Question

IF A+B+C=270then sin2A+sin2B−sin2C+4sinA  ​sinB  cosC=​

Answer 1

Explained

2 cos (A + B) sin (A – B) + 2 sinC cosC

2 cos (A + B) sin (A – B) + 2 sinC cosC

2 cos (A + B) sin (A – B) + 2 sinC cosC = 2 cos (270 – C) sin (A – B) + 2 sinC cosC

2 cos (A + B) sin (A – B) + 2 sinC cosC = 2 cos (270 – C) sin (A – B) + 2 sinC cosC = 2 sin C ( – sin (A – B) + cos (270 – ( A+B))

2 cos (A + B) sin (A – B) + 2 sinC cosC = 2 cos (270 – C) sin (A – B) + 2 sinC cosC = 2 sin C ( – sin (A – B) + cos (270 – ( A+B)) = 2 sin C ( – sin ( A – B) – sin (A + B))

2 cos (A + B) sin (A – B) + 2 sinC cosC = 2 cos (270 – C) sin (A – B) + 2 sinC cosC = 2 sin C ( – sin (A – B) + cos (270 – ( A+B)) = 2 sin C ( – sin ( A – B) – sin (A + B)) = – 4 sinA sinB sinC

2 cos (A + B) sin (A – B) + 2 sinC cosC = 2 cos (270 – C) sin (A – B) + 2 sinC cosC = 2 sin C ( – sin (A – B) + cos (270 – ( A+B)) = 2 sin C ( – sin ( A – B) – sin (A + B)) = – 4 sinA sinB sinC