If `cos^(2)A+cos^(2)B+cos^(2)C=1`, then `triangle ABC` is

Options

A. equilateral
B. isosceles
C. right angled
D. none of these.

Explanation

To solve the problem, we need to analyze the given equation \( \cos^2 A + \cos^2 B + \cos^2 C = 1 \) and determine the type of triangle \( ABC \) represents.

### Step-by-step Solution:

1. **Start with the given equation:**
\[ \cos^2 A + \cos^2 B + \cos^2 C = 1 \]

2. **Rearranging the equation:**
Bring \( 1 \) to the left-hand side:
\[ \cos^2 A + \cos^2 B + \cos^2 C - 1 = 0 \]

3. **Using the identity for cosine:**
We can express \( \cos^2 C \) using the identity \( 1 - \cos^2 C = \sin^2 C \):
\[ \cos^2 A + \cos^2 B - (1 - \cos^2 C) = 0 \]
This simplifies to:
\[ \cos^2 A + \cos^2 B - 1 + \cos^2 C = 0 \]

4. **Rearranging again:**
\[ \cos^2 A + \cos^2 B = 1 - \cos^2 C \]

5. **Using the cosine addition formula:**
We know that \( \cos^2 A + \cos^2 B \) can be expressed using the cosine of the sum and difference:
\[ \cos^2 A + \cos^2 B = \cos(A + B) \cos(A - B) \]
However, we will focus on the triangle properties instead.

6. **Using the triangle angle sum property:**
In a triangle, we have:
\[ A + B + C = 180^\circ \quad \text{(or } \pi \text{ radians)} \]
Therefore, we can express \( C \) as:
\[ C = 180^\circ - (A + B) \]

7. **Substituting for \( C \):**
Thus, we can write:
\[ \cos^2 C = \cos^2(180^\circ - (A + B)) = \cos^2(A + B) \]
Using the identity \( \cos(180^\circ - x) = -\cos x \):
\[ \cos^2 C = \cos^2(A + B) \]

8. **Analyzing the implications:**
The equation \( \cos^2 A + \cos^2 B + \cos^2 C = 1 \) suggests that at least one of the angles must be \( 90^\circ \) for the equality to hold true. This is because the sum of squares of the cosines of angles in a triangle can only equal 1 if one angle is a right angle.

9. **Conclusion:**
Therefore, we conclude that triangle \( ABC \) must be a right-angled triangle.

### Final Answer:
Triangle \( ABC \) is a right-angled triangle.

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