Question

plz answer my question number 45 ​

Answer 1

ANSWER: PLEASE FOLLOW

Sin2A+sin2B+sin2C

2×sin(2A+2B)/2×cos(2A-2B)/2+sin2C

=>2sin(180°-(A+B)×cos(A-B)+2sinC×cosC

=>2sinC×cos(A-B)+2sinC×cosC

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

=>2sinC[cos(A-B)×cos(180°-(A+B)]

=>2sinC[2sin(A-B+A+B)/2×sin(A+B-A+B)]

=>2sinC[2sinA×sinB]

=>4sinA×sinB×sinC

HENCE PROVED

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Answer 2

Answer:beautiful answer for beautiful girl

Step-by-step explanation:

sin2A + sin 2B +sin2C

= 2 sin(A+B)cos(A-B) + 2sinC cosC

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

=2sinC (cos(A-b) + cos C)

=2sin C(cos(A-B) - cos(A+B) )

= 2sinC . 2sin A sin B

=4 sinA sin B sin C