If in triangle ABC,frac{a^2 - b^2}{a^2 + b^2} | vedclass

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(C) Given $\frac{a^2 - b^2}{a^2 + b^2} = \frac{\sin(A - B)}{\sin(A + B)}$.Using the sine rule,$a = 2R \sin A$ and $b = 2R \sin B$,we have $\frac{\sin^

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Right angled or isosceles

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(C) Given $\frac{a^2 - b^2}{a^2 + b^2} = \frac{\sin(A - B)}{\sin(A + B)}$.Using the sine rule,$a = 2R \sin A$ and $b = 2R \sin B$,we have $\frac{\sin^2 A - \sin^2 B}{\sin^2 A + \sin^2 B} = \frac{\sin(A - B)}{\sin(A + B)}$.Since $\sin^2 A - \sin^2 B = \sin(A - B)\sin(A + B)$,the equation becomes $\frac{\sin(A - B)\sin(A + B)}{\sin^2 A + \sin^2 B} = \frac{\sin(A - B)}{\sin(A + B)}$.This implies $\sin(A - B) = 0$ or $\frac{\sin(A + B)}{\sin^2 A + \sin^2 B} = \frac{1}{\sin(A + B)}$.If $\sin(A - B) = 0$,then $A = B$,so the triangle is isosceles.If $\sin^2(A + B) = \sin^2 A + \sin^2 B$,then $\sin^2 C = \sin^2 A + \sin^2 B$,which implies $c^2 = a^2 + b^2$,so the triangle is right-angled.Thus,the triangle is right-angled or isosceles.

If in triangle $ABC$,$\frac{a^2 - b^2}{a^2 + b^2} = \frac{\sin(A - B)}{\sin(A + B)}$,then the triangle is

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