In a Delta ABC,frac{cos^2 left( frac{B - C}{2 | vedclass

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(A) We have to find the value of $\frac{\cos^2 \left( \frac{B - C}{2} \right)}{(b + c)^2} + \frac{\sin^2 \left( \frac{B - C}{2} \right)}{(b - c)^2}$.U

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(A) We have to find the value of $\frac{\cos^2 \left( \frac{B - C}{2} \right)}{(b + c)^2} + \frac{\sin^2 \left( \frac{B - C}{2} \right)}{(b - c)^2}$.Using the sine rule,$b = 2R \sin B$ and $c = 2R \sin C$,where $R$ is the circumradius.Substituting these into the expression:$\frac{\cos^2 \left( \frac{B - C}{2} \right)}{(2R(\sin B + \sin C))^2} + \frac{\sin^2 \left( \frac{B - C}{2} \right)}{(2R(\sin B - \sin C))^2}$$= \frac{1}{4R^2} \left[ \frac{\cos^2 \left( \frac{B - C}{2} \right)}{(2 \sin \frac{B+C}{2} \cos \frac{B-C}{2})^2} + \frac{\sin^2 \left( \frac{B - C}{2} \right)}{(2 \cos \frac{B+C}{2} \sin \frac{B-C}{2})^2} \right]$$= \frac{1}{4R^2} \left[ \frac{1}{4 \sin^2 \frac{B+C}{2}} + \frac{1}{4 \cos^2 \frac{B+C}{2}} \right]$$= \frac{1}{16R^2} \left[ \frac{\cos^2 \frac{B+C}{2} + \sin^2 \frac{B+C}{2}}{\sin^2 \frac{B+C}{2} \cos^2 \frac{B+C}{2}} \right]$$= \frac{1}{16R^2} \left[ \frac{1}{\sin^2 \frac{B+C}{2} \cos^2 \frac{B+C}{2}} \right] = \frac{1}{4R^2} \left[ \frac{1}{\sin^2 (B+C)} \right]$Since $A+B+C = \pi$,$\sin(B+C) = \sin(\pi - A) = \sin A$.Thus,the expression becomes $\frac{1}{4R^2 \sin^2 A} = \frac{1}{(2R \sin A)^2} = \frac{1}{a^2}$.

In a $\Delta ABC$,$\frac{\cos^2 \left( \frac{B - C}{2} \right)}{(b + c)^2} + \frac{\sin^2 \left( \frac{B - C}{2} \right)}{(b - c)^2} = $ (in $/ a^2$)

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