Question

In $△ABC$,
$tan\left(\frac{B+C}{2}\right)=$ _______.

• A. $cot\left(\frac{A}{2}\right)$
• B. $-cot\left(\frac{A}{2}\right)$
• C. $-tan\left(\frac{A}{2}\right)$
• D. $-tan\left(\frac{A}{2}\right)$

Answer 1

The correct option is A $cot\left(\frac{A}{2}\right)$

In a triangle, sum of all the internal angles is 180${}^{\circ }$.
$A+B+C={180}^{\circ }$
$⟹\frac{A+B+C}{2}={90}^{\circ }$
$⟹\frac{B+C}{2}={90}^{\circ }-\frac{A}{2}$
$⟹tan\left(\frac{B+C}{2}\right)=tan({90}^{\circ }-\frac{A}{2})$

Since
$tan({90}^{\circ }-\theta )=cot\theta$

$⟹tan\left(\frac{B+C}{2}\right)=tan({90}^{\circ }-\frac{A}{2})$


$\because tan({90}^{\circ }-\frac{A}{2})=cot\frac{A}{2}$

$⟹tan\left(\frac{B+C}{2}\right)=cot\frac{A}{2}$