In a triangle ABC if `sum tan^(2)A= sum tan A tan B`, then largest angle of the triangle in radian will be :

Options

A. `(2pi)/(3)`
B. `(pi)/(3)`
C. `(pi)/(2)`
D. `(3pi)/(4)`

Answer

To solve the problem, we start with the given equation in triangle ABC:

\[ \sum \tan^2 A = \sum \tan A \tan B \]

This can be expanded as:

\[ \tan^2 A + \tan^2 B + \tan^2 C = \tan A \tan B + \tan B \tan C + \tan C \tan A \]

### Step 1: Rearranging the Equation
Rearranging the equation gives us:

\[ \tan^2 A + \tan^2 B + \tan^2 C - (\tan A \tan B + \tan B \tan C + \tan C \tan A) = 0 \]

### Step 2: Recognizing the Condition
This condition is satisfied if all angles \(A\), \(B\), and \(C\) are equal. In a triangle, if all angles are equal, then:

\[ A = B = C \]

### Step 3: Finding the Angles
Since the sum of angles in a triangle is \(180^\circ\), we have:

\[ 3A = 180^\circ \implies A = 60^\circ \]

### Step 4: Converting to Radians
To find the largest angle in radians, we convert \(60^\circ\) to radians:

\[ A = 60^\circ = \frac{60 \times \pi}{180} = \frac{\pi}{3} \]

### Conclusion
Thus, the largest angle of the triangle in radians is:

\[ \frac{\pi}{3} \]