In any `Delta ABC,` prove the following : `r_1=rcot(B/2)cot(C/2)`

Answer

To prove that in any triangle \( ABC \), the relationship \( r_1 = r \cot \left( \frac{B}{2} \right) \cot \left( \frac{C}{2} \right) \) holds, we will start with the right-hand side (RHS) and manipulate it to show that it equals the left-hand side (LHS).

### Step-by-Step Solution:

1. **Identify the Right-Hand Side (RHS)**:
\[ \text{RHS} = r \cot \left( \frac{B}{2} \right) \cot \left( \frac{C}{2} \right) \]

2. **Use the Cotangent Half-Angle Formulas**:
The cotangent of half angles can be expressed in terms of the semi-perimeter \( s \) and the sides of the triangle:
\[ \cot \left( \frac{B}{2} \right) = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}} \]
\[ \cot \left( \frac{C}{2} \right) = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}} \]

3. **Substitute the Cotangent Formulas into the RHS**:
Now substitute these formulas into the RHS:
\[ \text{RHS} = r \cdot \sqrt{\frac{s(s-b)}{(s-a)(s-c)}} \cdot \sqrt{\frac{s(s-c)}{(s-a)(s-b)}} \]

4. **Combine the Square Roots**:
Combine the square roots:
\[ \text{RHS} = r \cdot \sqrt{\frac{s^2(s-b)(s-c)}{(s-a)^2(s-b)(s-c)}} \]

5. **Simplify the Expression**:
Notice that \( (s-b) \) and \( (s-c) \) cancel out:
\[ \text{RHS} = r \cdot \sqrt{\frac{s^2}{(s-a)^2}} = r \cdot \frac{s}{s-a} \]

6. **Recall the Definition of \( r \)**:
The circumradius \( r \) is defined as:
\[ r = \frac{\Delta}{s} \]
where \( \Delta \) is the area of triangle \( ABC \).

7. **Substitute \( r \) into the Expression**:
Substitute \( r \) into the simplified RHS:
\[ \text{RHS} = \frac{\Delta}{s} \cdot \frac{s}{s-a} = \frac{\Delta}{s-a} \]

8. **Recognize the Definition of \( r_1 \)**:
The inradius \( r_1 \) is defined as:
\[ r_1 = \frac{\Delta}{s-a} \]

9. **Conclude the Proof**:
Thus, we have shown that:
\[ \text{RHS} = r_1 \]
Therefore, we conclude that:
\[ r_1 = r \cot \left( \frac{B}{2} \right) \cot \left( \frac{C}{2} \right) \]
Hence proved.