Question
For $\triangle A B C$, prove that :
$
\tan \left(\frac{B+C}{2}\right)=\cot \frac{A}{2}
$
$
\tan \left(\frac{B+C}{2}\right)=\cot \frac{A}{2}
$
✓
Answer
$
\tan \left(\frac{B+C}{2}\right)=\cot \frac{A}{2}
$
We know that for a triangle $\triangle A B C$
$
\begin{aligned}
& <A+<B+<C=180^{\circ} \\
& <B+<C=180^{\circ}-<A \\
& \frac{<B+<C}{2}=90^{\circ}-\frac{<A}{2} \\
& \tan \left(\frac{B+C}{2}\right)=\tan \left(90^{\circ}-\frac{A}{2}\right) \\
& =\cot \left(\frac{A}{2}\right)
\end{aligned}
$
\tan \left(\frac{B+C}{2}\right)=\cot \frac{A}{2}
$
We know that for a triangle $\triangle A B C$
$
\begin{aligned}
& <A+<B+<C=180^{\circ} \\
& <B+<C=180^{\circ}-<A \\
& \frac{<B+<C}{2}=90^{\circ}-\frac{<A}{2} \\
& \tan \left(\frac{B+C}{2}\right)=\tan \left(90^{\circ}-\frac{A}{2}\right) \\
& =\cot \left(\frac{A}{2}\right)
\end{aligned}
$
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