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If $A+C=2 B$, then $\frac{\cos C-\cos A}{\sin A-\sin C}$ is equal to
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Verified Answer
The correct answer is:
$\cot 2 B$
Given that
$\begin{aligned} & \text { Now, } \frac{\cos C-\cos A}{\sin A-\sin C} \\ &= \frac{2 \sin \left(\frac{A+C}{2}\right) \sin \left(\frac{A-C}{2}\right)}{2 \cos \left(\frac{A+C}{2}\right) \sin \left(\frac{A-C}{2}\right)}\end{aligned}$
$\begin{array}{ll}\left.=\frac{2 \sin B}{2 \cos B} \quad \text { [from (i) }\right]\end{array}$
$=\tan B$
$\begin{aligned} & \text { Now, } \frac{\cos C-\cos A}{\sin A-\sin C} \\ &= \frac{2 \sin \left(\frac{A+C}{2}\right) \sin \left(\frac{A-C}{2}\right)}{2 \cos \left(\frac{A+C}{2}\right) \sin \left(\frac{A-C}{2}\right)}\end{aligned}$
$\begin{array}{ll}\left.=\frac{2 \sin B}{2 \cos B} \quad \text { [from (i) }\right]\end{array}$
$=\tan B$
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