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If in a $\Delta \mathrm{ABC}, \angle \mathrm{C}=90^{0}$, then prove that $\sin (\mathrm{A}-\mathrm{B})=\frac{\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)}{\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right)}$
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]
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