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If angles $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are in A.P., then $\frac{\mathrm{a}+\mathrm{c}}{\mathrm{b}}$ is equal to
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Verified Answer
The correct answer is:
$2 \cos \frac{A-C}{2}$
Hints: $2 \mathrm{~B}=\mathrm{A}+\mathrm{C}$
$$
=\frac{\sin A+\sin C}{\sin B}=\frac{2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right)}{\sin B}=\frac{2 \sin B}{\sin B} \cos \left(\frac{A-C}{2}\right)=2 \cos \left(\frac{A-C}{2}\right)
$$
$$
=\frac{\sin A+\sin C}{\sin B}=\frac{2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right)}{\sin B}=\frac{2 \sin B}{\sin B} \cos \left(\frac{A-C}{2}\right)=2 \cos \left(\frac{A-C}{2}\right)
$$
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