In any triangle,if $\frac{a^{2} - b^{2}}{a^{2} + b^{2}} = \frac{sin (A - B)}{sin (A + B)}$ then prove that the triangle is either right angled or isosceles.

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Solution

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

using side and angle relation

$\frac{4 R^{2} {sin}^{2} A - 4 R^{2} {sin}^{2} B}{4 R^{2} {sin}^{2} A - 4 R^{2} {sin}^{2} B} = \frac{sin (A - B)}{sin (A + B)}$
$\frac{sin (A + B) sin (A - B)}{{sin}^{2} A + {sin}^{2} B} = \frac{sin (A - B)}{sin (A + B)}$

$\Rightarrow sin (A - B) = 0 o r \frac{sin (π - C)}{{sin}^{2} A + {sin}^{2} B} = \frac{1}{sin (π - C)}$
$A = B$ or $s i n^{2} C = {sin}^{2} A + {sin}^{2} B$

$A = B$ or $c^{2} = a^{2} + b^{2}$
[from the sine rule]
Therefore, the triangle is isosceles or right angled.

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