Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

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

• A. $2\sin \left(\frac{A-C}{2}\right)$
• B. $2\cos \left(\frac{A-C}{2}\right)$
• C. $\cos \left(\frac{A-C}{2}\right)$
• D. $\sin \left(\frac{A-C}{2}\right)$

Answer 1

The correct option is B $2\cos \left(\frac{A-C}{2}\right)$

Since, $A+B+C=\pi$

But $2B=A+C$

$\therefore 3B=\pi \Rightarrow B=\pi /3$

$\frac{a+c}{b}=\frac{\sin A+\sin C}{\sin B}$

$=\frac{2\sin \left(\frac{A+C}{2}\right)\cos \left(\frac{A-C}{2}\right)}{\sin \frac{\pi }{3}}$

$=\frac{2\sin \frac{\pi }{3}\cos \left(\frac{A-C}{2}\right)}{\sin \frac{\pi }{3}}$

$=2\cos \left(\frac{A-C}{2}\right)$