Q61 / 160 · AP EAMCET 2022 (05 Jul Shift 2) · Mathematics

In a triangle $\mathrm{ABC}, \tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}=$

  • A
    $0$
  • Correct answer
    B
    $1$
  • C
    $\frac{1}{2}$
  • D
    $\pi$

Solution

$A+B+C=\pi$ $\begin{aligned} & \frac{A+B}{2}=\frac{\pi-C}{2} \\ & \frac{A}{2}+\frac{B}{2}=\frac{\pi}{2}-\frac{C}{2}\end{aligned}$ $\begin{aligned} \tan \left(\frac{A}{2}+\frac{B}{2}\right) & =\tan \left(\frac{\pi}{2}-\frac{C}{2}\right) \\ & =\cot \frac{C}{2} \ldots(\mathrm{i})\end{aligned}$ Also, $\tan \left(\frac{A}{2}+\frac{B}{2}\right)=\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}$ ...(ii) From (i) and (ii) $\cot \frac{C}{2}=\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}$ $\frac{1}{\tan \frac{c}{2}}=\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}$ $1-\tan \frac{A}{2} \tan \frac{B}{2}=\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}$ $\therefore \tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}=1$

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

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