Question

plzz ans this. very important ​

Answer 1

Answer:

good I dnot sorry I dear

Answer 2

Answer:

2) 4 sin A sin B sin C

Step-by-step explanation:

Given,

A + B + C = 180°

=> A+ B = 180° - C

$(sin2A +sin2B) + sin2C = 2sin( \frac{2A + 2B}{2} )cos(\frac{2A - 2B}{2} ) + 2sinCcosC \\ \: \: \: \: \: \: \: \: \: \: = 2sin(A + B)cos(A - B) + 2sinCcosC \\ \: \: \: \: \: \: \: \: \: \: = 2sin(180 {}^{0} - C )cos(A - B) + 2sinCcosC \\ \: \: \: \: \: \: \: \: = 2sinCcos(A - B) + 2sinCcosC \\ \: \: \: \: \: \: \: \: \: \: = 2sinC(cos(A - B) + cosC ) \\ = 2sinC(cos(A + B) - cos(A + B) ) \\ = 2sinC(2sinA \: sinB) \\ = 4sinA \: sinB \: sinC$

If u have any doubt in this answer, plz follow me and u can ask .....