Question

If a+b+c=2s then prove that (s-a)(s-b)+s(s-c)=ab

Answer 1

Step-by-step explanation:

Now sin(s - A) sin(s − B) = 1/ -

{cos[(s - A)

(s – B)] - cos[(s − A) + (s − B)]} - - - = cos(8-A-8+ B)- cos[2s- (A+B)] {cos(A - B)- cos C'}

=

[..cos(A - B) = cos(B − A)] -

Again

sin s sins - C = [cos C - cos(A+B)]

So,

LHS={cos(AB) - cosC+cosC cos(A+B)} -

=[cos(A - B) cos(A + B)] = = [2 sin A sin B] = sin Asin B = RHS

Answer 2

Answer:

(s-a)(s-b)+s(s-c)=ab

s²-sb-sa+ab+s²-sc=ab

2s²=ab-ab+sb+sa+sc

2s×s=s(a+b+c)

2s×s÷s=a+b+c

2s=a+b+c