MCQ
If $A + B + C = {180^o},$ then $\sum {\tan \frac{A}{2}\tan \frac{B}{2} = } $
- A$0$
- ✓$1$
- C$2$
- D$3$
==> $\frac{A}{2} = \frac{\pi }{2} - \left( {\frac{{B + C}}{2}} \right)$
$\therefore$ $\cot \frac{A}{2} = \tan \left( {\frac{B}{2} + \frac{C}{2}} \right)$
==> $\frac{1}{{\tan \frac{A}{2}}} = \frac{{\tan \frac{B}{2} + \tan \frac{C}{2}}}{{1 - \tan \frac{B}{2}\tan \frac{C}{2}}}$
==> $1 - \tan \frac{B}{2}\tan \frac{C}{2} $
$= \tan \frac{A}{2}.\tan \frac{B}{2} + \tan \frac{A}{2}.\tan \frac{C}{2}$
$\tan \frac{A}{2}.\tan \frac{B}{2} + \tan \frac{B}{2}\tan \frac{C}{2} + \tan \frac{A}{2}\tan \frac{C}{2} = 1$
$i.e.$, $\sum {\tan \frac{A}{2}\tan \frac{B}{2} = 1} $.
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($A$) $\left(\frac{1}{3}, \frac{1}{\sqrt{3}}\right)$ ($B$) $\left(\frac{1}{4}, \frac{1}{2}\right)$ ($C$) $\left(\frac{1}{3},-\frac{1}{\sqrt{3}}\right)$ ($D$) $\left(\frac{1}{4},-\frac{1}{2}\right)$