If $\Delta=a^2-(b-c)^2$, is the area of the $ riangle A B C$, then $ an A$ is equal to

Question

If $\Delta=a^2-(b-c)^2$, is the area of the $	riangle A B C$, then $	an A$ is equal to

Aicharya · Accepted Answer

$\Delta=a^2-(b-c)^2$ $\Rightarrow \quad \Delta=(a+b-c)(a-b+c)$ $\Rightarrow \quad \Delta=(2 s-c-c)(2 s-b-b)$ $(\because 2 s=a+b+c)$ $\Rightarrow \quad \Delta=(2 s-2 c)(2 s-2 b)$ $\Rightarrow \quad \Delta=4(s-b)(s-c)$ $\because \Delta=\sqrt{s(s-a)(s-b)(s-c)}$ $\Rightarrow \sqrt{s(s-a)(s-b)(s-c)}=4(s-b)(s-c)$ $\Rightarrow \quad \sqrt{s(s-a)}=4 \sqrt{(s-b)(s-c)}$ $\Rightarrow \quad \frac{1}{4}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ $\because \quad 	an \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ $\Rightarrow \quad 	an \frac{A}{2}=\frac{1}{4}$ $	herefore 	an A=\frac{2 	an \frac{A}{2}}{1-	an ^2 \frac{A}{2}}=\frac{2 	imes \frac{1}{4}}{1-\frac{1}{16}} \Rightarrow \frac{\frac{1}{2}}{\frac{15}{16}}$ $\Rightarrow \quad 	an A=\frac{8}{15}$

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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