If $ A, B, C $ are the interior angles of a $ \triangle A B C $, show that:$ \tan \frac{B+C}{2}=\cot \frac{A}{2} $
Given:
\( A, B, C \), are the interior angles of a triangle \( A B C \).
To do:
We have to prove that \( \tan \left(\frac{B+C}{2}\right)=\cot \frac{A}{2} \).
Solution:
We know that,
$tan\ (90^{\circ}- \theta) = cot\ \theta$
Sum of the angles in a triangle is $180^{\circ}$.
This implies,
$\angle A+\angle B+\angle C=180^{\circ}$
$\Rightarrow \frac{\angle A+\angle B+\angle C}{2}=\frac{180^{\circ}}{2}$
$\Rightarrow \frac{\angle A}{2}+ \frac{\angle B}{2}+ \frac{\angle C}{2}=90^{\circ}$
Therefore,
$\tan \left(\frac{B+C}{2}\right)=\tan (\frac{B}{2}+\frac{C}{2})$
$=\tan (90^{\circ}-\frac{A}{2})$
$=\cot \frac{A}{2}$
Hence proved.
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