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Question

If tanA and tanB are the roots of the equation x2−ax+b=0, then the value of sin2(A+B) is

A
a2a2+(1b)2
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B
a2a2+b2
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C
a2(a2+b2)
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D
a2b2+(1a)2
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Solution

The correct option is B a2a2+(1b)2

tanA and tanB are the roots of the equation x2ax+b=0


tanA+tanB=a,tanAtanB=b

tanA+tanB=a


sinAcosA+sinBcosB=a


sinAcosB+sinBcosAcosAcosB=a


sin(A+B)=acosAcosB


sin2(A+B)=a2cos2acos2B


sin2(A+B)=a2sec2Asec2B


sin2(A+B)=a2(1+tan2A)(1+tan2B)


sin2(A+B)=a21+tan2A+tan2B+tan2Atan2B


sin2(A+B)=a21+(tanA+tanB)22tanAtanB+tan2Atan2B


sin2(A+B)=a21+a22b+b2


sin2(A+B)=a2a2+(1b)2


So option A is correct.

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