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 in a triangle
Since \( A + B + C = 180^\circ \), we can express one angle in terms of the others. For example, we can write:
\[ C = 180^\circ - A - B \]

### Step 2: Apply the tangent half-angle formula
Using the tangent half-angle formula, we have:
\[ \tan\left(\frac{C}{2}\right) = \tan\left(\frac{180^\circ - A - B}{2}\right) = \cot\left(\frac{A + B}{2}\right) \]

### Step 3: Use the cotangent identity
The cotangent can be expressed in terms of tangent:
\[ \cot\left(\frac{A + B}{2}\right) = \frac{1}{\tan\left(\frac{A + B}{2}\right)} \]

### Step 4: Establish a relationship
Using the identity \( \tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right) + \tan\left(\frac{C}{2}\right) = \tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right) + \cot\left(\frac{A + B}{2}\right) \), we can derive:
\[ \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right) + \tan\left(\frac{B}{2}\right) \tan\left(\frac{C}{2}\right) + \tan\left(\frac{C}{2}\right) \tan\left(\frac{A}{2}\right) = 1 \]

### Step 5: Use the Cauchy-Schwarz inequality
Applying 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:
\[ 3(\tan^2\left(\frac{A}{2}\right) + \tan^2\left(\frac{B}{2}\right) + \tan^2\left(\frac{C}{2}\right)) \geq 1 \]

### Step 6: Solve for the minimum value
From the above inequality, we can conclude:
\[ \tan^2\left(\frac{A}{2}\right) + \tan^2\left(\frac{B}{2}\right) + \tan^2\left(\frac{C}{2}\right) \geq \frac{1}{3} \]
However, since we are looking for the minimum value of the sum of squares, we can find that the minimum occurs when \( A = B = C = 60^\circ \):
\[ \tan^2\left(\frac{60^\circ}{2}\right) = \tan^2(30^\circ) = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \]
Thus, the minimum value is:
\[ 3 \times \frac{1}{3} = 1 \]

### Final Answer
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 \( \boxed{1} \).