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in Trigonometry by (65.3k points)

If A + B + C = π. Prove that: cos2A + cos2B – cos2C = 1 – 4sinA . sinB · cosC

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L.H.S. = cos2A + cos2B – cos2C 

= 2cos(A + B) cos(A – B) – (2 cos2C – 1) 

= -2cosCCos(A – B) – 2 cos2C + 1 [∵ cosC=-cos(A+B)] 

= 1 – 2cos C[cos(A – B) – cos (A + B)] 

= 1 – 2cos C[-2 sinA . sin(- B)] 

= 1 – 4 sin A sin B cos C = R.H.S.

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