Question:medium

If \( \tan \text{A} = \frac{1}{\sqrt{x(x^2+x+1) \), \( \tan \text{B} = \frac{\sqrt{x}}{\sqrt{x^2+x+1}} \) and \( \tan \text{C} = \sqrt{x^{-1} + x^{-2} + x^{-3}} \) then}

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Look for common terms like \( \sqrt{x^2+x+1} \) to simplify complex algebraic trigonometric expressions.
Updated On: May 14, 2026
  • \( \text{A} + \text{B} = \text{C} \)
  • \( A + B = 2C \)
  • \( A + B = 3C \)
  • \( A + B = 4C \)
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
We are given the tangents of three angles A, B, and C in terms of a variable \( x \). We need to find a relationship between the angles.
A good approach is to calculate \( \tan(A+B) \) and compare it with the expression for \( \tan C \).
Step 2: Key Formula or Approach:
Use the trigonometric identity:
\[ \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \] Step 3: Detailed Explanation:
Let's simplify \( \tan C \) first to see our target:
\[ \tan C = \sqrt{\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}} = \sqrt{\frac{x^2 + x + 1}{x^3}} = \frac{\sqrt{x^2+x+1}}{x\sqrt{x}} \] Now let's compute \( \tan(A+B) \):
Given \( \tan A = \frac{1}{\sqrt{x}\sqrt{x^2+x+1}} \) and \( \tan B = \frac{\sqrt{x}}{\sqrt{x^2+x+1}} \).
Substitute these into the formula:
\[ \tan(A+B) = \frac{\frac{1}{\sqrt{x}\sqrt{x^2+x+1}} + \frac{\sqrt{x}}{\sqrt{x^2+x+1}}}{1 - \left(\frac{1}{\sqrt{x}\sqrt{x^2+x+1}}\right)\left(\frac{\sqrt{x}}{\sqrt{x^2+x+1}}\right)} \] To simplify the numerator, factor out the common denominator term \( \frac{1}{\sqrt{x^2+x+1}} \):
Numerator \( = \frac{1}{\sqrt{x^2+x+1}} \left( \frac{1}{\sqrt{x}} + \sqrt{x} \right) = \frac{1}{\sqrt{x^2+x+1}} \left( \frac{1 + x}{\sqrt{x}} \right) = \frac{x+1}{\sqrt{x}\sqrt{x^2+x+1}} \)
Now simplify the denominator:
Denominator \( = 1 - \frac{\sqrt{x}}{\sqrt{x}(x^2+x+1)} = 1 - \frac{1}{x^2+x+1} = \frac{x^2+x+1 - 1}{x^2+x+1} = \frac{x^2+x}{x^2+x+1} = \frac{x(x+1)}{x^2+x+1} \)
Now, divide the numerator by the denominator:
\[ \tan(A+B) = \frac{\frac{x+1}{\sqrt{x}\sqrt{x^2+x+1}}}{\frac{x(x+1)}{x^2+x+1}} \] \[ \tan(A+B) = \frac{x+1}{\sqrt{x}\sqrt{x^2+x+1}} \cdot \frac{x^2+x+1}{x(x+1)} \] Cancel the common term \( (x+1) \):
\[ \tan(A+B) = \frac{x^2+x+1}{\sqrt{x} \cdot x \cdot \sqrt{x^2+x+1}} \] Simplify \( \frac{x^2+x+1}{\sqrt{x^2+x+1}} \) to \( \sqrt{x^2+x+1} \):
\[ \tan(A+B) = \frac{\sqrt{x^2+x+1}}{x\sqrt{x}} \] Comparing this result with our simplified expression for \( \tan C \):
\[ \tan(A+B) = \tan C \] Taking the inverse tangent (assuming principal values or the simplest relation):
\[ A + B = C \] Step 4: Final Answer:
The relation is \( A + B = C \).
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