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 analyze the conditions under which the triangle formed by points A, B, and C on the circle \(x^2 + y^2 = a^2\) is equilateral. The points A, B, and C can be represented in terms of their angles \(\alpha\), \(\beta\), and \(\gamma\) on the circle.

### Step-by-Step Solution:

1. **Identify the Points on the Circle**:
The points A, B, and C on the circle can be expressed in Cartesian coordinates as:
- \(A(a \cos \alpha, a \sin \alpha)\)
- \(B(a \cos \beta, a \sin \beta)\)
- \(C(a \cos \gamma, a \sin \gamma)\)

2. **Centroid of the Triangle**:
The centroid (G) of triangle ABC can be calculated using the coordinates of the vertices:
\[ G\left(\frac{a \cos \alpha + a \cos \beta + a \cos \gamma}{3}, \frac{a \sin \alpha + a \sin \beta + a \sin \gamma}{3}\right) \]
Since the triangle is equilateral and symmetric about the origin, the centroid must be at the origin (0, 0).

3. **Setting Up the Equations**:
For the centroid to be at the origin, we set the x and y coordinates of the centroid equal 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. **Simplifying the Equations**:
From the above equations, we can simplify:
\[ \cos \alpha + \cos \beta + \cos \gamma = 0 \]
\[ \sin \alpha + \sin \beta + \sin \gamma = 0 \]

5. **Conclusion**:
The conditions derived indicate that for the triangle ABC to be equilateral, the sum of the cosines of the angles must equal zero, and the sum of the sines of the angles must also equal zero.

### Final Result:
Thus, the conditions that must hold for the angles \(\alpha\), \(\beta\), and \(\gamma\) are:
\[ \cos \alpha + \cos \beta + \cos \gamma = 0 \quad \text{and} \quad \sin \alpha + \sin \beta + \sin \gamma = 0 \]