If frac{sin A - sin C}{cos C - cos A} = cot B | vedclass

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(A) Given: $\frac{\sin A - \sin C}{\cos C - \cos A} = \cot B$Using the sum-to-product formulas:$\sin A - \sin C = 2 \cos \frac{A + C}{2} \sin \frac{A

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$A.P.$

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(A) Given: $\frac{\sin A - \sin C}{\cos C - \cos A} = \cot B$Using the sum-to-product formulas:$\sin A - \sin C = 2 \cos \frac{A + C}{2} \sin \frac{A - C}{2}$$\cos C - \cos A = 2 \sin \frac{A + C}{2} \sin \frac{A - C}{2}$Substituting these into the equation:$\frac{2 \cos \frac{A + C}{2} \sin \frac{A - C}{2}}{2 \sin \frac{A + C}{2} \sin \frac{A - C}{2}} = \cot B$$\cot \frac{A + C}{2} = \cot B$$\frac{A + C}{2} = B$$A + C = 2B$This condition implies that $A, B, C$ are in $A.P.$

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

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