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Question

If 2(sina+sinb)x2sin(ab)y=3 and 2(cosa+cosb)x+2cos(ab)y=5 are perpendicular, then sin2a+sin2b=

A
sin(ab)2sin(a+b)=sin(2a)+sin(2b)
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B
sin2(ab)2sin(a+b)=sin(2a)+sin(2b)
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C
2sin(ab)sin(a+b)=sin(2a)+sin(2b)
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D
sin2(ab)sin(a+b)=sin(2a)+sin(2b)
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Solution

The correct option is A sin2(ab)2sin(a+b)=sin(2a)+sin(2b)
Given equations are
2(sina+sinb)x2sin(ab)y=3(1)
and 2(cosa+cosb)x+2cos(ab)y=5(2)
Let the slope of the equation (1) be m1
and equation (2) be m2
m1=sina+sinbsin(ab)(3)
m2=cosa+cosbcos(ab)(4)
Since both the lines are mutually perpendicular (m1)(m2)=1(5)
From (3), (4) and (5)
Using the identities, cos2α=cos2(α)sin2(α)
sin(α+β)=sinαcosβ+cosαsinβ
We get m1m2=1
sinacosa+sinacosb+sinbcosa+sinbcosbsin(ab)cos(ab)=1
sin(a+b)+12[sin2b+sin2a]sin(ab)cos(ab)=1
sin(2a)+sin(2b)=2[12sin(2a2b)sin(a+b)]
=sin2(ab)2sin(a+b)
We get, sin2a+sin2b=sin2(ab)2sin(a+b)

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