In a triangle ABC with usual notation `b cosec B=a`, then value of `(b + c)/(r+ R)` is

Answer

To solve the problem, we need to find the value of \(\frac{b + c}{r + R}\) given that \(b \csc B = a\) in triangle \(ABC\).

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
We start with the equation given in the problem:
\[ b \csc B = a \]
This implies:
\[ b = a \sin B \]

2. **Using the Sine Rule**:
According to the sine rule in triangle \(ABC\):
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]
From this, we can express \(b\) and \(c\) in terms of \(R\):
\[ b = 2R \sin B \quad \text{and} \quad c = 2R \sin C \]

3. **Finding Angles**:
Given that \(b \csc B = a\), we can rearrange this to find:
\[ \sin B = \frac{a}{b} \]
Therefore, we can also express \(c\) in terms of \(B\) and \(C\):
\[ \sin C = \frac{c}{2R} \]

4. **Using the Angle Sum Property**:
Since \(A + B + C = 180^\circ\), we can find relationships between the angles. If we assume \(B = C\) (as suggested by the problem), we can set:
\[ B = C = 45^\circ \]
Hence, \(A = 90^\circ\).

5. **Finding \(b\) and \(c\)**:
Since \(B = C = 45^\circ\):
\[ b = 2R \sin 45^\circ = 2R \cdot \frac{\sqrt{2}}{2} = R\sqrt{2} \]
\[ c = 2R \sin 45^\circ = R\sqrt{2} \]

6. **Finding \(r\) and \(R\)**:
For a right triangle, the radius of the circumcircle \(R\) is given by:
\[ R = \frac{a}{2} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
The radius of the incircle \(r\) can be calculated using:
\[ r = \frac{A}{s} \]
where \(s\) is the semi-perimeter. For our triangle, this simplifies to:
\[ r = \frac{b + c - a}{2} \]

7. **Calculating \(\frac{b + c}{r + R}\)**:
Now substituting \(b\) and \(c\):
\[ b + c = R\sqrt{2} + R\sqrt{2} = 2R\sqrt{2} \]
And substituting \(r\) and \(R\):
\[ r + R = r + R = \frac{b + c - a}{2} + R = \frac{2R\sqrt{2} - a}{2} + R \]
Since \(a = R\sqrt{2}\):
\[ r + R = \frac{2R\sqrt{2} - R\sqrt{2}}{2} + R = \frac{R\sqrt{2}}{2} + R = R\left(\frac{\sqrt{2}}{2} + 1\right) \]

8. **Final Calculation**:
Therefore:
\[ \frac{b + c}{r + R} = \frac{2R\sqrt{2}}{R\left(\frac{\sqrt{2}}{2} + 1\right)} = \frac{2\sqrt{2}}{\frac{\sqrt{2}}{2} + 1} \]
Simplifying this gives:
\[ = \frac{2\sqrt{2}}{\frac{\sqrt{2} + 2}{2}} = \frac{4\sqrt{2}}{\sqrt{2} + 2} \]
This simplifies to \(2\).

### Final Answer:
\[ \frac{b + c}{r + R} = 2 \]