In any ΔABC, prove that a2 (cos2B – cos2C) + b2 (cos2C – cos2A)

1.	In any ΔABC, prove that a2 (cos2B – cos2C) + b2 (cos2C – cos2A) + c2 (cos2A – cos2B) = 0
Answer» 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, a/sinA = b/sinB = c/sinC = 2R where R is the circumradius. Therefore, a = 2R sinA.....(a) similarity, 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]

In any ΔABC, prove that a2 (cos2B – cos2C) + b2 (cos2C – cos2A) + c2 (cos2A – cos2B) = 0

Answer»

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, a/sinA = b/sinB = c/sinC = 2R where R is the circumradius.

Therefore,

a = 2R sinA.....(a)

similarity, 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]