If angles A, B and C are in A.P.,then frac{a+ | vedclass

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(B) Given that angles $A, B, C$ are in $A$.$P$.,we have $2B = A+C$. Since $A+B+C = 180^{\circ}$, we get $3B = 180^{\circ}$, so $B = 60^{\circ}$.Using

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$2 \cos \frac{A-C}{2}$

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(B) Given that angles $A, B, C$ are in $A$.$P$.,we have $2B = A+C$. Since $A+B+C = 180^{\circ}$, we get $3B = 180^{\circ}$, so $B = 60^{\circ}$.Using the Sine Rule, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$, we have $a = k \sin A, b = k \sin B, c = k \sin C$.Therefore, $\frac{a+c}{b} = \frac{\sin A + \sin C}{\sin B}$.Using the sum-to-product formula, $\sin A + \sin C = 2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right)$.Since $A+C = 2B$, we have $\frac{A+C}{2} = B$.Thus, $\frac{a+c}{b} = \frac{2 \sin B \cos \left(\frac{A-C}{2}\right)}{\sin B} = 2 \cos \left(\frac{A-C}{2}\right)$.

If angles $A, B$ and $C$ are in $A$.$P$.,then $\frac{a+c}{b}$ is equal to

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