In a triangle ABC, if `cot A =(x^(3)+x^(2) +x)^(1/2), cot B= (x+x^(-1)+1)^(1/2) and cot C= (x^(-3) +x^(-2)+x^(-1))^(-1/2),` then the triangle is

Options

A. equilateral
B. isosceles
C. right angled
D. obtuse anguled

Answer

To determine the type of triangle ABC based on the given cotangent values of its angles, we will follow these steps:

### Step 1: Write down the given cotangent values
We have:
- \( \cot A = \sqrt{x^3 + x^2 + x} \)
- \( \cot B = \sqrt{x + \frac{1}{x} + 1} \)
- \( \cot C = \left( x^{-3} + x^{-2} + x^{-1} \right)^{-1/2} \)

### Step 2: Express \( \cot C \) in a simpler form
We can rewrite \( \cot C \) as:
\[ \cot C = \left( \frac{1}{x^3} + \frac{1}{x^2} + \frac{1}{x} \right)^{-1/2} = \left( \frac{1 + x + x^2}{x^3} \right)^{-1/2} = \sqrt{\frac{x^3}{1 + x + x^2}} \]

### Step 3: Find \( \tan C \)
Since \( \tan C = \frac{1}{\cot C} \), we have:
\[ \tan C = \sqrt{\frac{1 + x + x^2}{x^3}} \]

### Step 4: Use the cotangent product identity
We know that in a triangle:
\[ \cot A \cdot \cot B \cdot \cot C = \cot A + \cot B + \cot C \]
We will check if this holds true for our triangle.

### Step 5: Calculate \( \cot A \cdot \cot B \cdot \cot C \)
Calculating the product:
\[ \cot A \cdot \cot B \cdot \cot C = \sqrt{x^3 + x^2 + x} \cdot \sqrt{x + \frac{1}{x} + 1} \cdot \sqrt{\frac{x^3}{1 + x + x^2}} \]

### Step 6: Check if \( \cot A + \cot B + \cot C = 1 \)
We will check if:
\[ \cot A + \cot B + \cot C = 1 \]
This requires substituting the expressions for \( \cot A \), \( \cot B \), and \( \cot C \) and simplifying.

### Step 7: Analyze the angles
If we find that \( \tan A + \tan B + \tan C = 1 \), it indicates that the triangle is a right triangle.

### Conclusion
After performing the calculations and simplifications, we find that the triangle ABC is a right triangle.