Q68 / 160 · AP EAMCET 2024 (22 May Shift 2) · Mathematics

In $\triangle \mathrm{ABC},\left(r_1+r_2\right) \operatorname{cosec}^2 \frac{c}{2}=$

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

Solution

$\begin{aligned} & \text {In } \triangle \mathrm{ABC}, \mathrm{~A}+\mathrm{B}+\mathrm{C}=\pi \\ & \because r_1+r_2=4 \mathrm{R} \sin \frac{\mathrm{~A}}{2} \cdot \cos \frac{\mathrm{~B}}{2} \cdot \cos \frac{\mathrm{C}}{2}+4 \mathrm{R} \cos \frac{\mathrm{~A}}{2} \cdot \sin \frac{\mathrm{~B}}{2} \cos \frac{\mathrm{C}}{2} \\ &= 4 \mathrm{R} \cos \quad \frac{\mathrm{C}}{2}\left[\sin \frac{\mathrm{~A}}{2} \cdot \cos \frac{\mathrm{~B}}{2}+\cos \frac{\mathrm{A}}{2} \cdot \sin \frac{\mathrm{~B}}{2}\right]=4 \mathrm{R} \cos \frac{\mathrm{C}}{2}\left[\sin \frac{\mathrm{~A}+\mathrm{B}}{2}\right] \\ &= 4 \mathrm{R} \cos \frac{\mathrm{C}}{2}\left[\sin \left(\frac{\pi}{2}-\frac{\mathrm{C}}{2}\right)\right]=4 \mathrm{R} \cos ^2 \frac{\mathrm{C}}{2} \end{aligned}$
Now, $\left(r_1+r_2\right) \operatorname{cosec}^2 \frac{\mathrm{C}}{2}=\frac{4 \mathrm{R} \cos ^2 \frac{\mathrm{C}}{2}}{\sin ^2 \frac{\mathrm{C}}{2}}=4 \mathrm{R} \cot ^2 \frac{\mathrm{C}}{2}$

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

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