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

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Question

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

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$A.P.$

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(A) Given the equation: $\frac{\sin A - \sin C}{\cos C - \cos A} = \cot B$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$This implies $\frac{A + C}{2} = B$,or $A + C = 2B$.This is the condition for $A, B, C$ to be in $A.P.$ (Arithmetic Progression).

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

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