Question

The area of a $\Delta$ $ABC$ is given by $=a^2-(b-c{)}^2$ then $\tan \frac{A}{2}$ is equal to

• A. $-1$
• B. $0$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

$\frac{1}{4}$

Given $a,b,c$ are sides of a triangle

Formulae used :

$s=\frac{a+b+c}{2}$

$\Delta =\sqrt{s\times (s-a)\times (s-b)\times (s-c)}$

$\tan \left(\frac{A}{2}\right)=\frac{\Delta }{s\times (s-a)}$

But $\Delta =a^2-(b-c{)}^2$ .....(Given)

$=4\times (s-b)(s-c)$

$\Rightarrow$ $\sqrt{s\times (s-a)}=4\times \sqrt{(s-b)\times (s-c)}$

Thus, $\Delta =\frac{1}{4}\times s\times (s-a)$

Therefore, $\tan \left(\frac{A}{2}\right)=\frac{\Delta }{s\times (s-a)}=\frac{1}{4}$