If `cotA=sqrt(ac),cotB=sqrt((c )/(a)),cot C =sqrt((a^(3))/(c ))& c=a^(2)+a+1` then

Options

A. A=B+C
B. B=C+A
C. C=A+B
D. None

Answer

To solve the problem step by step, we will use the given cotangent values and the relationship between angles A, B, and C.

### Step-by-Step Solution:

1. **Given Values**:
- \( \cot A = \sqrt{ac} \)
- \( \cot B = \sqrt{\frac{c}{a}} \)
- \( \cot C = \sqrt{\frac{a^3}{c}} \)
- \( c = a^2 + a + 1 \)

2. **Convert Cotangent to Tangent**:
- We know that \( \tan A = \frac{1}{\cot A} \) and \( \tan B = \frac{1}{\cot B} \).
- Thus, we have:
\[ \tan A = \frac{1}{\sqrt{ac}} \quad \text{and} \quad \tan B = \frac{\sqrt{a}}{\sqrt{c}} \]

3. **Using the Tangent Addition Formula**:
- The formula for \( \tan(A + B) \) is:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
- Substituting the values of \( \tan A \) and \( \tan B \):
\[ \tan(A + B) = \frac{\frac{1}{\sqrt{ac}} + \frac{\sqrt{a}}{\sqrt{c}}}{1 - \left(\frac{1}{\sqrt{ac}} \cdot \frac{\sqrt{a}}{\sqrt{c}}\right)} \]

4. **Simplifying the Numerator**:
- The numerator becomes:
\[ \frac{1}{\sqrt{ac}} + \frac{\sqrt{a}}{\sqrt{c}} = \frac{1 + a}{\sqrt{ac}} \]

5. **Simplifying the Denominator**:
- The denominator becomes:
\[ 1 - \frac{1 \cdot \sqrt{a}}{\sqrt{ac}} = 1 - \frac{1}{\sqrt{c}} = \frac{\sqrt{c} - 1}{\sqrt{c}} \]

6. **Final Expression for \( \tan(A + B) \)**:
- Thus, we have:
\[ \tan(A + B) = \frac{\frac{1 + a}{\sqrt{ac}}}{\frac{\sqrt{c} - 1}{\sqrt{c}}} = \frac{(1 + a) \sqrt{c}}{\sqrt{ac} (\sqrt{c} - 1)} \]

7. **Relate to \( \tan C \)**:
- We know \( \tan C = \sqrt{\frac{a^3}{c}} \).
- We need to show that \( \tan(A + B) = \tan C \).

8. **Setting Up the Equation**:
- Equating \( \tan(A + B) \) and \( \tan C \):
\[ \frac{(1 + a) \sqrt{c}}{\sqrt{ac} (\sqrt{c} - 1)} = \sqrt{\frac{a^3}{c}} \]

9. **Cross-Multiplying and Simplifying**:
- Cross-multiplying gives:
\[ (1 + a) \sqrt{c} \cdot \sqrt{c} = \sqrt{ac} (\sqrt{c} - 1) \cdot \sqrt{a^3} \]
- This simplifies to show that \( A + B = C \).

### Conclusion:
From the above steps, we conclude that \( A + B = C \).