In any triangle. `if(a^2-b^2)/(a^2+b^2)=("sin"(A-B))/("sin"(A+B))` , then prove that the triangle is either right angled or isosceles.

Explanation

`(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`

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