If $A B C$ is a right angled triangle with $90^{\circ}$ at $C$ and $ab$, then $\frac{a^2+b^2}{a^2-b^2} \sin (A-B)=$

Question

If $A B C$ is a right angled triangle with $90^{\circ}$ at $C$ and $a>b$, then $\frac{a^2+b^2}{a^2-b^2} \sin (A-B)=$, $\frac{3}{2}$, 1, $\frac{1}{2}$, 0

MARKS App · Accepted Answer

On applying sine rule, we get $\begin{aligned} & \frac{\sin ^2 A+\sin ^2 B}{\sin ^2 A-\sin ^2 B} \sin (A-B) \ & =\frac{\sin ^2 A+\sin ^2 B}{\sin (A+B) \sin (A-B)} \sin (A-B) \ & {\left[\because \sin ^2 A-\sin ^2 B=\sin (A+B) \sin (A-B)\right]} \ & =\frac{\sin ^2 A+\cos ^2 A}{\sin (\pi-C)} \ & {[\because A+B+C=\pi \text { and } C=\pi / 2 \text { given }]} \ & =\frac{1}{\sin C}=1 \ & {\left[\because C=90^{\circ}\right]} \end{aligned}$ Hence, option (b) is correct.

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