Question
For triangle ABC, show that:
$\frac{\tan (B+C)}{2}=\frac{\cot A}{2}$
$\frac{\tan (B+C)}{2}=\frac{\cot A}{2}$
✓
Answer
We know that for a triangle $\triangle A B C$
$\angle A +\angle B +\angle C =180^{\circ}$
$\angle B +\angle C =180^{\circ}-\angle A $
$ \Rightarrow \frac{\angle B +\angle C }{2}=90-\frac{\angle A }{2} $
$ \Rightarrow \tan \left(\frac{ B + C }{2}\right)=\tan \left(90^{\circ}-\frac{ A }{2}\right)$
$=\cot \left(\frac{ A }{2}\right)$
$\angle A +\angle B +\angle C =180^{\circ}$
$\angle B +\angle C =180^{\circ}-\angle A $
$ \Rightarrow \frac{\angle B +\angle C }{2}=90-\frac{\angle A }{2} $
$ \Rightarrow \tan \left(\frac{ B + C }{2}\right)=\tan \left(90^{\circ}-\frac{ A }{2}\right)$
$=\cot \left(\frac{ A }{2}\right)$
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