In Delta ABC,frac{sin(A - B)}{sin(A + B)} = | vedclass

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(A) We know that in $\Delta ABC$,$A + B + C = 180^{\circ}$,so $\sin(A + B) = \sin(180^{\circ} - C) = \sin C$. Thus,$\frac{\sin(A - B)}{\sin(A + B)} =

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) We know that in $\Delta ABC$,$A + B + C = 180^{\circ}$,so $\sin(A + B) = \sin(180^{\circ} - C) = \sin C$. Thus,$\frac{\sin(A - B)}{\sin(A + B)} = \frac{\sin A \cos B - \sin B \cos A}{\sin C}$. Using the sine rule,$\sin A = \frac{a}{2R}$,$\sin B = \frac{b}{2R}$,and $\sin C = \frac{c}{2R}$. Substituting these,we get $\frac{\frac{a}{2R} \cos B - \frac{b}{2R} \cos A}{\frac{c}{2R}} = \frac{a \cos B - b \cos A}{c}$. Using the projection formula,$a = b \cos C + c \cos B$ and $b = a \cos C + c \cos A$,we have $a \cos B - b \cos A = (b \cos C + c \cos B) \cos B - (a \cos C + c \cos A) \cos A$. Alternatively,using the cosine rule: $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$ and $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$. Substituting these: $\frac{a(\frac{a^2 + c^2 - b^2}{2ac}) - b(\frac{b^2 + c^2 - a^2}{2bc})}{c} = \frac{\frac{a^2 + c^2 - b^2}{2c} - \frac{b^2 + c^2 - a^2}{2c}}{c} = \frac{a^2 + c^2 - b^2 - b^2 - c^2 + a^2}{2c^2} = \frac{2a^2 - 2b^2}{2c^2} = \frac{a^2 - b^2}{c^2}$.

In $\Delta ABC$,$\frac{\sin(A - B)}{\sin(A + B)} = $

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