If in a triangle $\frac{a^{2} - b^{2}}{a^{2} + b^{2}} = \frac{sin (A - B)}{sin (A + B)}$ ,
Prove that it is either a right angled or an isosceles triangle.

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Solution

$\frac{sin (A - B)}{sin (A + B)} = \frac{k^{2} ({sin}^{2} A - {sin}^{2} B)}{k^{2} ({sin}^{2} A + {sin}^{2} B)}$ by sine formula
or $\frac{sin (A - B)}{sin C} = \frac{sin (A - B) sin (A - B)}{{sin}^{2} A + {sin}^{2} B}$
or $sin (A - B) [\frac{1}{sin C} - \frac{sin C}{{sin}^{2} A + {sin}^{2} B}] = 0$
$∴$ Either $sin (A - B) = 0$ $∴ A = B$ i.e $△$ is isosceles
or ${sin}^{2} A + {sin}^{2} B = {sin}^{2} C$ or $a^{2} + b^{2} = c^{2}$
$∴ △$ is right angled

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