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In a triangle \(\mathrm{ABC}, \angle \mathrm{C}=90^{\circ}\), then \(\frac{a^2-b^2}{a^2+b^2}\) is equal to :
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The correct answer is:
\(\sin (A-B)\)
\(\begin{aligned}
& \mathrm{A}+\mathrm{B}=180^{\circ}-\mathrm{C}=90^{\circ} \\
& \mathrm{a}=2 \mathrm{R} \sin \mathrm{A}, b=2 \mathrm{R} \sin \mathrm{B}, c=2 \mathrm{R} \sin \mathrm{C} \\
& \begin{array}{l}
\therefore \frac{a^2-b^2}{a^2+b^2}=\frac{\sin ^2 \mathrm{~A}-\sin ^2 \mathrm{~B}}{\sin ^2 \mathrm{~A}+\sin ^2 \mathrm{~B}} \\
\quad=\frac{\sin (\mathrm{A}+\mathrm{B}) \sin (\mathrm{A}-\mathrm{B})}{\sin ^2 \mathrm{~A}+\sin ^2\left(90^{\circ}-\mathrm{A}\right)} \\
{\left[\because \mathrm{A}+\mathrm{B}=90^{\circ}\right]} \\
=\frac{\sin 90^{\circ} \sin (\mathrm{A}-\mathrm{B})}{\sin ^2 \mathrm{~A}+\cos ^2 \mathrm{~A}}=\sin (\mathrm{A}-\mathrm{B})
\end{array}
\end{aligned}\)
& \mathrm{A}+\mathrm{B}=180^{\circ}-\mathrm{C}=90^{\circ} \\
& \mathrm{a}=2 \mathrm{R} \sin \mathrm{A}, b=2 \mathrm{R} \sin \mathrm{B}, c=2 \mathrm{R} \sin \mathrm{C} \\
& \begin{array}{l}
\therefore \frac{a^2-b^2}{a^2+b^2}=\frac{\sin ^2 \mathrm{~A}-\sin ^2 \mathrm{~B}}{\sin ^2 \mathrm{~A}+\sin ^2 \mathrm{~B}} \\
\quad=\frac{\sin (\mathrm{A}+\mathrm{B}) \sin (\mathrm{A}-\mathrm{B})}{\sin ^2 \mathrm{~A}+\sin ^2\left(90^{\circ}-\mathrm{A}\right)} \\
{\left[\because \mathrm{A}+\mathrm{B}=90^{\circ}\right]} \\
=\frac{\sin 90^{\circ} \sin (\mathrm{A}-\mathrm{B})}{\sin ^2 \mathrm{~A}+\cos ^2 \mathrm{~A}}=\sin (\mathrm{A}-\mathrm{B})
\end{array}
\end{aligned}\)
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