Question

If $A + B + C = {180^ \circ }$, then $\sum\limits_{}^{} {\tan \dfrac{A}{2}\tan \dfrac{B}{2}} $ is equal to:

A.0
B.1
C.2
D.3

Answer 1

Hint: Divide the equation $A + B + C = {180^ \circ }$by 2 and then take tangent on both sides of the equation and solve the question then.

Complete step-by-step answer:

Given in the question, $A + B + C = {180^ \circ }$
Dividing the given equation by 2 on both sides, we get-
$
  \dfrac{A}{2} + \dfrac{B}{2} + \dfrac{C}{2} = {90^ \circ } \\
   \Rightarrow \dfrac{A}{2} + \dfrac{B}{2} = {90^ \circ } - \dfrac{C}{2} \\
 $

Taking tangent both sides, we get-
$
   \Rightarrow \tan \left( {\dfrac{A}{2} + \dfrac{B}{2}} \right) = \tan \left( {{{90}^ \circ } - \dfrac{C}{2}} \right) \\
   \Rightarrow \tan \left( {\dfrac{A}{2} + \dfrac{B}{2}} \right) = \cot \left( {\dfrac{C}{2}} \right) \\
 $
{Since, $\tan \left( {{{90}^ \circ } - \dfrac{C}{2}} \right) = \cot \dfrac{C}{2}$}
Now using the formula, $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$, we get-
$\dfrac{{\tan \dfrac{A}{2} + \tan \dfrac{B}{2}}}{{1 - \tan \dfrac{A}{2}\tan \dfrac{B}{2}}} = \dfrac{1}{{\tan \dfrac{C}{2}}}$ $\left\{ {\because \cot \dfrac{C}{2} = \dfrac{1}{{\tan \dfrac{C}{2}}}} \right\}$

Therefore, this implies that, $\sum\limits_{}^{} {\tan \dfrac{A}{2}\tan \dfrac{B}{2}} = 1$.

Hence, the correct option is (B).

Note- On solving such types of questions, always focus on the conditions given. As mentioned in the solution, using the equation, $A + B + C = {180^ \circ }$, and then modifying it further, we have solved the question by proceeding step by step and by using standard results like, $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$