In $\triangle A B C, \tan \frac{A}{2}+\tan \frac{B}{2}=$

In $\triangle A B C, \tan \frac{A}{2}+\tan \frac{B}{2}=$
  1. $\frac{\cot \frac{C}{2}}{4 s}$
  2. $\frac{2 c \cot \frac{c}{2}}{a+b+c}$
  3. $\frac{2 c \tan \frac{C}{2}}{s}$
  4. $\frac{c \tan \frac{c}{2}}{a+b+c}$

Solution

$ \begin{aligned} & \text { }\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)=\frac{\Delta}{s(s-a)}+\frac{\Delta}{s(s-b)} \\ & =\frac{\Delta}{s}\left\{\frac{1}{s-a}+\frac{1}{s-b}\right\}=\frac{c \Delta}{s(s-a)(s-b)} \\ & =\frac{2 c \Delta}{(a+b+c)(s-a)(s-b)} \\ & =\frac{2 c}{(a+b+c)} \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}=\frac{2 c \cot \frac{c}{2}}{a+b+c} \end{aligned} $

Asked in: AP EAMCET 2018 (22 Apr Shift 1)

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