Q. 135.0( 3 Votes )

# In any ΔABC, prove that

a^{2}(cos^{2}B – cos^{2}C) + b^{2}(cos^{2}C – cos^{2}A) + c^{2}(cos^{2}A – cos^{2}B) = 0

Answer :

Need to prove: a^{2}(cos^{2}B – cos^{2}C) + b^{2}(cos^{2}C – cos^{2}A) + c^{2}(cos^{2}A – cos^{2}B) = 0

From left hand side,

a^{2}(cos^{2}B – cos^{2}C) + b^{2}(cos^{2}C – cos^{2}A) + c^{2}(cos^{2}A – cos^{2}B)

a^{2}((1 - sin^{2}B) – (1 - sin^{2}C)) + b^{2}((1 - sin^{2}C) – (1 - sin^{2}A)) + c^{2}((1 - sin^{2}A) – (1 - sin^{2}B))

a^{2}( - sin^{2}B + sin^{2}C) + b^{2}( - sin^{2}C + sin^{2}A) + c^{2}( - sin^{2}A + sin^{2}B)

We know that, where R is the circumradius.

Therefore,

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

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

So,

4R^{2}[ sin^{2}A( - sin^{2}B + sin^{2}C) + sin^{2}B( - sin^{2}C + sin^{2}A) + sin^{2}C( - sin^{2}A + sin^{2}B)

4R^{2}[ - sin^{2}Asin^{2}B + sin^{2}Asin^{2}C – sin^{2}Bsin^{2}C + sin^{2}Asin^{2}B – sin^{2}Asin^{2}C + sin^{2}Bsin^{2}C ]

4R^{2} [0]

0 [Proved]

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