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

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Question

(A) Given $\angle A = 90^{\circ}$, then $B + C = 90^{\circ}$, so $C = 90^{\circ} - B$.Using the sine rule, $\frac{b}{\sin B} = \frac{c}{\sin C} = a$,

Vedclass · Accepted Answer

$\frac{b^2 - c^2}{a^2}$

Vedclass · Answer

(A) Given $\angle A = 90^{\circ}$, then $B + C = 90^{\circ}$, so $C = 90^{\circ} - B$.Using the sine rule, $\frac{b}{\sin B} = \frac{c}{\sin C} = a$, where $a$ is the hypotenuse.Thus, $b = a \sin B$ and $c = a \sin C = a \sin(90^{\circ} - B) = a \cos B$.Now, $\sin(B - C) = \sin B \cos C - \cos B \sin C$.Since $C = 90^{\circ} - B$, $\cos C = \sin B$ and $\sin C = \cos B$.So, $\sin(B - C) = \sin^2 B - \cos^2 B$.From $b = a \sin B$ and $c = a \cos B$, we have $\sin B = \frac{b}{a}$ and $\cos B = \frac{c}{a}$.Substituting these, $\sin(B - C) = (\frac{b}{a})^2 - (\frac{c}{a})^2 = \frac{b^2 - c^2}{a^2}$.

In $\triangle ABC$, if $\angle A = 90^{\circ}$, then $\sin(B - C) =$

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