Value of `tan^(-1) ""a/(b+c)+tan^(-1)"" b/(c+a), " if " angleC=90^@ " in " DeltaABC=`

Options

A. `pi/4`
B. `pi/3`
C. `pi/2`
D. `pi`

Answer

To find the value of \( \tan^{-1} \left( \frac{a}{b+c} \right) + \tan^{-1} \left( \frac{b}{c+a} \right) \) given that \( \angle C = 90^\circ \) in triangle ABC, we can use the properties of right triangles and the tangent addition formula.

### Step-by-Step Solution:

1. **Understanding the Triangle Configuration**:
Since \( \angle C = 90^\circ \), we can label the sides of the triangle as follows:
- Let \( a \) be the length of side opposite angle A.
- Let \( b \) be the length of side opposite angle B.
- Let \( c \) be the length of the hypotenuse (the side opposite the right angle).

2. **Using the Tangent Function**:
In a right triangle, the tangent of an angle is the ratio of the opposite side to the adjacent side. Therefore:
- \( \tan A = \frac{a}{b} \)
- \( \tan B = \frac{b}{a} \)

3. **Applying the Tangent Addition Formula**:
We can use the formula for the sum of two inverse tangents:
\[ \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right) \quad \text{if } xy < 1 \]
Here, let \( x = \frac{a}{b+c} \) and \( y = \frac{b}{c+a} \).

4. **Finding the Sum**:
We need to calculate:
\[ \tan^{-1} \left( \frac{a}{b+c} \right) + \tan^{-1} \left( \frac{b}{c+a} \right) = \tan^{-1} \left( \frac{\frac{a}{b+c} + \frac{b}{c+a}}{1 - \frac{a}{b+c} \cdot \frac{b}{c+a}} \right) \]

5. **Calculating the Numerator**:
The numerator becomes:
\[ \frac{a(c+a) + b(b+c)}{(b+c)(c+a)} = \frac{ac + a^2 + b^2 + bc}{(b+c)(c+a)} \]

6. **Calculating the Denominator**:
The denominator becomes:
\[ 1 - \frac{ab}{(b+c)(c+a)} = \frac{(b+c)(c+a) - ab}{(b+c)(c+a)} = \frac{bc + ac + c^2}{(b+c)(c+a)} \]

7. **Final Expression**:
Thus, we have:
\[ \tan^{-1} \left( \frac{ac + a^2 + b^2 + bc}{bc + ac + c^2} \right) \]

8. **Using Pythagorean Theorem**:
Since \( c^2 = a^2 + b^2 \) (from the Pythagorean theorem), we can substitute \( c^2 \) into our expression.

9. **Conclusion**:
After simplification, we find that:
\[ \tan^{-1} \left( \frac{a}{b+c} \right) + \tan^{-1} \left( \frac{b}{c+a} \right) = \frac{\pi}{4} \]