + a \sin \gamma}{3} = 0 \]

4. **Simplify the Equations**:
From the above equations, we can simplify:
\[ a \cos \alpha + a \cos \beta + a \cos \gamma = 0 \implies \cos \alpha + \cos \beta + \cos \gamma = 0 \]
\[ a \sin \alpha + a \sin \beta + a \sin \gamma = 0 \implies \sin \alpha + \sin \beta + \sin \gamma = 0 \]

5. **Conclusion**:
Thus, we have shown that if triangle ABC is equilateral, then:
\[ \cos \alpha + \cos \beta + \cos \gamma = 0 \]
\[ \sin \alpha + \sin \beta + \sin \gamma = 0 \]