Question

a+b+c=180 sin2a+sin2b-sin2c=4cosacosbcosc

Answer 1

Heya friends....✅

Here is ur answer...
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Question is wrong .. question should be,A+B+C÷180° ,sin2A+sin2B-Sin2C=4cosA*cosB*sinC

from lhs ...

=Sin2A+sin2B+sin2C

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

=2sin(A+B)*cos(A-B)-sin2C

=2sin(π-c)*cos(A-B)-2sinC*cosC

=2sinC*[cos(A-B)+cosC]

=2sinC[cos(A-B)+cos(A+B)

=2sinC[cos(A-B)+cos(A+B) 【° • °cos(π-¢)=-cos¢】

=2sinC[cos(A-B)+cos(A+B)]>>

[• ° •cos(π-¢)=-cos¢]

cosC+CosD=2cos(C+D)*cos(C-D) here appling

then ....2sinC*2cosA*cosB

=>4cosA*cosB×sinC..Rhs ...
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hope it helps you..

@Rajukumar☺☺
now,,,,.....2sinC*cos