If tan of the angles A , B , C are the solutions of the equations `tan^(3)x-3ktan^(2)x-3tanx+k=0` , then the triangle ABC is

Options

A. isosceles
B. equilateral
C. acute angled
D. none of these

Explanation

To solve the given problem, we need to analyze the cubic equation and the properties of the angles in triangle ABC.

### Step-by-Step Solution

1. **Understanding the Problem:**
We are given that the tangent of the angles A, B, and C of triangle ABC are the solutions to the equation:
\[ \tan^3 x - 3k \tan^2 x - 3 \tan x + k = 0 \]
We need to determine the type of triangle ABC based on the solutions of this equation.

2. **Using the Sum of Angles in a Triangle:**
In any triangle, the sum of the angles is:
\[ A + B + C = 180^\circ \]
From trigonometric identities, we know:
\[ \tan A + \tan B + \tan C = \tan A \tan B \tan C \]
This holds true when \( A + B + C = 180^\circ \).

3. **Identifying Coefficients of the Cubic Equation:**
The given cubic equation can be rewritten as:
\[ \tan^3 x - 3k \tan^2 x - 3 \tan x + k = 0 \]
Here, we identify the coefficients:
- \( a = 1 \) (coefficient of \( \tan^3 x \))
- \( b = -3k \) (coefficient of \( \tan^2 x \))
- \( c = -3 \) (coefficient of \( \tan x \))
- \( d = k \) (constant term)

4. **Applying Vieta's Formulas:**
According to Vieta's formulas for the roots of a cubic equation:
- The sum of the roots (solutions) is given by:
\[ \tan A + \tan B + \tan C = -\frac{b}{a} = -\frac{-3k}{1} = 3k \]
- The product of the roots is given by:
\[ \tan A \tan B \tan C = \frac{d}{a} = \frac{k}{1} = k \]

5. **Setting Up the Equation:**
From the properties of the angles in triangle ABC, we have:
\[ \tan A + \tan B + \tan C = 3k \]
and
\[ \tan A \tan B \tan C = k \]
By substituting these into the identity, we find:
\[ 3k = k \]

6. **Solving for k:**
Rearranging the equation \( 3k = k \) gives:
\[ 3k - k = 0 \implies 2k = 0 \implies k = 0 \]

7. **Interpreting the Result:**
If \( k = 0 \), then:
\[ \tan A \tan B \tan C = 0 \]
This implies that at least one of the angles \( A, B, \) or \( C \) must be \( 0^\circ \), which is not possible in a triangle.

8. **Conclusion:**
Since having an angle of \( 0^\circ \) means that triangle ABC cannot exist, we conclude that there is no triangle satisfying the given conditions.

Thus, the answer is that the triangle ABC is **none of the types** (option D).