Q68 / 160 · AP EAMCET 2019 (20 Apr Shift 2) · Mathematics

In $\triangle A B C,\left(r_2+r_3\right) \cot \left(\frac{B+C}{2}\right)=$

  • A
    a+b+c
  • Correct answer
    B
    a
  • C
    b
  • D
    c

Solution

Given, in $\triangle A B C$ $$ \begin{aligned} & \left(r_2+r_3\right) \cot \left(\frac{B+c}{2}\right)=\left(\frac{\Delta}{s-b}+\frac{\Delta}{s-c}\right) \tan \frac{A}{2} \\ & =\frac{\Delta(s-c+s-b)}{(s-b)(s-c)} \frac{\Delta}{s(s-a)} \end{aligned} $$ $$ =2 s-b-c=a $$ Hence, option (b) is correct.

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

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