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Question

If sin1a+sin1b+sin1c=π, then the value of a(1a2)+b(1b2)+c(1c2) will be

A
2abc
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B
abc
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C
12abc
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D
13abc
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Solution

The correct option is C 2abc
The above expression can be re-written as
sin1(sinA)+sin1(sinB)+sin1(sinC)=π
Where
a=sinA, b=sinB and c=sinC
Hence A,B and C can be assumed as angles of the triangle.
Therefore.
a1a2+b1b2+c1c2
Simplifies as
sinAcosA+sinBcosB+sinCcosC.
Multiplying and dividing by 2, we get
sin2A+sin2B+sin2C2
=2sin(A+B)cos(AB)+sin2C2
=2sinCcos(AB)+2sinCcosC2 ...(A+B=πC)
=sinC(cos(AB)+cos(π(A+B)))
=sinC(cos(AB)cos(A+B))
=2sinCsinAsinB
=2abc

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