If A, B, C are in A.P.,


If $A, B, C$ are in A.P., then $\frac{\sin A-\sin C}{\cos C-\cos A}=$

(a) tan B

(b) cot B

(c) tan 2 B

(d) None of these


(b) cot B

Since A,B and C are in A.P,



or, $2 B=A+C$

$\frac{\sin A-\sin C}{\cos C-\cos A}$


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

$\left[\because \sin A-\sin B=2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)\right.$ and $\left.\cos A-\cos B=-2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$

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

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


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

$=\frac{\cos B}{\sin B}$


$=\cot B$

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