Question

If the area of a triangle $ABC$ is given by $△=a^2-{(b-c)}^2,$ then $\tan \left(\frac{A}{2}\right)$ is equal to

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

Answer 1

The correct option is C $\frac{1}{4}$

$\because △=a^2-{(b-c)}^2$
$=(a+b-c)(a-b+c)$
We know that $a+b+c=2s$
$△=(2s-c-c)(2s-b-b)$
$=(2s-2c)(2s-2b)$
$=4(s-b)(s-c)$
We have $\sqrt{s(s-a)(s-b)(s-c)}=4(s-b)(s-c)$
Squaring both sides we get
$s(s-a)(s-b)(s-c)={\left(4(s-b)(s-c)\right)}^2$
$s(s-a)=16(s-b)(s-c)$
$\Rightarrow \frac{1}{16}=\frac{(s-b)(s-c)}{s(s-a)}$
$\Rightarrow \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}=\frac{1}{16}$
$\Rightarrow \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}=\frac{1}{4}$
$\Rightarrow \tan \frac{A}{2}=\frac{1}{4}$
$\therefore \tan \frac{A}{2}=\frac{1}{4}$