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Question

lf α,β are solutions of sin2x+asinx+b=0 and cos2x+ccosx+d=0 then sin(α+β) equals:-

A
2aca2+c2
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B
a2+c22ac
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C
2bdb2+d2
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D
b2+d22bd
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Solution

The correct option is A 2aca2+c2
Let α and β are roots of the given equation sin2x+asinx+b=0. Then,
sinα+sinβ=a1=a .......... (A)

sinαsinβ=b1=b

Let α and β are roots of the given equation cos2x+ccosx+d=0. Then,

cosα+cosβ=c1=c .......... (B)

cosαcosβ=d1=d

a2+c2=(sinα+sinβ)2+(cosα+cosβ)2

=sin2α+sin2β+2sinαsinβ+cos2α+cos2β+2cosαcosβ

=1+1+2(sinαsinβ+cosαcosβ)

=2[1+cos(αβ)]

=2×2cos2(αβ2)

=4cos2(αβ2)

a2+c2 =4cos2(αβ2) ---- ( 1 )

Now, from A and B
2sinα+β2.cosαβ2=a --- ( 2 )

2cosα+β2.cosαβ2=c ---- ( 3 )

Multiplying ( 2 ) and ( 3 ),

4sinα+β2.cosα+β2.cos2(αβ2)=ac

2sin(α+β)(a2+c24)=ac [ From ( 1 ) ]

sin(α+β)=2aca2+c2

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