In Delta ABC,if 2s = a + b + c,then the value | vedclass

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(B) We know the half-angle formulas for a triangle:$\cos^2 \frac{A}{2} = \frac{s(s - a)}{bc}$$\sin^2 \frac{A}{2} = \frac{(s - b)(s - c)}{bc}$Substitut

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$\cos A$

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(B) We know the half-angle formulas for a triangle:$\cos^2 \frac{A}{2} = \frac{s(s - a)}{bc}$$\sin^2 \frac{A}{2} = \frac{(s - b)(s - c)}{bc}$Substituting these into the given expression:$\frac{s(s - a)}{bc} - \frac{(s - b)(s - c)}{bc} = \cos^2 \frac{A}{2} - \sin^2 \frac{A}{2}$Using the identity $\cos(2	heta) = \cos^2 	heta - \sin^2 	heta$,we get:$\cos^2 \frac{A}{2} - \sin^2 \frac{A}{2} = \cos \left( 2 	imes \frac{A}{2} ight) = \cos A$

In $\Delta ABC$,if $2s = a + b + c$,then the value of $\frac{s(s - a)}{bc} - \frac{(s - b)(s - c)}{bc} = $

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