$\tan α and \tan β are roots of the equation x^{2} + a x + b = 0, then the$
$value of \sin^{2} (α + β) + a \sin (α + β) \cdot \cos (α + β) + b \cos^{2} (α + β) is equal to$

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$a b$

$b$

$\frac{a}{b}$

$a$

answer is B.

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Detailed Solution

$\begin{array}{l} \tan α + \tan β = - a, \tan α \cdot \tan β = b \\ \tan (α + β) = \frac{\tan α + \tan β}{1 - \tan α \cdot \tan β} = \frac{- a}{1 - b} = \frac{a}{b - 1} \\ \Rightarrow \sin^{2} (α + β) + a \cdot \sin (α + β) \cos (α + β) + b \cos^{2} (α + β) \\ = \cos^{2} (α + β) (\tan^{2} (α + β) + b + a \tan (α + β)) \\ = \frac{\tan^{2} (α + β) + b + a \tan (α + β)}{1 + \tan^{2} (α + β)} [\cos^{2} (α + β) = \frac{1}{s e c^{2} (α + β)} = \frac{1}{1 + \tan^{2} (α + β)}] \\ = \frac{\frac{a}{b - 1} (a + \frac{a}{b - 1}) + b}{1 + \frac{a^{2}}{(b - 1)^{2}}} = \frac{\frac{a^{2}}{b - 1} + \frac{a^{2}}{{(b - 1)}^{2}} + b}{1 + \frac{a^{2}}{(b - 1)^{2}}} = \frac{\frac{a^{2} (b - 1) + a^{2}}{{(b - 1)}^{2}} + b}{1 + \frac{a^{2}}{(b - 1)^{2}}} \\ \frac{\frac{a^{2} b}{{(b - 1)}^{2}} + b}{1 + \frac{a^{2}}{(b - 1)^{2}}} = b \end{array}$