Question

If $A+C=2B$, then $\frac{\cos C-\cos A}{\sin A-\sin C}$ is equal to

• A. $cotB$
• B. $cot2B$
• C. $\tan 2B$
• D. $\tan B$

Answer 1

The correct option is D

$\tan B$

Explanation for the correct option:

Step 1. Find the value of $\frac{\cos C-\cos A}{\sin A-\sin C}$:

Given,

$A+C=2B$

Now,

$\begin{array}{rcl}\frac{\cos C-\cos A}{\sin A-\sin C}& =& \frac{-(\cos A–\cos C)}{\sin A–\sin C}\end{array}$

$\begin{array}{rcl}& =& \frac{-\left[-2\sin {\displaystyle \frac{(A+C)}{2}}\sin {\displaystyle \frac{(A–C)}{2}}\right]}{2\cos \frac{(A+C)}{2}\sin \frac{(A–C)}{2}}\end{array}$

Step 2. Use the identities:$\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\theta }\mathbf{}\mathbf{-}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\varphi }\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{2}\mathit{s}\mathit{i}\mathit{n}\frac{(\theta +\varphi )}{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\frac{(\theta -\varphi )}{\mathbf{2}}\mathbf{,}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\theta }\mathbf{}\mathbf{-}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\varphi }\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathit{c}\mathit{o}\mathit{s}\frac{(\theta +\varphi )}{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\frac{(\theta -\varphi )}{\mathbf{2}}$

$\begin{array}{rcl}\frac{\cos C-\cos A}{\sin A-\sin C}& =& \frac{\sin {\displaystyle \frac{(A+C)}{2}}}{\cos \frac{(A+C)}{2}}\\ & =& \frac{\sin \left({\displaystyle \frac{2B}{2}}\right)}{\cos \left(\frac{2B}{2}\right)}\\ & =& \frac{\sin B}{\cos B}\\ & =& \tan B\end{array}$

$\therefore$$\frac{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{C}\mathbf{}\mathbf{-}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{A}}{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{A}\mathbf{}\mathbf{-}\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{C}}\mathbf{}\mathbf{=}\mathbf{}\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{B}$

Hence, Option ‘D’ is Correct.