If A+C=2B,then frac{cos C-cos A}{sin A-sin C} | vedclass

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(D) Given that $A+C=2B$  ... $(i)$ We need to evaluate the expression: $\frac{\cos C-\cos A}{\sin A-\sin C}$ Using the trigonometric identities: $\cos

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$	an B$

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(D) Given that $A+C=2B$  ... $(i)$ We need to evaluate the expression: $\frac{\cos C-\cos A}{\sin A-\sin C}$ Using the trigonometric identities: $\cos C - \cos A = 2 \sin \left(\frac{A+C}{2}ight) \sin \left(\frac{A-C}{2}ight)$ $\sin A - \sin C = 2 \cos \left(\frac{A+C}{2}ight) \sin \left(\frac{A-C}{2}ight)$ Substituting these into the expression: $\frac{2 \sin \left(\frac{A+C}{2}ight) \sin \left(\frac{A-C}{2}ight)}{2 \cos \left(\frac{A+C}{2}ight) \sin \left(\frac{A-C}{2}ight)}$ Canceling the common terms $2$ and $\sin \left(\frac{A-C}{2}ight)$: $= \frac{\sin \left(\frac{A+C}{2}ight)}{\cos \left(\frac{A+C}{2}ight)}$ Since $A+C=2B$,we have $\frac{A+C}{2} = B$: $= \frac{\sin B}{\cos B} = 	an B$

If $A+C=2B$,then $\frac{\cos C-\cos A}{\sin A-\sin C}$ is equal to

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