Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

The value of $ABC,\frac{\tan \frac{A}{2}-\tan \frac{B}{2}}{\tan \frac{A}{2}+\tan \frac{B}{2}}$ is?

• A. $\frac{a-b}{a+b}$
• B. $\frac{a-b}{c}$
• C. $\frac{a-b}{a+b+c}$
• D. $\frac{c}{a+b}$

Answer 1

Answer: (B)

$\frac{a-b}{c}$

$\frac{\tan \frac{A}{2}-\tan \frac{B}{2}}{\tan \frac{A}{2}+\tan \frac{B}{2}}=\frac{\sin (\frac{A}{2}-\frac{B}{2})}{\sin (\frac{A}{2}+\frac{B}{2})}$

$=\frac{\sin (\frac{A}{2}-\frac{B}{2})\sin (\frac{A}{2}+\frac{B}{2})}{{\sin }^2(\frac{A}{2}+\frac{B}{2})}=\frac{{\sin }^2\frac{A}{2}-{\sin }^2\frac{B}{2}}{{\sin }^2\frac{C}{2}}$

$=\frac{\frac{(s-b)(s-c)}{bc}-\frac{(s-a)(s-c)}{ac}}{\frac{s(s-c)}{ab}}$

$=\frac{(s-c)}{c}\left[\frac{a(s-b)-b(s-a)}{ab}\right]\times \frac{ab}{s(s-c)}=\frac{a-b}{c}$