If A,B,C are angles of a triangle ,then the minimum value of `tan^(2)(A/2)+tan^(2)(B/2)+tan^(2)(C/2)` , is

Options

A. 0
B. 1
C. `1//2`
D. none of these

Answer

To find the minimum value of \( \tan^2\left(\frac{A}{2}\right) + \tan^2\left(\frac{B}{2}\right) + \tan^2\left(\frac{C}{2}\right) \) where \( A, B, C \) are the angles of a triangle, we can follow these steps:

### Step 1: Use the identity for angles of a triangle
Since \( A + B + C = 180^\circ \), we can express the angles in terms of each other.

### Step 2: Apply the tangent half-angle formula
We know that:
\[ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{1 - \cos A}{1 + \cos A}} \]
Similarly for \( B \) and \( C \).

### Step 3: Use the Cauchy-Schwarz inequality
We can apply the Cauchy-Schwarz inequality:
\[ (\tan^2\left(\frac{A}{2}\right) + \tan^2\left(\frac{B}{2}\right) + \tan^2\left(\frac{C}{2}\right))(1 + 1 + 1) \geq (\tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right) + \tan\left(\frac{C}{2}\right))^2 \]
This simplifies to:
\[ 3\left(\tan^2\left(\frac{A}{2}\right) + \tan^2\left(\frac{B}{2}\right) + \tan^2\left(\frac{C}{2}\right)\right) \geq \left(\tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right) + \tan\left(\frac{C}{2}\right)\right)^2 \]

### Step 4: Analyze the equality condition
The equality in Cauchy-Schwarz holds when:
\[ \tan\left(\frac{A}{2}\right) = \tan\left(\frac{B}{2}\right) = \tan\left(\frac{C}{2}\right) \]
This occurs when \( A = B = C = 60^\circ \).

### Step 5: Calculate the minimum value
If \( A = B = C = 60^\circ \):
\[ \tan\left(\frac{60^\circ}{2}\right) = \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
Thus:
\[ \tan^2\left(\frac{60^\circ}{2}\right) = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \]
So:
\[ \tan^2\left(\frac{A}{2}\right) + \tan^2\left(\frac{B}{2}\right) + \tan^2\left(\frac{C}{2}\right) = 3 \cdot \frac{1}{3} = 1 \]

### Conclusion
Therefore, the minimum value of \( \tan^2\left(\frac{A}{2}\right) + \tan^2\left(\frac{B}{2}\right) + \tan^2\left(\frac{C}{2}\right) \) is:
\[ \boxed{1} \]