In a triangle ABc , if `r^2cot(A/2) cot(B/2) cot(C/2)=`

Options

A. acute angled
B. obtuse angled
C. right angled
D. none of these

Explanation

To solve the problem, we need to find the value of \( r^2 \cot\left(\frac{A}{2}\right) \cot\left(\frac{B}{2}\right) \cot\left(\frac{C}{2}\right) \) in triangle ABC. Let's break it down step by step.

### Step 1: Write the formulas for cotangent of half angles
In a triangle, the cotangent of half angles can be expressed as:
\[ \cot\left(\frac{A}{2}\right) = \frac{s(s-a)}{\Delta}, \quad \cot\left(\frac{B}{2}\right) = \frac{s(s-b)}{\Delta}, \quad \cot\left(\frac{C}{2}\right) = \frac{s(s-c)}{\Delta} \]
where \( s \) is the semi-perimeter of the triangle, \( s = \frac{a+b+c}{2} \), and \( \Delta \) is the area of the triangle.

### Step 2: Substitute the cotangent values into the expression
Now we substitute these values into the expression:
\[ r^2 \cot\left(\frac{A}{2}\right) \cot\left(\frac{B}{2}\right) \cot\left(\frac{C}{2}\right) = r^2 \left(\frac{s(s-a)}{\Delta}\right) \left(\frac{s(s-b)}{\Delta}\right) \left(\frac{s(s-c)}{\Delta}\right) \]

### Step 3: Simplify the expression
This simplifies to:
\[ = r^2 \cdot \frac{s^3 (s-a)(s-b)(s-c)}{\Delta^3} \]

### Step 4: Use the formula for the area of the triangle
We know that:
\[ s(s-a)(s-b)(s-c) = \Delta^2 \]
Thus,
\[ r^2 \cdot \frac{s^3 \cdot \Delta^2}{\Delta^3} = r^2 \cdot \frac{s^3}{\Delta} \]

### Step 5: Substitute the value of \( r \)
The circumradius \( R \) is given by the formula:
\[ R = \frac{\Delta}{s} \]
So,
\[ R^2 = \frac{\Delta^2}{s^2} \]
Substituting this into our expression gives:
\[ = \frac{\Delta^2}{s^2} \cdot \frac{s^3}{\Delta} = \frac{s \Delta}{1} \]

### Step 6: Final simplification
Thus, we find that:
\[ r^2 \cot\left(\frac{A}{2}\right) \cot\left(\frac{B}{2}\right) \cot\left(\frac{C}{2}\right) = \Delta \]

### Conclusion
The final value is:
\[ \boxed{\Delta} \]