Question

Given $S=a^2-{(b-c)}^2$ in a triangle $ABC$ whose area is $S$, then the value of $\tan \frac{A}{2}$ is ..

Answer 1

Ans. $\frac{1}{4}$

$S=\frac{1}{2}bc\sin A=(a+b-c)(a-b+c)$

or $\sin \frac{A}{2}\cos \frac{A}{2}=\frac{2(s-b).2(s-c)}{bc}=4{\sin }^2\frac{A}{2}$

$\therefore \tan \frac{A}{2}=\frac{1}{4}$.