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If $\Delta=a^2-(b-c)^2$, is the area of the $\triangle A B C$, then $\tan A$ is equal to
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$\frac{8}{15}$
$\Delta=a^2-(b-c)^2$
$\Rightarrow \quad \Delta=(a+b-c)(a-b+c)$
$\Rightarrow \quad \Delta=(2 s-c-c)(2 s-b-b)$ $(\because 2 s=a+b+c)$
$\Rightarrow \quad \Delta=(2 s-2 c)(2 s-2 b)$
$\Rightarrow \quad \Delta=4(s-b)(s-c)$
$\because \Delta=\sqrt{s(s-a)(s-b)(s-c)}$
$\Rightarrow \sqrt{s(s-a)(s-b)(s-c)}=4(s-b)(s-c)$
$\Rightarrow \quad \sqrt{s(s-a)}=4 \sqrt{(s-b)(s-c)}$
$\Rightarrow \quad \frac{1}{4}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$
$\because \quad \tan \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$
$\Rightarrow \quad \tan \frac{A}{2}=\frac{1}{4}$
$\therefore \tan A=\frac{2 \tan \frac{A}{2}}{1-\tan ^2 \frac{A}{2}}=\frac{2 \times \frac{1}{4}}{1-\frac{1}{16}} \Rightarrow \frac{\frac{1}{2}}{\frac{15}{16}}$
$\Rightarrow \quad \tan A=\frac{8}{15}$
$\Rightarrow \quad \Delta=(a+b-c)(a-b+c)$
$\Rightarrow \quad \Delta=(2 s-c-c)(2 s-b-b)$ $(\because 2 s=a+b+c)$
$\Rightarrow \quad \Delta=(2 s-2 c)(2 s-2 b)$
$\Rightarrow \quad \Delta=4(s-b)(s-c)$
$\because \Delta=\sqrt{s(s-a)(s-b)(s-c)}$
$\Rightarrow \sqrt{s(s-a)(s-b)(s-c)}=4(s-b)(s-c)$
$\Rightarrow \quad \sqrt{s(s-a)}=4 \sqrt{(s-b)(s-c)}$
$\Rightarrow \quad \frac{1}{4}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$
$\because \quad \tan \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$
$\Rightarrow \quad \tan \frac{A}{2}=\frac{1}{4}$
$\therefore \tan A=\frac{2 \tan \frac{A}{2}}{1-\tan ^2 \frac{A}{2}}=\frac{2 \times \frac{1}{4}}{1-\frac{1}{16}} \Rightarrow \frac{\frac{1}{2}}{\frac{15}{16}}$
$\Rightarrow \quad \tan A=\frac{8}{15}$
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