Question

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

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

Answer 1

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

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

Since
$sec({90}^{\circ }-\theta )=cosec\theta$

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


$\because sec({90}^{\circ }-\frac{A}{2})=cosec\frac{A}{2}$

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