Quadratic EquationMedium
Question
Let $tanA,tanB$, 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 :
Options
A.$10 + \sqrt{10}$
B.$10 - 2\sqrt{10}$
C.$10 - 3\sqrt{10}$
D.$10 - \sqrt{10}$
Solution
Sol. [Key Concept: $tan(A + B) = \frac{tan\ A + tan\ B}{1 - tan\ Atan\ B}$; then $20\sin^{2}\frac{A + B}{2} = 10\ (1 - cos(A + B))$.]
$${x^{2} - 2x - 5 = 0 }{tanA + tanB = 2;tanAtanB = - 5 }{\therefore tan(A + B) = \frac{2}{1 - ( - 5)} = \frac{1}{3} }{\Rightarrow cos(A + B) = \frac{3}{\sqrt{10}} }{\therefore 20\left( \sin^{2}\left( \frac{\text{ }A + B}{2} \right) \right) = \frac{10}{2}(1 - cos(A + B)) }{= 10\left( 1 - \frac{3}{\sqrt{10}} \right) = (10 - 3\sqrt{10}) }$$
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