In a triangle ABC,if angle C = 90^{circ},then | vedclass

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(B) Given $\angle C = 90^{\circ}$,we have $A+B = 90^{\circ}$.Using the sine rule,$a = k \sin A$,$b = k \sin B$,and $c = k \sin C = k \sin 90^{\circ} =

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$\sin (A-B)$

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(B) Given $\angle C = 90^{\circ}$,we have $A+B = 90^{\circ}$.Using the sine rule,$a = k \sin A$,$b = k \sin B$,and $c = k \sin C = k \sin 90^{\circ} = k$.Then,$\frac{a^2-b^2}{a^2+b^2} = \frac{k^2 \sin^2 A - k^2 \sin^2 B}{k^2 \sin^2 A + k^2 \sin^2 B} = \frac{\sin^2 A - \sin^2 B}{\sin^2 A + \sin^2 B}$.Since $B = 90^{\circ} - A$,we have $\sin B = \cos A$ and $\cos B = \sin A$.Substituting these,$\frac{\sin^2 A - \cos^2 A}{\sin^2 A + \cos^2 A} = \frac{-(\cos^2 A - \sin^2 A)}{1} = -\cos 2A$.Alternatively,using the identity $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$:$\frac{\sin(A+B)\sin(A-B)}{\sin^2 A + \cos^2 A} = \sin(90^{\circ})\sin(A-B) = 1 \cdot \sin(A-B) = \sin(A-B)$.

In a $\triangle ABC$,if $\angle C = 90^{\circ}$,then $\frac{a^2-b^2}{a^2+b^2}$ is equal to

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