If the sides of a triangle a, b, c are in A.P | vedclass

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(C) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$. We need to evaluate $a \cos ^2 \frac{C}{2} + c \cos ^2 \frac{A}{2}$. Using the identity $

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$\frac{3b}{2}$

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(C) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$. We need to evaluate $a \cos ^2 \frac{C}{2} + c \cos ^2 \frac{A}{2}$. Using the identity $\cos ^2 	heta = \frac{1 + \cos 2	heta}{2}$,we get: $a \left( \frac{1 + \cos C}{2} ight) + c \left( \frac{1 + \cos A}{2} ight)$ $= \frac{a + a \cos C + c + c \cos A}{2}$ $= \frac{(a + c) + (a \cos C + c \cos A)}{2}$ By the projection formula,$b = a \cos C + c \cos A$. Substituting $a + c = 2b$ and $a \cos C + c \cos A = b$: $= \frac{2b + b}{2} = \frac{3b}{2}$.

If the sides of a triangle $a, b, c$ are in $A.P.$,then with usual notations,$a \cos ^2 \frac{C}{2} + c \cos ^2 \frac{A}{2}$ is

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