Question

Sin 2a+sin2b-sin2c=4cosacosbsinc

Answer 1

Hey !!!

from LHS

Sin2A + Sin2B - Sin2C

We know that ,

SinC + SinD = 2sin(C+D)/2 *cos ( C-D)/2

since ,

or, 2sin ( 2A + 2B)/2* cos (2A-2B)/2 - sin2C

or, 2sin(A + B) *cos(A-B) - 2sinC *cosC

or, 2sin( 180°-C) * cos(A - B) -2sinC * cosC

or, 2sinC * cos ( A - B) - 2sinC * cosC

or, 2sinC { cos(A-B) - cos(180°-( A + B) }

or, 2sinC { cos( A- B) +cos ( A+B) }

using formula

↘cosC +cosD = 2cos(C+ D)/2* cos ( C- D) /2

taking cos(A+B) as sinC and cos ( A- B) = cosD

or, 2sinC {2cos(A-B+A+B)/2 * cos ( A+B -A+ B)/2

or, 2sinC{2cos2A/2 * cos2B/2 }

or, 4sinC * cosA *cosB

, 4cosA*cosB*sinC

so ,

LHS = RHS prooved ♻
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Hope it helps you ;!!

@Rajukumar111

Answer 2

Answer:

4cosAcosBsinC

Step-by-step explanation:

Here is your answer.