Q: If α and β are solutions of sin2x + a sinx + b = 0 as well as that of cos2x + c cosx + d = 0, then sin(α + β) is equal to
(A) $\large \frac{2 b d}{b^2 + d^2} $
(B) $\large \frac{a^2 + c^2}{2 a c} $
(C) $\large \frac{b^2 + d^2}{2 b d} $
(D) $\large \frac{2 a c}{a^2 + c^2} $
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Ans: (D)
Sol: According to the given condition,
sinα + sinβ = –a and cosα + cosβ = -c
$\large 2sin\frac{\alpha + \beta}{2} cos\frac{\alpha – \beta}{2} = – a $ …(i)
and , $\large 2cos\frac{\alpha + \beta}{2} cos\frac{\alpha – \beta}{2} = – c $ …(ii)
On dividing ,
$\large tan\frac{\alpha + \beta}{2} = \frac{a}{c} $
$\large sin(\alpha + \beta) = \frac{2tan\frac{\alpha + \beta}{2}}{1+ tan^2\frac{\alpha + \beta}{2}} $
$\large = \frac{2 a c}{a^2 + c^2} $
Hence (D) is the correct answer.