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Solution:

Given: $\angle C=90^{\circ}$

Need to prove: $\sin (A-B)=\frac{\left(a^{2}-b^{2}\right)}{\left(a^{2}+b^{2}\right)}$

Here, $\angle C=90^{\circ} ; \sin C=1$

So, it is a Right-angled triangle.

And also, $a^{2}+b^{2}=c^{2}$

Now

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

We know that, $\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}}=2 \mathrm{R}$ where $\mathrm{R}$ is the circumradius.

Therefore,

$=\frac{4 R^{2} \sin ^{2} C}{4 R^{2} \sin ^{2} A-4 R^{2} \sin ^{2} B} \sin (A-B)=\frac{\sin (A-B)}{\sin ^{2} A-\sin ^{2} B}[A s, \sin C=1]$

$=\frac{\sin (A-B)}{(\sin A+\sin B)(\sin A-\sin B)}=\frac{\sin (A-B)}{\left[2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}\right]\left[2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}\right]}$

$=\frac{\sin (A-B)}{2 \sin \frac{A+B}{2} \cos \frac{A+B}{2} .2 \sin \frac{A-B}{2} \cos \frac{A-B}{2}}=\frac{\sin (A-B)}{\sin (A+B) \sin (A-B)}$

$=\frac{1}{\sin (\pi-C)}=\frac{1}{\sin C}=1$

Therefore,

$\Rightarrow \frac{a^{2}+b^{2}}{a^{2}-b^{2}} \sin (A-B)=1$

$\Rightarrow \sin (A-B)=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$ [Proved]