If `tan A = frac{1}{sqrt (x(x^2 + x + 1)}`, `tan B = frac{sqrt x}{sqrt (x^2 + x + 1)}`, and `tan C = (x^(-3) + x^(-2) + x^(-1))^(1/2)`, `0 < A, B, C < pi/2`, then A + B is equal to :

Options

A. C
B. `pi-C`
C. `2pi-C`
D. `pi/2-C`

Answer

To solve the problem, we need to find the value of \( A + B \) given the values of \( \tan A \), \( \tan B \), and \( \tan C \).

### Step 1: Write the expressions for \( A \) and \( B \)

Given:
\[ \tan A = \frac{1}{\sqrt{x(x^2 + x + 1)}} \]
\[ \tan B = \frac{\sqrt{x}}{\sqrt{x^2 + x + 1}} \]

### Step 2: Use the formula for \( A + B \)

We know from trigonometric identities that:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]

### Step 3: Substitute \( \tan A \) and \( \tan B \)

Substituting the values of \( \tan A \) and \( \tan B \):
\[ \tan(A + B) = \frac{\frac{1}{\sqrt{x(x^2 + x + 1)}} + \frac{\sqrt{x}}{\sqrt{x^2 + x + 1}}}{1 - \left(\frac{1}{\sqrt{x(x^2 + x + 1)}}\right)\left(\frac{\sqrt{x}}{\sqrt{x^2 + x + 1}}\right)} \]

### Step 4: Simplify the numerator

The numerator becomes:
\[ \frac{1 + x}{\sqrt{x(x^2 + x + 1)}} \]

### Step 5: Simplify the denominator

The denominator becomes:
\[ 1 - \frac{1 \cdot \sqrt{x}}{\sqrt{x(x^2 + x + 1)}} = 1 - \frac{1}{\sqrt{x^2 + x + 1}} \]

### Step 6: Combine the fractions

Now we can write:
\[ \tan(A + B) = \frac{\frac{1 + x}{\sqrt{x(x^2 + x + 1)}}}{1 - \frac{1}{\sqrt{x^2 + x + 1}}} \]

### Step 7: Simplify further

To simplify the expression, we can multiply the numerator and the denominator by \( \sqrt{x^2 + x + 1} \):
\[ \tan(A + B) = \frac{(1 + x)}{\sqrt{x(x^2 + x + 1)} \left( \sqrt{x^2 + x + 1} - 1 \right)} \]

### Step 8: Relate to \( \tan C \)

Given:
\[ \tan C = \sqrt{x^{-3} + x^{-2} + x^{-1}} = \sqrt{\frac{1}{x^3} + \frac{1}{x^2} + \frac{1}{x}} = \sqrt{\frac{1 + x + x^2}{x^3}} \]

### Step 9: Compare \( \tan(A + B) \) and \( \tan C \)

We can see that:
\[ \tan(A + B) = \tan C \]

### Conclusion

Thus, we conclude that:
\[ A + B = C \]

Therefore, the final answer is:
\[ \boxed{C} \]