If tan α, tan...

Question

If $t a n α, t a n β$ are the roots of the equation $x^{2}$ + ax + b = 0 then the value of $s i n^{2} (α + β) + a s i n (α + β) c o s (α + β) + b c o s^{2} (α + β) =$

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$\frac{1}{a} + b$

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Solution

The correct option is B
b

$t a n α + t a n β = - a, t a n α t a n β = b t a n (α + β) = \frac{t a n α + t a n β}{1 - t a n α t a n β} t a n (α + β) = \frac{- a}{1 - b} = \frac{a}{b - 1}$
$N o w s i n^{2} (α + β) + a s i n (α + β) c o s (α + β) + b c o s^{2} (α + β) = c o s^{2} (α + β) [t a n^{2} (α + β) + a t a n (α + β) + b] = \frac{t a n^{2} (α + β) + a t a n (α + β) + b}{s e c^{2} (α + β)} = \frac{t a n^{2} (α + β) + a t a n (α + β) + b}{1 + t a n^{2} (α + β)}$
$= \frac{\frac{a^{2}}{(b - 1)^{2}} + a . \frac{a}{b - 1} + b}{1 + \frac{a^{2}}{(b - 1)^{2}}} = \frac{\frac{a}{b - 1} (\frac{a}{b - 1} + a) + b}{1 + \frac{a^{2}}{(b - 1)^{2}}} = \frac{\frac{a}{b - 1} (\frac{a + a b - a}{b - 1}) + b}{1 + \frac{a^{2}}{(b - 1)^{2}}} = b$

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