In a triangle ABC,if frac{sin A - sin C}{cos | vedclass

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

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Arithmetic progression

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(D) Given the equation: $\frac{\sin A - \sin C}{\cos C - \cos A} = \cot B$ Applying the sum-to-product formulas: $\frac{2 \cos \left(\frac{A+C}{2}\right) \sin \left(\frac{A-C}{2}\right)}{2 \sin \left(\frac{A+C}{2}\right) \sin \left(\frac{A-C}{2}\right)} = \cot B$ Simplifying the expression: $\cot \left(\frac{A+C}{2}\right) = \cot B$ This implies: $\frac{A+C}{2} = B \implies A+C = 2B$ Since the sum of two angles is twice the third angle,$A, B, C$ are in Arithmetic Progression $(A.P.)$.

In a triangle $ABC$,if $\frac{\sin A - \sin C}{\cos C - \cos A} = \cot B$,then $A, B, C$ are in

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