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

Given: \[ \tan A = \frac{1}{\sqrt{x(x^2 + x + 1)}}, \quad \tan B = \frac{\sqrt{x}}{\sqrt{x^2 + x + 1}}, \quad \tan C = \left(x^{-3} + x^{-2} + x^{-1}\right)^{-\frac12}, \quad 0 < A, B, C < \frac{\pi}{2}. \] Find \( A + B \).

Concept Used:

Tangent addition formula: \[ \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \] Also, verify if \( A + B = C \) or \( A + B = \frac{\pi}{2} - C \) etc.

Step-by-Step Solution:

Step 1: Write down \(\tan A\) 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: Compute \(\tan(A+B)\) using the 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 numerator and 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 numerator by denominator.

\[ \tan(A+B) = \frac{\frac{1+x}{\sqrt{x(x^2+x+1)}}}{\frac{x(x+1)}{x^2+x+1}} = \frac{1+x}{\sqrt{x(x^2+x+1)}} \cdot \frac{x^2+x+1}{x(x+1)} \] Cancel \(1+x\) (since \(x>0\)): \[ = \frac{x^2+x+1}{x\sqrt{x(x^2+x+1)}} = \frac{\sqrt{x^2+x+1}}{x\sqrt{x}} = \frac{\sqrt{x^2+x+1}}{x^{3/2}} \]

Step 5: Compare with \(\tan C\).

Given: \[ \tan C = \left(x^{-3} + x^{-2} + x^{-1}\right)^{-1/2} \] Factor \(x^{-3}\): \[ x^{-3} + x^{-2} + x^{-1} = x^{-3}(1 + x + x^2) \] So: \[ \tan C = \left[x^{-3}(1+x+x^2)\right]^{-1/2} = \left[x^{-3}(x^2+x+1)\right]^{-1/2} \] \[ = x^{3/2} (x^2+x+1)^{-1/2} = \frac{x^{3/2}}{\sqrt{x^2+x+1}} \] Thus: \[ \tan C = \frac{x^{3/2}}{\sqrt{x^2+x+1}} \]

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

\[ \tan(A+B) = \frac{\sqrt{x^2+x+1}}{x^{3/2}}, \quad \tan C = \frac{x^{3/2}}{\sqrt{x^2+x+1}} \] So: \[ \tan(A+B) = \frac{1}{\tan C} = \cot C \] Since \(0 < A,B,C < \pi/2\), we have: \[ A+B = \frac{\pi}{2} - C \] But the problem likely expects \(A+B = C\) or \(A+B = \pi/2 - C\). Let's check: \(\tan(A+B) \cdot \tan C = 1\) ⇒ \(A+B + C = \pi/2\).

 

Step 7: Final answer.

\[ A + B = \frac{\pi}{2} - C \] But the options might just say \(A+B = \pi/2 - C\).

 

Given the symmetry, the correct relation is: \[ A + B + C = \frac{\pi}{2} \] So \(A+B = \frac{\pi}{2} - C\).

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

Answer 2

The problem requires us to find the value of the sum of two angles, \( A + B \), given the expressions for their tangents and the tangent of a third angle, C, in terms of a variable \(x\). All angles are acute.

Concept Used:

We will use the tangent addition formula to find an expression for \( \tan(A + B) \). The formula is:

\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]

We will also use the complementary angle identity for tangent and cotangent:

\[ \cot C = \tan\left(\frac{\pi}{2} - C\right) \]

Note: The expression for \( \tan C \) appears to contain a typographical error in many sources. A common version of this problem, which leads to a conclusive answer, uses \( \tan C = (x^{-1} + x^{-2} + x^{-3})^{-1/2} \). We will proceed assuming this intended form.

Step-by-Step Solution:

Step 1: Calculate the numerator of the tangent addition formula, \( \tan A + \tan B \).

Given:

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

Their sum is:

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

Step 2: Calculate the denominator of the tangent addition formula, \( 1 - \tan A \tan B \).

First, find the product \( \tan A \tan B \):

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

Now, subtract this from 1:

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

Step 3: Compute \( \tan(A + B) \).

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

Assuming \( x > 0 \), we can cancel the \( (x+1) \) term:

\[ \tan(A + B) = \frac{x^2 + x + 1}{x\sqrt{x}\sqrt{x^2 + x + 1}} = \frac{\sqrt{x^2 + x + 1}}{x\sqrt{x}} = \sqrt{\frac{x^2 + x + 1}{x^3}} \] \[ \tan(A+B) = \sqrt{\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}} = \sqrt{x^{-1} + x^{-2} + x^{-3}} \]

Step 4: Relate \( \tan(A + B) \) with \( \tan C \).

Using the intended expression for \( \tan C \):

\[ \tan C = (x^{-1} + x^{-2} + x^{-3})^{-1/2} = \frac{1}{\sqrt{x^{-1} + x^{-2} + x^{-3}}} \]

Comparing this with our result for \( \tan(A+B) \), we see that:

\[ \tan(A + B) = \frac{1}{\tan C} = \cot C \]

Step 5: Find the value of \( A + B \).

We have the relation \( \tan(A+B) = \cot C \). Using the identity \( \cot C = \tan(\frac{\pi}{2} - C) \), we get:

\[ \tan(A + B) = \tan\left(\frac{\pi}{2} - C\right) \]

Since \( 0 < A, B, C < \frac{\pi}{2} \), we know that \( 0 < A+B < \pi \) and \( 0 < \frac{\pi}{2} - C < \frac{\pi}{2} \). Since \( \tan(A+B) = \cot C > 0 \), \( A+B \) must be in the first quadrant. Therefore, we can equate the angles:

\[ A + B = \frac{\pi}{2} - C \]

The value of \( A + B \) is \( \frac{\pi}{2} - C \), which also implies that \( A+B+C = \frac{\pi}{2} \).