In a triangle ABC,with usual notations,if c=4 | vedclass

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(B) We are given the expression: $(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$ Expanding the terms,we get: $(a^2 - 2ab + b^2) \cos^2 \frac

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$16$

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(B) We are given the expression: $(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$ Expanding the terms,we get: $(a^2 - 2ab + b^2) \cos^2 \frac{C}{2} + (a^2 + 2ab + b^2) \sin^2 \frac{C}{2}$ Grouping the terms: $(a^2 + b^2) (\cos^2 \frac{C}{2} + \sin^2 \frac{C}{2}) - 2ab \cos^2 \frac{C}{2} + 2ab \sin^2 \frac{C}{2}$ Since $\cos^2 	heta + \sin^2 	heta = 1$,this simplifies to: $(a^2 + b^2) - 2ab (\cos^2 \frac{C}{2} - \sin^2 \frac{C}{2})$ Using the identity $\cos 2	heta = \cos^2 	heta - \sin^2 	heta$,we have $\cos C = \cos^2 \frac{C}{2} - \sin^2 \frac{C}{2}$: $= a^2 + b^2 - 2ab \cos C$ By the Law of Cosines,$c^2 = a^2 + b^2 - 2ab \cos C$,so: $= c^2$ Given $c = 4$,the value is $4^2 = 16$.

In a triangle $ABC$,with usual notations,if $c=4$,then the value of $(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$ is

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