In triangle ABC, if `cos^(2)A + cos^(2)B - cos^(2) C = 1`, then identify the type of the triangle

Explanation

To solve the problem, we start with the given equation:

\[ \cos^2 A + \cos^2 B - \cos^2 C = 1 \]

### Step 1: Rearranging the equation
We can rearrange the equation to isolate the cosine terms:

\[ \cos^2 A + \cos^2 B = \cos^2 C + 1 \]

### Step 2: Using the Pythagorean identity
We know from the Pythagorean identity that:

\[ \sin^2 A + \cos^2 A = 1 \quad \text{and} \quad \sin^2 B + \cos^2 B = 1 \quad \text{and} \quad \sin^2 C + \cos^2 C = 1 \]

Thus, we can express \(\cos^2 A\) and \(\cos^2 B\) in terms of sine:

\[ \cos^2 A = 1 - \sin^2 A \quad \text{and} \quad \cos^2 B = 1 - \sin^2 B \]

### Step 3: Substitute into the equation
Substituting these into our rearranged equation gives:

\[ (1 - \sin^2 A) + (1 - \sin^2 B) = \cos^2 C + 1 \]

This simplifies to:

\[ 2 - \sin^2 A - \sin^2 B = \cos^2 C + 1 \]

### Step 4: Rearranging further
Rearranging this equation leads to:

\[ \sin^2 A + \sin^2 B = \cos^2 C + 1 - 2 \]

This simplifies to:

\[ \sin^2 A + \sin^2 B = \cos^2 C - 1 \]

### Step 5: Using the identity for \(\sin^2 C\)
Recall that \(\sin^2 C = 1 - \cos^2 C\). Thus, we can write:

\[ \sin^2 A + \sin^2 B = -\sin^2 C \]

### Step 6: Analyzing the equation
The equation \(\sin^2 A + \sin^2 B = \sin^2 C\) indicates that the sum of the squares of the sines of angles A and B equals the square of the sine of angle C.

### Step 7: Applying the sine rule
Using the sine rule, we know that:

\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]

From our earlier equation, we can derive that:

\[ a^2 + b^2 = c^2 \]

### Conclusion
This is the Pythagorean theorem, which indicates that triangle ABC is a right-angled triangle.

### Final Answer
Thus, triangle ABC is a right-angled triangle.

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