In any ΔABC, prove that (cos^2B - cos^2C)/b + c

Question

In any ΔABC, prove that

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

(cos²B - cos²C)/b + c + (cos²C - cos²A)/c + a + (cos²A - cos²B)/a + b = 0

Answer 1

To prove

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

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

From left hand side,

= \(\cfrac{(cos^2B-cos^2C)}{b+c}\) + \(\cfrac{(cos^2C-cos^2A)}{c+a}\) + \(\cfrac{(cos^2A-cos^2B)}{a+b}\)

 

Now,

= 0 [proved]