Q. 135.0( 3 Votes )

In any ΔABC, prove that

a2(cos2B – cos2C) + b2(cos2C – cos2A) + c2(cos2A – cos2B) = 0

Answer :

Need to prove: a2(cos2B – cos2C) + b2(cos2C – cos2A) + c2(cos2A – cos2B) = 0


From left hand side,


a2(cos2B – cos2C) + b2(cos2C – cos2A) + c2(cos2A – cos2B)


a2((1 - sin2B) – (1 - sin2C)) + b2((1 - sin2C) – (1 - sin2A)) + c2((1 - sin2A) – (1 - sin2B))


a2( - sin2B + sin2C) + b2( - sin2C + sin2A) + c2( - sin2A + sin2B)


We know that, where R is the circumradius.


Therefore,


a = 2R sinA ---- (a)


Similarly, b = 2R sinB and c = 2R sinC


So,


4R2[ sin2A( - sin2B + sin2C) + sin2B( - sin2C + sin2A) + sin2C( - sin2A + sin2B)


4R2[ - sin2Asin2B + sin2Asin2C – sin2Bsin2C + sin2Asin2B – sin2Asin2C + sin2Bsin2C ]


4R2 [0]


0 [Proved]


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