Question

Let tan A, 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:

• A. $10+\sqrt{10}$
• B. $10-2\sqrt{10}$
• C. $10-3\sqrt{10}$
• D. $10-\sqrt{10}$

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a quadratic equation whose roots are tan A and tan B. We need to find the value of a trigonometric expression involving the sum of the angles A and B. 
Step 2: Key Formula or Approach: 
1. Use Vieta's formulas to find the sum and product of the roots (tan A + tan B and tan A tan B). 
2. Use the tangent addition formula: $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$. 
3. Use the half-angle identity in the form $2\sin^2(\theta) = 1 - \cos(2\theta)$. Here, $\theta = \frac{A+B}{2}$, so $2\theta = A+B$. 
Thus, $2\sin^2\left(\frac{A+B}{2}\right) = 1 - \cos(A+B)$. 
4. We can find $\cos(A+B)$ from $\tan(A+B)$ using the identity $\cos(X) = \pm \frac{1}{\sqrt{1+\tan^2(X)}}$. We'll need to determine the sign. 
Step 3: Detailed Explanation: 
The quadratic equation is $x^2 - 2x - 5 = 0$. The roots are $\tan A$ and $\tan B$. 
From Vieta's formulas: 
Sum of roots: $\tan A + \tan B = -(-2)/1 = 2$. 
Product of roots: $\tan A \tan B = -5/1 = -5$. 
Now, find $\tan(A+B)$: 
\[ \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{2}{1 - (-5)} = \frac{2}{6} = \frac{1}{3} \] Next, we need $\cos(A+B)$. 
\[ \cos(A+B) = \pm \frac{1}{\sqrt{1 + \tan^2(A+B)}} = \pm \frac{1}{\sqrt{1 + (1/3)^2}} = \pm \frac{1}{\sqrt{1 + 1/9}} = \pm \frac{1}{\sqrt{10/9}} = \pm \frac{3}{\sqrt{10}} \] To determine the sign, we need to find the quadrant of $A+B$. We are given $A, B \in (-\pi/2, \pi/2)$. 
Since $\tan A \tan B = -5 < 0$, one angle must have a positive tangent and the other a negative tangent. This means one angle is in $(0, \pi/2)$ and the other is in $(-\pi/2, 0)$. 
Also, $\tan A + \tan B = 2>0$. This means the positive tangent value is larger in magnitude than the negative tangent value. 
Let $\tan A>0$ and $\tan B < 0$. Then $A \in (0, \pi/2)$ and $B \in (-\pi/2, 0)$. 
The sum $A+B$ will lie in the interval $(-\pi/2, \pi/2)$. 
Since $\tan(A+B) = 1/3>0$, the angle $A+B$ must be in the first quadrant, i.e., $(0, \pi/2)$. 
Therefore, $\cos(A+B)$ must be positive. 
\[ \cos(A+B) = \frac{3}{\sqrt{10}} \] Now we can evaluate the required expression $20 \sin^2\left(\frac{A+B}{2}\right)$. 
\[ 20 \sin^2\left(\frac{A+B}{2}\right) = 10 \times \left[ 2\sin^2\left(\frac{A+B}{2}\right) \right] \] Using the identity $2\sin^2(\theta) = 1 - \cos(2\theta)$: 
\[ 20 \sin^2\left(\frac{A+B}{2}\right) = 10 [1 - \cos(A+B)] \] Substitute the value of $\cos(A+B)$: 
\[ 10 \left(1 - \frac{3}{\sqrt{10}}\right) = 10 - \frac{30}{\sqrt{10}} \] Rationalize the term $\frac{30}{\sqrt{10}}$: 
\[ \frac{30}{\sqrt{10}} = \frac{30\sqrt{10}}{10} = 3\sqrt{10} \] So, the final expression is: 
\[ 10 - 3\sqrt{10} \] Step 4: Final Answer: 
The value of $20 \sin^2\left(\frac{A+B}{2}\right)$ is $10-3\sqrt{10}$.