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 C -

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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)$$\sin A - \sin C = 2 \cos\left(\frac{A+C}{2}ight) \sin\left(\frac{A-C}{2}ight)$Substituting these into the expression:$\frac{\cos C-\cos A}{\sin A-\sin C} = \frac{2 \sin\left(\frac{A+C}{2}ight) \sin\left(\frac{A-C}{2}ight)}{2 \cos\left(\frac{A+C}{2}ight) \sin\left(\frac{A-C}{2}ight)}$Canceling the common terms $2$ and $\sin\left(\frac{A-C}{2}ight)$:$= \frac{\sin\left(\frac{A+C}{2}ight)}{\cos\left(\frac{A+C}{2}ight)}$$= 	an\left(\frac{A+C}{2}ight)$Since $A+C=2B$,we have $\frac{A+C}{2} = B$.Therefore,the expression equals $	an B$.

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

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