If in a ` A B C ,(a^2-b^2)/(a^2+b^2)=(sin(A-B))/(sin(A+B)`, prove that it is either a right angled or an isosceles triangle.

If in a ` A B C ,(a^2-b^2)/(a^2+b^2)=(sin(A-B))/(sin(A+B)`, prove that

1.	If in a ` A B C ,(a^2-b^2)/(a^2+b^2)=(sin(A-B))/(sin(A+B)`, prove that it is either a right angled or an isosceles triangle.
Answer» `(a^2-b^2)/(a^2+b^2) = (sin(A-B))/(sin(A+B))` `=>(a^2+b^2)/(a^2-b^2) = (sin(A+B))/(sin(A-B))` Using componendo and dividendo, `=>(2a^2)/(2b^2) = (sin(A+B)+sin(A-B))/(sin(A+B)-sin(A-B))` `=>a^2/b^2 = (2sinAcosB)/(2sinBcosA)` `=>a^2/b^2 = (sinAcosB)/(sinBcosA)` Now, from sine law, `a/sinA = b/sinB = c/sinC = k` `=> a = ksinA, b = ksin B, c = ksinC` So, our equation becomes, `a^2/b^2 = (akcosB)/(bkcosA)` `=>cosB/cosA = a/b` `=>bcosB = acosA` `=>b((a^2+c^2-b^2)/(ac)) = a((b^2+c^2-a^2)/(bc))` `=>b^2a^2+b^2c^2-b^4 = a^2b^2+a^2c^2-a^4` `=>c^2(b^2-a^2) = b^4-a^4` `=>c^2(b^2-a^2) -(b^2+a^2)(b^2-a^2) = 0` `=>(b^2-a^2)(c^2-(b^2+a^2)) = 0` `=>(b^2-a^2) = 0 or (c^2-(b^2+a^2)) = 0` `=> b^2 = a^2 or c^2= (b^2+a^2)` `=>b = a or c^2= (b^2+a^2)` It means, either the `Delta ABC` is a right angle triangle or a isoceles triangle.