If $A, B$ and $C$ are interior angles of a triangle $ABC$, then the value of $\tan \bigg(\large\frac{B+C}{2}\bigg)$

Question

$\begin{array}{1 1}(A) \; \tan \large\frac{A}{2}\\(B)\;\sin \large\frac{A}{2} \\(C)\;\cot \large\frac{A}{2} \\(D)\; none\;of\;these \end{array}$

Answer 1

Solution :

According to question

$A+B+C= 180^{\circ}$

=> $ \large\frac{A+B+C}{2} $$=90^{\circ}$

=> $ \large\frac{A}{2}+\frac{B+C}{2} =90^{\circ}$

=> $ \tan \bigg( \large\frac{B+C}{2} \bigg) =$$ \tan \bigg(90^{\circ} - \frac{A}{2}\bigg)$

=> $\tan \bigg( \large\frac{B+C}{2}\bigg) =\cot \large\frac{A}{2}$