with usual notations, if triangle $\mathrm{ABC}$ is right angled at $\mathrm{C}$, then $\left(\frac{\mathrm{a}^{2}+\mathrm{b}^{2}}{\mathrm{a}^{2}-\mathrm{b}^{2}}\right) \sin (\mathrm{A}-\mathrm{B})=$

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Aicharya · Accepted Answer

Given that in $	riangle \mathrm{ABC}, \mathrm{m} \angle \mathrm{C}=90^{\circ} \Rightarrow \mathrm{A}+\mathrm{B}=90^{\circ} \ldots(1)$ Then $\begin{aligned} & \frac{a^{2}+b^{2}}{a^{2}-b^{2}} \sin (A-B) \ =& \frac{(k \sin A)^{2}+(k \sin B)^{2}}{(k \sin A)^{2}-(k \sin B)^{2}} \sin (A-B) \quad \ldots\left[\because \frac{a}{\sin A}=\frac{b}{\sin B}=kight] \ =& \frac{\sin ^{2} A+\sin ^{2} B}{\sin ^{2} A-\sin ^{2} B} \sin (A-B)=\frac{1-\cos 2 A+1-\cos 2 B}{1-\cos 2 A-1+\cos 2 B} \sin (A-B) \ =& \frac{2-(\cos 2 A+\cos 2 B)}{\cos 2 B-\cos 2 A} \sin (A-B) \ =& \frac{2-[2 \cos (A+B) \cdot \cos (A-B)]}{-2 \sin (A+B) \sin (B-A)} \cdot \sin (A-B) \ =& \frac{2-[(0) \cos (A-B)]}{2 \sin (A+B) \sin (A-B)} \cdot \sin (A-B) \ =& \frac{2}{2 \sin (A+B)}=\frac{1}{\sin 90^{\circ}}=1 \end{aligned}$

with usual notations, if triangle $\mathrm{ABC}$ is right angled at $\mathrm{C}$, then…

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