Question
यदि $A + B + C = {180^o},$ तब $\sum {\tan \frac{A}{2}\tan \frac{B}{2} = } $
==> $\frac{A}{2} = \frac{\pi }{2} - \left( {\frac{{B + C}}{2}} \right)$
$\therefore$ $\cot \frac{A}{2} = \tan \left( {\frac{B}{2} + \frac{C}{2}} \right)$
==> $\frac{1}{{\tan \frac{A}{2}}} = \frac{{\tan \frac{B}{2} + \tan \frac{C}{2}}}{{1 - \tan \frac{B}{2}\tan \frac{C}{2}}}$
==> $1 - \tan \frac{B}{2}\tan \frac{C}{2} $
$= \tan \frac{A}{2}.\tan \frac{B}{2} + \tan \frac{A}{2}.\tan \frac{C}{2}$
$\tan \frac{A}{2}.\tan \frac{B}{2} + \tan \frac{B}{2}\tan \frac{C}{2} + \tan \frac{A}{2}\tan \frac{C}{2} = 1$
अर्थात्, $\sum {\tan \frac{A}{2}\tan \frac{B}{2} = 1} $.
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कथन$-1$ $: \operatorname{Tr}(A)=0$
कथन$-2$ $:|A|=1$