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).^@


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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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