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

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Aicharya · Accepted Answer

Given, $\begin{aligned} & \text { Area of a triangle }=b^2-(c-a)^2 \ \Delta & =b^2-(c-a)^2 \ & =(b+c-a)(b-c+a) \ & =(2 s-2 a)(2 s-2 c) \ & =4(s-a)(s-c) \end{aligned}$ Now $\sqrt{s(s-a)(s-b)(s-c)}=4(s-a)(s-c)$ $\begin{aligned} & \Rightarrow \quad \sqrt{s(s-b)}=4 \sqrt{(s-a)(s-c)} \ & \Rightarrow \quad \frac{1}{4}=\sqrt{\frac{(s-a)(s-c)}{s(s-b)}} \Rightarrow \frac{1}{4}=\tan \left(\frac{B}{2}\right) \ & \therefore \tan B=\frac{2 \tan \left(\frac{B}{2}\right)}{1-\tan ^2\left(\frac{B}{2}\right)}=\frac{2 \times \frac{1}{4}}{1-\frac{1}{16}}=\frac{\frac{1}{2}}{\frac{15}{16}} \ & =\frac{1}{2} \times \frac{16}{15}=\frac{8}{15} \end{aligned}$

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

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