(a-b)+(b-c)+(c-a)

Show that: (a-b)(a+b)+ (b-c) (b+c) +(c-a) (c+a) =0

Add the product:a(a-b), b(b-c), c(c-a).

Show that (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a) =0

Simplify the (a+b)(a-b)+(b+c)(b-c)+(c+a)(c-a)

The value of (a-b)(a+b)+(b-c)(b+c)+(c+a)(c-a) is :

If in a triangle A B C angle B=90^0 then tan^2(A/2) is (b-c)/a (b) (b-c)/(b+c) (c) (b+c)/(b-c) (d) (b+c)/a

If in a triangle A B C angle B=90^0 then tan^2(A/2) is (b-c)/a (b) (b-c)/(b+c) (c) (b+c)/(b-c) (d) (b+c)/a

Prove: |(a, b-c,c-b),( a-c, b, c-a),( a-b,b-a, c)|=(a+b-c)(b+c-a)(c+a-b)

Prove: |(a, b-c,c-b),( a-c, b, c-a),( a-b,b-a, c)|=(a+b-c)(b+c-a)(c+a-b)