If A+B+C+D=2 pi,then cos A-cos B+cos C-cos D= | vedclass

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(C) Given $A+B+C+D=2 \pi$. We have $\cos A - \cos B + \cos C - \cos D = (\cos A - \cos B) + (\cos C - \cos D)$. Using the formula $\cos X - \cos Y = -

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$-4 \sin \frac{A+B}{2} \sin \frac{A+C}{2} \sin \frac{A+D}{2}$

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(C) Given $A+B+C+D=2 \pi$. We have $\cos A - \cos B + \cos C - \cos D = (\cos A - \cos B) + (\cos C - \cos D)$. Using the formula $\cos X - \cos Y = -2 \sin \frac{X+Y}{2} \sin \frac{X-Y}{2}$,we get: $= -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2} - 2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$. Since $A+B+C+D=2 \pi$,we have $\frac{C+D}{2} = \pi - \frac{A+B}{2}$,so $\sin \frac{C+D}{2} = \sin \frac{A+B}{2}$. Also,$\frac{C-D}{2} = \frac{(2 \pi - A - B) - 2D}{2} = \pi - \frac{A+B+2D}{2}$. Using the identity,the expression simplifies to: $-4 \sin \frac{A+B}{2} \sin \frac{A+C}{2} \sin \frac{A+D}{2}$.

If $A+B+C+D=2 \pi$,then $\cos A-\cos B+\cos C-\cos D=$

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