Q117 / 180 · COMEDK 2013 · Mathematics

If $A+B+C=180^{\circ}$, then $\Sigma \tan \frac{A}{2} \tan \frac{B}{2}$ is equal to

  • A
    $0$
  • Correct answer
    B
    $1$
  • C
    $2$
  • D
    $3$

Solution

We have, $A+B+C=\pi$ $\Rightarrow \quad A+B=\pi-C$ $\Rightarrow \quad \frac{A}{2}+\frac{B}{2}=\frac{\pi}{2}-\frac{C}{2}$ $\Rightarrow \quad \tan \left(\frac{A}{2}+\frac{B}{2}\right)=\tan \left(\frac{\pi}{2}-\frac{C}{2}\right)$ $\Rightarrow \quad \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}=\cot \frac{C}{2}$ $\Rightarrow \quad \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}=\frac{1}{\tan \frac{C}{2}}$ $\Rightarrow \quad \tan \frac{C}{2} \tan \frac{A}{2}+\tan \frac{B}{2} \tan \frac{C}{2}$ $=1-\tan \frac{A}{2} \tan \frac{B}{2}$ $\Rightarrow \quad \tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}=1$ $\Rightarrow \quad \Sigma \tan \frac{A}{2} \tan \frac{B}{2}=1$

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

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