Question

if A+B+C=270 degrees, then sinA-sin2B+sin2C=​

Answer 1

$\bigstar \: \boxed{\sf{\color{purple}{Answer:}}}$

L.H.S. = cos

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

cos(A – B) + cos 2C = 2cos

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

(A – B) + 1 – 2 sin2C = 1 – 2 sinC

[cos(A – B) + sinC] = 1 – 2 sin C [cos(A – B) + sin(270° – (A + B))] = 1 – 2 sinC [cos(A – B) – cos(A + B)] = 1 – 2 sinC

[2 sinA sinB] = 1 – 4 sinA sinB = R.H.S

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