In a Delta ABC,evaluate frac{b sin(C - A)}{c^ | vedclass

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(B) Given the expression: $E = \frac{b \sin(C - A)}{c^2 - a^2} + \frac{c \sin(A - B)}{a^2 - b^2}$.Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\si

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$\frac{1}{R}$

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(B) Given the expression: $E = \frac{b \sin(C - A)}{c^2 - a^2} + \frac{c \sin(A - B)}{a^2 - b^2}$.Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.Substituting these into the denominator $c^2 - a^2 = 4R^2(\sin^2 C - \sin^2 A) = 4R^2 \sin(C - A) \sin(C + A)$.Since $A + B + C = \pi$,$\sin(C + A) = \sin(\pi - B) = \sin B$.Thus,$c^2 - a^2 = 4R^2 \sin(C - A) \sin B$.Substituting this into the first term: $\frac{b \sin(C - A)}{4R^2 \sin(C - A) \sin B} = \frac{2R \sin B \sin(C - A)}{4R^2 \sin(C - A) \sin B} = \frac{1}{2R}$.Similarly,for the second term: $\frac{c \sin(A - B)}{a^2 - b^2} = \frac{2R \sin C \sin(A - B)}{4R^2 \sin(A - B) \sin C} = \frac{1}{2R}$.Adding them together: $\frac{1}{2R} + \frac{1}{2R} = \frac{1}{R}$.

In a $\Delta ABC$,evaluate $\frac{b \sin(C - A)}{c^2 - a^2} + \frac{c \sin(A - B)}{a^2 - b^2}$.

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