Question

In any triangle, prove that (Exs. 1-5)
1. (a) a (b cos C-c cos B) = b²-c².

Answer 1

Explanation:

By cosine rule, we have

`cos B = (a^(2) +c^(2) -b^(2) )/(2ac) \\\\cos C = (a^(2) +b^(2) -c^(2) )/(2ab)`

a(b cos C-c cos B) = ab cos C - ac cos B

substituting cosine formula,

`ab(a^(2) +b^(2) -c^(2) )/(2ab) -ac(a^(2) +c^(2) -b^(2) )/(2ac)\\\\= (a^(2) +b^(2) -c^(2) )/(2) -(a^(2) +c^(2) -b^(2) )/(2)\\\\= (2b^(2) -2c^(2) )/(2) \\`

= b²-c²