Question

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

• A. \( \frac{\pi}{2} - C \)
• B. \( \pi - C \)
• C. \( 2\pi - C \)
• D. \( C \)

Answer 1

The Correct Option is A

Provided: \( \tan A = \frac{1}{\sqrt{x(x^2 + x + 1)}} \), \( \tan B = \frac{\sqrt{x}}{\sqrt{x^2 + x + 1}} \), \( \tan C = \left(x^{-3} + x^{-2} + x^{-1}\right)^{-\frac12} \), with \( 0 < A, B, C < \frac{\pi}{2} \). Determine \( A + B \).

Concept Used:

The tangent addition formula is \( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \). We also need to consider potential relationships like \( A + B = C \) or \( A + B = \frac{\pi}{2} - C \).

Step-by-Step Solution:

Step 1: State the given values for \( \tan A \) and \( \tan B \).

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

Step 2: Calculate \( \tan(A+B) \) using the tangent addition formula.

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

Step 3: Simplify the numerator and the denominator separately.

Numerator: \[ \frac{1}{\sqrt{x(x^2+x+1)}} + \frac{\sqrt{x}}{\sqrt{x^2+x+1}} = \frac{1 + x}{\sqrt{x(x^2+x+1)}} \] Denominator: \[ 1 - \frac{1}{\sqrt{x(x^2+x+1)}} \cdot \frac{\sqrt{x}}{\sqrt{x^2+x+1}} = 1 - \frac{1}{x^2+x+1} = \frac{x^2+x+1 - 1}{x^2+x+1} = \frac{x^2+x}{x^2+x+1} = \frac{x(x+1)}{x^2+x+1} \]

Step 4: Divide the simplified numerator by the simplified denominator.

\[ \tan(A+B) = \frac{\frac{1+x}{\sqrt{x(x^2+x+1)}}}{\frac{x(x+1)}{x^2+x+1}} = \frac{1+x}{\sqrt{x(x^2+x+1)}} \cdot \frac{x^2+x+1}{x(x+1)} \] Given \(x>0\), we can cancel \(1+x\): \[ = \frac{x^2+x+1}{x\sqrt{x(x^2+x+1)}} = \frac{\sqrt{x^2+x+1}}{x\sqrt{x}} = \frac{\sqrt{x^2+x+1}}{x^{3/2}} \]

Step 5: Express \( \tan C \) in a comparable form.

Given: \[ \tan C = \left(x^{-3} + x^{-2} + x^{-1}\right)^{-1/2} \] Factor out \(x^{-3}\) from the expression in the parenthesis: \[ x^{-3} + x^{-2} + x^{-1} = x^{-3}(1 + x + x^2) \] Therefore: \[ \tan C = \left[x^{-3}(1+x+x^2)\right]^{-1/2} = \left[x^{-3}(x^2+x+1)\right]^{-1/2} \] \[ = x^{3/2} (x^2+x+1)^{-1/2} = \frac{x^{3/2}}{\sqrt{x^2+x+1}} \] Thus: \[ \tan C = \frac{x^{3/2}}{\sqrt{x^2+x+1}} \]

Step 6: Compare \( \tan(A+B) \) with \( \tan C \).

We found: \[ \tan(A+B) = \frac{\sqrt{x^2+x+1}}{x^{3/2}} \] And: \[ \tan C = \frac{x^{3/2}}{\sqrt{x^2+x+1}} \] It is evident that \( \tan(A+B) = \frac{1}{\tan C} = \cot C \). Since \(0 < A,B,C < \pi/2\), this implies \( A+B = \frac{\pi}{2} - C \).

Step 7: State the final answer.

The relationship \( A+B = \frac{\pi}{2} - C \) leads to \( A + B + C = \frac{\pi}{2} \). Therefore, \( A+B = \mathbf{\frac{\pi}{2} - C} \).