Let $$\tan A$$ and $$\tan B$$, where $$A, B \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$, be the roots of the quadratic equation $$x^2 - 2x - 5 = 0$$. Then $$20\sin^2\left(\frac{A+B}{2}\right)$$ is equal to :

To solve this, we'll combine properties of quadratic roots with trigonometric identities.

1. Extract Information from the Quadratic Equation

Given $$x^2 - 2x - 5 = 0$$ has roots $$\tan A$$ and $$\tan B$$. Using Vieta's formulas:

• Sum of roots: $$\tan A + \tan B = 2$$
• Product of roots: $$\tan A \cdot \tan B = -5$$

Now, find $$\tan(A+B)$$:

$$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{2}{1 - (-5)} = \frac{2}{6} = \frac{1}{3}$$

2. Relate $$\tan(A+B)$$ to $$\sin^2\left(\frac{A+B}{2}\right)$$

Let $$\theta = A+B$$. We know $$\tan \theta = \frac{1}{3}$$. We need to find $$20 \sin^2(\frac{\theta}{2})$$.

Recall the half-angle identity:

$$\sin^2\left(\frac{\theta}{2}\right) = \frac{1 - \cos \theta}{2}$$

So, our target is: $$20 \left( \frac{1 - \cos \theta}{2} \right) = 10(1 - \cos \theta)$$.

3. Find $$\cos \theta$$

Since $$\tan \theta = \frac{1}{3}$$, we can visualize a right triangle where the opposite side is $$1$$ and the adjacent side is $$3$$.

• Hypotenuse $$= \sqrt{1^2 + 3^2} = \sqrt{10}$$
• $$\cos \theta$$ $$= \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{3}{\sqrt{10}}$$

Note on Signs: Since $$\tan \theta$$ is positive and $$A, B \in (-\frac{\pi}{2}, \frac{\pi}{2})$$, $$\theta$$ must be in the first quadrant, making $$\cos \theta$$ positive.

4. Final Calculation

Substitute $$\cos \theta$$ back into the expression:

$$\text{Result} = 10 \left( 1 - \frac{3}{\sqrt{10}} \right)$$

$$\text{Result} = 10 - \frac{30}{\sqrt{10}}$$

Rationalize the second term: $$\frac{30}{\sqrt{10}} \times \frac{\sqrt{10}}{\sqrt{10}} = \frac{30\sqrt{10}}{10} = 3\sqrt{10}$$.

$$\text{Result} = 10 - 3\sqrt{10}$$

Correct Answer: C ($$10 - 3\sqrt{10}$$)