In a `DeltaABC`, If `(Sin^(2)A+Sin^(2)B+Sin^(2)C)/(Cos^(2)A+Cos^(2)B+Cos^(2)C)=2` then the triangle is

Options

A. A) right angled
B. B) equilateral
C. C) isosceles
D. D) obtuse angled

Answer

To solve the problem, we need to analyze the given expression and determine the type of triangle based on the provided condition.

### Step-by-Step Solution:

1. **Given Expression**:
\[ \frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C} = 2 \]

2. **Cross Multiply**:
\[ \sin^2 A + \sin^2 B + \sin^2 C = 2 (\cos^2 A + \cos^2 B + \cos^2 C) \]

3. **Use the Identity**:
We know that for any angle \(X\):
\[ \sin^2 X + \cos^2 X = 1 \]
Therefore, we can express \(\sin^2 A\), \(\sin^2 B\), and \(\sin^2 C\) in terms of \(\cos^2 A\), \(\cos^2 B\), and \(\cos^2 C\):
\[ \sin^2 A = 1 - \cos^2 A, \quad \sin^2 B = 1 - \cos^2 B, \quad \sin^2 C = 1 - \cos^2 C \]

4. **Substituting the Identities**:
Substitute these into the equation:
\[ (1 - \cos^2 A) + (1 - \cos^2 B) + (1 - \cos^2 C) = 2 (\cos^2 A + \cos^2 B + \cos^2 C) \]
Simplifying this gives:
\[ 3 - (\cos^2 A + \cos^2 B + \cos^2 C) = 2 (\cos^2 A + \cos^2 B + \cos^2 C) \]

5. **Rearranging the Equation**:
Combine like terms:
\[ 3 = 3 (\cos^2 A + \cos^2 B + \cos^2 C) \]
Dividing both sides by 3:
\[ 1 = \cos^2 A + \cos^2 B + \cos^2 C \]

6. **Using the Cosine Identity**:
We can express \(\cos^2 A + \cos^2 B + \cos^2 C\) using the cosine double angle identity:
\[ \cos^2 A = \frac{1 + \cos 2A}{2}, \quad \cos^2 B = \frac{1 + \cos 2B}{2}, \quad \cos^2 C = \frac{1 + \cos 2C}{2} \]
Substituting these into the equation gives:
\[ \frac{3 + \cos 2A + \cos 2B + \cos 2C}{2} = 1 \]

7. **Solving for Cosine Terms**:
Multiply through by 2:
\[ 3 + \cos 2A + \cos 2B + \cos 2C = 2 \]
Rearranging gives:
\[ \cos 2A + \cos 2B + \cos 2C = -1 \]

8. **Analyzing the Angles**:
Since \(A + B + C = \pi\) in any triangle, we can express \(C\) as \(C = \pi - A - B\). Therefore, we can analyze the angles:
\[ \cos 2C = \cos(2(\pi - A - B)) = -\cos(2A + 2B) \]
This leads us to conclude that the angles must satisfy certain relationships.

9. **Conclusion**:
Given that one of the angles must be \(90^\circ\) (as derived from the cosine relationships), we conclude that triangle \(ABC\) is a right triangle.

### Final Answer:
The triangle is a **right triangle**.