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 equation given:

\[ \tan^{-1}\left(\frac{b}{c+a}\right) + \tan^{-1}\left(\frac{c}{a+b}\right) = \frac{\pi}{4} \]

We will use the property of the tangent inverse function, which states:

\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \quad \text{if } xy < 1 \]

### Step 1: Apply the tangent addition formula

Let \( x = \frac{b}{c+a} \) and \( y = \frac{c}{a+b} \). According to the formula:

\[ \tan^{-1}\left(\frac{b}{c+a}\right) + \tan^{-1}\left(\frac{c}{a+b}\right) = \tan^{-1}\left(\frac{\frac{b}{c+a} + \frac{c}{a+b}}{1 - \frac{b}{c+a} \cdot \frac{c}{a+b}}\right) \]

### Step 2: Simplify the numerator

The numerator becomes:

\[ \frac{b}{c+a} + \frac{c}{a+b} = \frac{b(a+b) + c(c+a)}{(c+a)(a+b)} = \frac{ba + b^2 + c^2 + ac}{(c+a)(a+b)} \]

### Step 3: Simplify the denominator

The denominator becomes:

\[ 1 - \frac{bc}{(c+a)(a+b)} = \frac{(c+a)(a+b) - bc}{(c+a)(a+b)} = \frac{(c+a)(a+b) - bc}{(c+a)(a+b)} \]

### Step 4: Set the equation equal to \(\tan^{-1}(1)\)

Since \(\tan^{-1}(1) = \frac{\pi}{4}\), we have:

\[ \tan^{-1}\left(\frac{ba + b^2 + c^2 + ac}{(c+a)(a+b) - bc}\right) = \tan^{-1}(1) \]

This implies:

\[ \frac{ba + b^2 + c^2 + ac}{(c+a)(a+b) - bc} = 1 \]

### Step 5: Cross-multiply and simplify

Cross-multiplying gives:

\[ ba + b^2 + c^2 + ac = (c+a)(a+b) - bc \]

Expanding the right side:

\[ (c+a)(a+b) = ac + ab + bc + a^2 \]

Thus, we have:

\[ ba + b^2 + c^2 + ac = ac + ab + bc + a^2 - bc \]

### Step 6: Rearranging terms

Rearranging gives:

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

### Step 7: Recognizing the triangle condition

The equation \( b^2 + c^2 = a^2 \) indicates that triangle ABC is a right triangle (specifically, angle A is 90 degrees).

### Conclusion

Thus, the triangle ABC is a right triangle.

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