MCQ
$\tan 100^\circ + \tan 125^\circ + \tan 100^\circ \tan 125^\circ = $
- A$0$
- B$1/2$
- C$-1$
- ✓$1$
$\therefore $ $\tan \,{225^o} = \frac{{\tan \,\,{{100}^o} + \tan \,\,{{125}^o}}}{{1 - \tan \,{{100}^o}\,\tan \,{{125}^o}}}$
$i.e.$, $1 = \frac{{\tan \,{{100}^o} + \tan \,\,{{125}^o}}}{{1 - \tan \,\,{{100}^o}\,\tan \,\,{{125}^o}}}$
$i.e.$,$\tan {100^o} + \tan {125^o} + \tan {100^o}\tan {125^o} = 1.$
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