Question

If \( \tan A \) and \( \tan B \) are roots of the equation \( x^2 - 2x - 5 = 0 \), then the value of \( 10\left(\sin^2\left(\frac{A+B}{2}\right)\right) \) is:

• A. \( 5 + \frac{3}{2}\sqrt{10} \)
• B. \( 10 + \frac{3}{2}\sqrt{10} \)
• C. \( 5 - \frac{3}{2}\sqrt{10} \)
• D. \( 10 - \frac{3}{2}\sqrt{10} \)

Hint:

The expression \( 1 - \cos \theta \) is the standard bridge between trigonometric functions of a full angle and the squared sine of its half-angle.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We first find \( \tan(A+B) \) using the sum and product of the roots. Then, we use the half-angle formula for \( \sin^2 \theta \).

Step 2: Key Formula or Approach:
1. \( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \) 2. \( \sin^2(\theta/2) = \frac{1 - \cos \theta}{2} \) 3. \( \cos \theta = \frac{1}{\pm \sqrt{1 + \tan^2 \theta}} \)

Step 3: Detailed Explanation:
1. From the equation: \( \tan A + \tan B = 2 \) and \( \tan A \tan B = -5 \). 2. \( \tan(A+B) = \frac{2}{1 - (-5)} = \frac{2}{6} = \frac{1}{3} \). 3. Let \( \theta = A+B \). We need \( 10 \cdot \frac{1 - \cos \theta}{2} = 5(1 - \cos \theta) \). 4. Since \( \tan \theta = 1/3 \), \( \cos^2 \theta = \frac{1}{1 + (1/3)^2} = \frac{9}{10} \implies \cos \theta = \frac{3}{\sqrt{10}} \). 5. Substituting: \( 5(1 - \frac{3}{\sqrt{10}}) = 5 - \frac{15}{\sqrt{10}} = 5 - \frac{15\sqrt{10}}{10} = 5 - \frac{3}{2}\sqrt{10} \).

Step 4: Final Answer:
The value is \( 5 - \frac{3}{2}\sqrt{10} \).