If `tan^(-1) . b/(c+a) + tan^(-1) . (c)/(a + b) = pi/4`where a, b, c , are the sides of `Delta ABC",then" Delta ABC` is

Options

A. Acute - angled triangle
B. Obtuse - angled triangle
C. Right- angled triangle
D. Equilateral triangle

Answer

To solve the problem, we need to analyze the given equation involving inverse tangent functions and determine the type of triangle formed by the sides \(a\), \(b\), and \(c\).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ \tan^{-1}\left(\frac{b}{c+a}\right) + \tan^{-1}\left(\frac{c}{a+b}\right) = \frac{\pi}{4} \]

2. **Use the formula for the sum of inverse tangents:**
The formula states that:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \]
Here, let \(x = \frac{b}{c+a}\) and \(y = \frac{c}{a+b}\).

3. **Apply the formula:**
\[ \tan^{-1}\left(\frac{\frac{b}{c+a} + \frac{c}{a+b}}{1 - \frac{b}{c+a} \cdot \frac{c}{a+b}}\right) = \frac{\pi}{4} \]

4. **Set the argument of the tangent inverse equal to 1:**
Since \(\tan\left(\frac{\pi}{4}\right) = 1\), we have:
\[ \frac{\frac{b}{c+a} + \frac{c}{a+b}}{1 - \frac{bc}{(c+a)(a+b)}} = 1 \]

5. **Cross-multiply to eliminate the fraction:**
\[ \frac{b}{c+a} + \frac{c}{a+b} = 1 - \frac{bc}{(c+a)(a+b)} \]

6. **Multiply through by \((c+a)(a+b)\) to clear the denominators:**
\[ b(a+b) + c(c+a) = (c+a)(a+b) - bc \]

7. **Expand both sides:**
- Left side: \(ba + b^2 + c^2 + ac\)
- Right side: \(ca + ab + a^2 + ac - bc\)

8. **Set the equation:**
\[ ba + b^2 + c^2 + ac = ca + ab + a^2 + ac - bc \]

9. **Cancel out common terms:**
\[ b^2 + c^2 = a^2 + bc \]

10. **Rearranging gives us:**
\[ a^2 = b^2 + c^2 \]

11. **Conclusion:**
The equation \(a^2 = b^2 + c^2\) follows the Pythagorean theorem, which indicates that triangle \(ABC\) is a right-angled triangle.

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