Q72 / 160 · TS EAMCET 2007 · Mathematics

In $\triangle A B C,(a+b+c)\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)$ is equal to

  • Correct answer
    A
    $2 c \cot \frac{C}{2}$
  • B
    $2 a \cot \frac{A}{2}$
  • C
    $2 b \cot \frac{\bar{B}}{2}$
  • D
    $\tan \frac{C}{2}$

Solution

We have, $(a+b+c)\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)$ $\begin{aligned} & =2 s\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right) \\ & =2\left(s \tan \frac{A}{2}+s \tan \frac{B}{2}\right) \\ & =2\left(\frac{\Delta}{s-a}+\frac{\Delta}{s-b}\right) \\ & =2 \Delta\left(\frac{s-b+s-a}{(s-a)(s-b)}\right) \\ & =2 \Delta \frac{2 s-(a+b)}{(s-a)(s-b)} \\ & =2 \Delta\left(\frac{c}{(s-a)(s-b)}\right) \\ & =2 c \cot \frac{C}{2}\end{aligned}$

Hence, the correct answer is option: 'A'.

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