If Delta = a^2 - (b - c)^2,where Delta is the | vedclass

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(B) Given $\Delta = a^2 - (b - c)^2 = a^2 - b^2 - c^2 + 2bc$. Since $a^2 = b^2 + c^2 - 2bc \cos A$,we have $a^2 - b^2 - c^2 = -2bc \cos A$. Substituti

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$\frac{8}{15}$

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(B) Given $\Delta = a^2 - (b - c)^2 = a^2 - b^2 - c^2 + 2bc$. Since $a^2 = b^2 + c^2 - 2bc \cos A$,we have $a^2 - b^2 - c^2 = -2bc \cos A$. Substituting this into the expression for $\Delta$: $\Delta = 2bc - 2bc \cos A = 2bc(1 - \cos A) = 2bc(2 \sin^2 \frac{A}{2}) = 4bc \sin^2 \frac{A}{2}$ .....$(i)$ Also,the area of the triangle is given by $\Delta = \frac{1}{2} bc \sin A = bc \sin \frac{A}{2} \cos \frac{A}{2}$ .....$(ii)$ Dividing $(i)$ by $(ii)$: $\frac{\Delta}{\Delta} = \frac{4bc \sin^2 \frac{A}{2}}{bc \sin \frac{A}{2} \cos \frac{A}{2}} \implies 1 = 4 	an \frac{A}{2}$. Thus,$	an \frac{A}{2} = \frac{1}{4}$. Using the formula $	an A = \frac{2 	an \frac{A}{2}}{1 - 	an^2 \frac{A}{2}}$: $	an A = \frac{2(\frac{1}{4})}{1 - (\frac{1}{4})^2} = \frac{\frac{1}{2}}{1 - \frac{1}{16}} = \frac{\frac{1}{2}}{\frac{15}{16}} = \frac{1}{2} 	imes \frac{16}{15} = \frac{8}{15}$.

If $\Delta = a^2 - (b - c)^2$,where $\Delta$ is the area of triangle $ABC$,then $\tan A$ is equal to

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