Q68 / 160 · AP EAMCET 2024 (23 May Shift 1) · Mathematics

In $\triangle \mathrm{ABC},\left(r_2+r_3\right) \operatorname{cosec}^2 \frac{A}{2}=$

  • A
    4 R
  • Correct answer
    B
    $4 R \cot ^2 \frac{A}{2}$
  • C
    $4 R \tan ^2 \frac{A}{2}$
  • D
    $R \tan ^2 \frac{A}{2}$

Solution

$\left(r_2+r_3\right) \operatorname{cosec}^2 \frac{A}{2}=\left(\frac{\Delta}{s-b}+\frac{\Delta}{s-c}\right) \times \frac{1}{\sin ^2 \frac{A}{2}}$ $\begin{aligned} & =\frac{\Delta(2 s-b-c)}{(s-b)(s-c)} \times \frac{b c}{(s-b)(s-c)} \\ & =\frac{\Delta a b c}{(s-b)^2(s-c)^2}=\frac{4 R \Delta^2}{(s-b)^2(s-c)^2} \quad\left[\because R=\frac{a b c}{4 \Delta}\right] \\ & =4 R\left[\cdot \frac{\Delta}{(s-b)(s-c)}\right]^2=4 R \times\left[\frac{1}{\tan \frac{A}{2}}\right]^2=4 R \cot ^2 \frac{A}{2} .\end{aligned}$

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

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