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Question: If α and β are the solutions of sin<sup>2</sup>x + asinx + b = 0 as well as that of cos<sup>2</sup>x...

If α and β are the solutions of sin2x + asinx + b = 0 as well as that of cos2x + ccosx + d = 0, then sin(α + β) is equal to

A

2bdb2+d2\frac{2bd}{b^{2} + d^{2}}

B

a2+c22ac\frac{a^{2} + c^{2}}{2ac}

C

b2+d22bd\frac{b^{2} + d^{2}}{2bd}

D

2aca2+c2\frac{2ac}{a^{2} + c^{2}}

Answer

2aca2+c2\frac{2ac}{a^{2} + c^{2}}

Explanation

Solution

According to the given condition,

sinα + sinβ = –a and cos α + cosβ = – c .

2sin⁡α+β2cos⁡α−β2=−aand2cos⁡α+β2cos⁡α−β2=−c2\sin\frac{\alpha + \beta}{2}\cos\frac{\alpha - \beta}{2} = - aand2\cos\frac{\alpha + \beta}{2}\cos\frac{\alpha - \beta}{2} = - c ⇒ tan⁡α+β2=ac\tan\frac{\alpha + \beta}{2} = \frac{a}{c}

⇒ sin⁡(α+β)=2tan⁡α+β21+tan⁡2α+β2=2aca2+c2\sin(\alpha + \beta) = \frac{2\tan\frac{\alpha + \beta}{2}}{1 + \tan^{2}\frac{\alpha + \beta}{2}} = \frac{2ac}{a^{2} + c^{2}}