Question
If A, B and C are angles of triangle, then sin2A + sin2B + sin2C– 2cosA cosB cosC
Solution
Let A = B = 45º, C = 90º
then, sin2A, + sin2B + sin2C – 2 cosA cosB cosC
= (1/√2) 2 + ( 1/√2) 2 + 1 – 0 (as cosC = 0)
= 1/2 + 1/2 + 1 = 2
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