Q68 / 160 · TS EAMCET 2022 (19 Jul Shift 1) · Mathematics

In any triangle $\mathrm{ABC}, r^2 \cot \frac{\mathrm{A}}{2} \cot \frac{\mathrm{B}}{2} \cot \frac{\mathrm{C}}{2}=$

  • Correct answer
    A
    $\Delta$
  • B
    $2 \Delta$
  • C
    $\Delta^2$
  • D
    $5 \Delta$

Solution

Given $r^2 \cot \left(\frac{A}{2}\right) \cot \left(\frac{B}{2}\right) \cot \left(\frac{C}{2}\right)$ We have $\cot \left(\frac{\mathrm{A}}{2}\right)=\left(\frac{s-a}{r}\right), \cot \left(\frac{\mathrm{B}}{2}\right)=\left(\frac{s-b}{r}\right)$ formula Then, $r^2\left(\frac{s-a}{r}\right) \cdot\left(\frac{s-b}{r}\right) \cdot \cot \left(\frac{\mathrm{c}}{2}\right)$ Now, Apply $\cot \left(\frac{\mathrm{C}}{2}\right)=\sqrt{\frac{\mathrm{s}(\mathrm{s}-\mathrm{C})}{(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})}}$ So, $r^2 \frac{(s-a)(s-b)}{r^2} \times \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$ $\Rightarrow \sqrt{s(s-a)(s-b)(s-c)}=\Delta$ Therefore, correct option is (a).

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

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