2015/100)n any triangle ABC, prove that

$a^2(\cos^2 B - \cos^2 C) + b^2(\cos^2 C -\cos^2 A) + c^2(\cos^2 A - \cos^2 B) = 0$

proof:

$a^2(\cos^2 B - \cos^2 C) + b^2(\cos^2 C - \cos^2 A) + c^2(\cos^2 A - \cos^2 B)$
= $a^2(\sin^2 C - \sin^2 B) + b^2 (\sin^2 A - \sin^2 C) + c^2( \sin^2 B - \sin^2 A)$

using law of sines we have

let $\dfrac{a}{\sin A} = \dfrac{b}{sin B} = \dfrac{c}{\sin C} = k$ (say)

we get

$a^2(\sin^2 C - \sin^2 B) + b^2 (\sin^2 A- \sin^2 C) + c^2( \sin^2 B - \sin^2 A)$

=$k^2 \sin ^2 A(\sin^2 C - \sin^2 B) + k^2 \sin ^2 B (\sin^2 A -\sin^2 C) + k^2 \sin ^2 C( \sin^2 B - \sin^2 A)$

= $k^2( \sin ^2 A \sin ^2 C – \sin ^2 A \sin ^2 B + \sin ^2 B \sin ^2 A – \sin ^2B \sin^2 C + \sin ^2 C \sin ^2 B – \sin ^2C \sin ^2 A)$

= 0