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Question

IfA,B,C are angles of a triangle, then sin2A+sin2B-sin2C is equal to


A

4sinAcosBcosC

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B

4cosA

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C

4sinAcosA

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D

4cosAcosBsinC

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Solution

The correct option is D

4cosAcosBsinC


Finding the value of sin 2 A plus sin 2 B minus sin 2 C:

Given that, A,B,C are the angle of triangle.

A+B+C=π2A+2B+2C=2π.(i)

Now,

sin2A+sin2B-sin2C[sinx+siny=2sin(x+y)2cos(xy)2,]=2sin(2A+2B)2cos(2A2B)2sin[2π2(A+B)]from(i)=2sin(A+B)cos(AB)+sin2(A+B)=2sin(A+B)cos(AB)+2sin(A+B)cos(A+B)since,sin2x=2sinxcosx=2sin(A+B)[cos(AB)+cos(A+B)]=2sin(πC)(2cosAcosB)=4cosAcosBsinC

Hence, correct option is (D).


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