Question

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

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

Answer 1

The correct option is B

$2\cos \frac{(A–C)}{2}$

Explanation for the correct option:

Step 1. Find the value of $\frac{(a+c)}{b}$:

Given, $A,B$ and $C$ are in AP.

$\Rightarrow$$2B=A+C$

$\Rightarrow$ $B=\frac{(A+C)}{2}$ …(i)

As We know,

$\left(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\right)=k$

$a=k\sin A$

$b=k\sin B$

$c=k\sin C$

Step 2. Put the values of $a,b,c$ in given expression:

$\begin{array}{rcl}\therefore \frac{(a+c)}{b}& =& \frac{(k\sin A+k\sin C)}{k\sin B}\\ & =& \frac{(\sin A+\sin C)}{\sin B}\\ & =& \frac{2\sin {\displaystyle \frac{(A+C)}{2}}\cos {\displaystyle \frac{(A–C)}{2}}}{\sin B}\\ & =& \frac{2\sin {\displaystyle \frac{(A+C)}{2}}\cos {\displaystyle \frac{(A–C)}{2}}}{\sin {\displaystyle \frac{(A+C)}{2}}}\\ & =& \mathbf{2}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\frac{\mathbf{(}\mathbf{A}\mathbf{}\mathbf{–}\mathbf{}\mathbf{C}\mathbf{)}}{\mathbf{2}}\end{array}$ $\left[\mathbf{\because }\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\varphi }\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\frac{\left(\theta +\varphi \right)}{\mathbf{2}}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\frac{\left(\theta +\varphi \right)}{\mathbf{2}}\right]$

Hence, Option ‘B’ is Correct.