### If $A$, $B$, and $C$ are the angles of $\Delta ABC$, show that $cot \bigg( \dfrac { B + C } { 2 } \bigg)$ = $tan \bigg( \dfrac { A } { 2 } \bigg).$

Answer:

Step by Step Explanation:
1. We know that the sum of the angles of a triangle is $180^\circ.$ \begin{aligned} \therefore{\space} & A + B + C = 180^\circ \\ {\implies} &\dfrac { B + C } { 2 } = 90^\circ - \dfrac { A } { 2 } \\ {\implies}&cot \bigg(\dfrac { B + C } { 2 }\bigg) = cot \bigg(90^\circ - \dfrac { A } { 2 }\bigg ) \\ {\implies}&cot \bigg(\dfrac { B + C } { 2 }\bigg) = tan \dfrac { A } { 2 } &&[{\space}\because cot (90^\circ - \theta) = tan \theta {\space}] \\ \end{aligned}
2. Thus, $cot \bigg(\dfrac { B + C } { 2 }\bigg) = tan \dfrac { A } { 2 }$.

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