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Question

In △ABC the value of sin2A+sin2B−sin2CsinAsinBsinC=?

A
2cotB
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B
12cosA
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C
sinB
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D
2sinB
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Solution

The correct option is A 2cotB
sin2A+sin2Bsin2CsinAsinBsinC
cos2A=12sin2A
=1cos2A+1cos2B+1+cos2C2sinAsinBsinC
=1(cos2A+cos2Bcos2C)2sinAsinBsinC
=1(cos2A2sin(2B+2C)/2sin(2B2C)/2)2sinAsinBsinC.
=1(cos2A2sin(B+C)sin(BC))2sinAsinBsinC
A+B+C=100o
=1(cos2A2sin(180A)sin(BC))2sinAsinBsinC
=1(cos2A+2sinAsin(BC))2sinAsinBsinC
=1(12sin2A+2sinAsin(BC))2sinAsinBsinC
=1(1+2sin2A2sinAsin(BC)2sinAsinBsinC
=2sinA(sinAsin(BC))2sinAsinBsinC
=2cos(A+BC)2sin(AB+C)2sinBsinC
=2cos(1802C2)sin(1802B2)sinBsinC
=2sinCcosBsinBsinC
=2cosBsinB=2cotB.

1154447_697937_ans_198c673527ea4676895d47737dee4b9e.jpg

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