If a cosA=b cosB, then `triangleABC` is

Options

A. Isosceles
B. Right angled
C. Equilateral
D. Either Right angled or Isosceles

Answer

To solve the problem, we start with the given equation \( A \cos A = B \cos B \) and analyze the implications for triangle \( ABC \).

### Step-by-Step Solution:

1. **Given Relation**:
We start with the equation:
\[ A \cos A = B \cos B \]

2. **Using the Sine Rule**:
According to the sine rule in a triangle:
\[ \frac{A}{\sin A} = \frac{B}{\sin B} \]
From this, we can derive:
\[ A \sin B = B \sin A \quad \text{(Equation 1)} \]

3. **Dividing the Two Equations**:
Now, we divide the two equations:
\[ \frac{A \cos A}{A \sin B} = \frac{B \cos B}{B \sin A} \]
This simplifies to:
\[ \frac{\cos A}{\sin B} = \frac{\cos B}{\sin A} \]

4. **Cross Multiplying**:
Cross multiplying gives us:
\[ \sin A \cos A = \sin B \cos B \]

5. **Using Double Angle Identity**:
We can use the double angle identity for sine:
\[ 2 \sin A \cos A = \sin 2A \quad \text{and} \quad 2 \sin B \cos B = \sin 2B \]
Therefore, we have:
\[ \sin 2A = \sin 2B \]

6. **Setting Up the Equation**:
This leads to:
\[ \sin 2A - \sin 2B = 0 \]
Using the sine difference identity:
\[ 2 \cos\left(\frac{2A + 2B}{2}\right) \sin\left(\frac{2A - 2B}{2}\right) = 0 \]

7. **Finding Conditions**:
This gives us two conditions:
- \( \cos(A + B) = 0 \) which implies \( A + B = 90^\circ \) (i.e., \( C = 90^\circ \)), indicating a right triangle.
- \( \sin(A - B) = 0 \) which implies \( A = B \), indicating that the triangle is isosceles.

8. **Conclusion**:
Therefore, we conclude that triangle \( ABC \) can either be:
- Right-angled (if \( C = 90^\circ \))
- Isosceles (if \( A = B \))

Thus, the correct answer is that triangle \( ABC \) is either right-angled or isosceles.

### Final Answer:
The triangle \( ABC \) is either right-angled or isosceles.

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