If A, B, C are the angles of a Delta ABC,then | vedclass

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(D) Using the cosine rule,we have $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ and $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.Thus,$b^2 + c^2 - a^2 = 2bc \cos A

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$\frac{	an B}{	an A}$

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(D) Using the cosine rule,we have $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ and $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.Thus,$b^2 + c^2 - a^2 = 2bc \cos A$ and $a^2 + c^2 - b^2 = 2ac \cos B$.Substituting these into the given expression:$\frac{c^2 - a^2 + b^2}{a^2 - b^2 + c^2} = \frac{2bc \cos A}{2ac \cos B} = \frac{b \cos A}{a \cos B}$.By the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = 2R$,so $a = 2R \sin A$ and $b = 2R \sin B$.Substituting these values:$= \frac{(2R \sin B) \cos A}{(2R \sin A) \cos B} = \frac{\sin B \cos A}{\sin A \cos B} = \frac{	an B}{	an A}$.

If $A, B, C$ are the angles of a $\Delta ABC$,then with usual notations,$\frac{c^{2}-a^{2}+b^{2}}{a^{2}-b^{2}+c^{2}} = $

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