If the area of triangle ABC is b^2-(c-a)^2,th | vedclass

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(D) Given,the area of triangle $\Delta = b^2-(c-a)^2$. Using the identity $x^2-y^2 = (x-y)(x+y)$,we have $\Delta = (b-c+a)(b+c-a)$. Since $2s = a+b+c$

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$\frac{8}{15}$

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(D) Given,the area of triangle $\Delta = b^2-(c-a)^2$. Using the identity $x^2-y^2 = (x-y)(x+y)$,we have $\Delta = (b-c+a)(b+c-a)$. Since $2s = a+b+c$,we have $b-c+a = 2s-2c$ and $b+c-a = 2s-2a$. Thus,$\Delta = (2s-2c)(2s-2a) = 4(s-a)(s-c)$. By Heron's formula,$\Delta = \sqrt{s(s-a)(s-b)(s-c)}$. Equating the two expressions: $\sqrt{s(s-a)(s-b)(s-c)} = 4(s-a)(s-c)$. Dividing both sides by $\sqrt{(s-a)(s-c)}$,we get $\sqrt{s(s-b)} = 4\sqrt{(s-a)(s-c)}$. Therefore,$	an(\frac{B}{2}) = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}} = \frac{1}{4}$. Using the double angle formula $	an B = \frac{2	an(B/2)}{1-	an^2(B/2)}$,we get $	an B = \frac{2(1/4)}{1-(1/4)^2} = \frac{1/2}{1-1/16} = \frac{1/2}{15/16} = \frac{8}{15}$.

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

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