If Delta = a^2 - (b - c)^2 is the area of the | vedclass

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(B) Given $\Delta = a^2 - (b - c)^2$. Using the identity $x^2 - y^2 = (x - y)(x + y)$,we have $\Delta = (a - (b - c))(a + (b - c)) = (a - b + c)(a + b

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$\frac{8}{15}$

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(B) Given $\Delta = a^2 - (b - c)^2$. Using the identity $x^2 - y^2 = (x - y)(x + y)$,we have $\Delta = (a - (b - c))(a + (b - c)) = (a - b + c)(a + b - c)$. Since $2s = a + b + c$,we have $a + b - c = 2s - 2c$ and $a - b + c = 2s - 2b$. Thus,$\Delta = (2s - 2b)(2s - 2c) = 4(s - b)(s - c)$. We know that $\Delta = \sqrt{s(s - a)(s - b)(s - c)}$. Equating the two expressions: $\sqrt{s(s - a)(s - b)(s - c)} = 4(s - b)(s - c)$. Dividing both sides by $\sqrt{(s - b)(s - c)}$,we get $\sqrt{s(s - a)} = 4\sqrt{(s - b)(s - c)}$. This implies $\sqrt{\frac{(s - b)(s - c)}{s(s - a)}} = \frac{1}{4}$. Since $	an \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}}$,we have $	an \frac{A}{2} = \frac{1}{4}$. Using the formula $	an A = \frac{2 	an \frac{A}{2}}{1 - 	an^2 \frac{A}{2}}$,we get $	an A = \frac{2 	imes \frac{1}{4}}{1 - (\frac{1}{4})^2} = \frac{\frac{1}{2}}{1 - \frac{1}{16}} = \frac{\frac{1}{2}}{\frac{15}{16}} = \frac{8}{15}$.

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

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