Question

In \[\Delta ABC\],find \[1 - \tan \dfrac{A}{2}\tan \dfrac{B}{2}\].
A. \[\dfrac{{2c}}{{a + b + c}}\]
B. \[\dfrac{a}{{a + b + c}}\]
C. \[\dfrac{2}{{a + b + c}}\]
D. \[\dfrac{{4a}}{{a + b + c}}\]

Answer 1

Hint: First we will break tangent into cosine and sine. Then apply the half angle formulas of the triangle to get the desired result.

Formula used:
\[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
\[\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b\]
\[\cos \dfrac{A}{2} = \sqrt {\dfrac{{s\left( {s - a} \right)}}{{bc}}} \]
\[\cos \dfrac{B}{2} = \sqrt {\dfrac{{s\left( {s - b} \right)}}{{ca}}} \]
\[\cos \dfrac{C}{2} = \sqrt {\dfrac{{s\left( {s - c} \right)}}{{ab}}} \]
\[\sin \dfrac{A}{2} = \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{bc}}} \]
\[\sin \dfrac{B}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{ac}}} \]
\[\sin \dfrac{C}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{ab}}} \]

Complete step by step solution:
The given expression is
\[1 - \tan \dfrac{A}{2}\tan \dfrac{B}{2}\]
Then apply the formula \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
\[ = 1 - \dfrac{{\sin \dfrac{A}{2}}}{{\cos \dfrac{A}{2}}} \cdot \dfrac{{\sin \dfrac{B}{2}}}{{\cos \dfrac{B}{2}}}\]
\[ = \dfrac{{\cos \dfrac{A}{2}\cos \dfrac{B}{2} - \sin \dfrac{A}{2}\sin \dfrac{B}{2}}}{{\cos \dfrac{A}{2}\cos \dfrac{B}{2}}}\]
Applying the sum of angle of cosine is \[\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b\]
\[ = \dfrac{{\cos \dfrac{A}{2}\cos \dfrac{B}{2} - \sin \dfrac{A}{2}\sin \dfrac{B}{2}}}{{\cos \dfrac{A}{2}\cos \dfrac{B}{2}}}\]
\[ = \dfrac{{\cos \left( {\dfrac{A}{2} + \dfrac{B}{2}} \right)}}{{\cos \dfrac{A}{2}\cos \dfrac{B}{2}}}\]
Since ABC is a triangle, thus \[A + B + C = \pi \] \[ \Rightarrow A + B = \pi - C\]
\[ = \dfrac{{\cos \left( {\dfrac{\pi }{2} - \dfrac{C}{2}} \right)}}{{\cos \dfrac{A}{2}\cos \dfrac{B}{2}}}\]
\[ = \dfrac{{\sin \dfrac{C}{2}}}{{\cos \dfrac{A}{2}\cos \dfrac{B}{2}}}\]
Applying half angle of triangle formula
\[ = \dfrac{{\sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{ab}}} }}{{\sqrt {\dfrac{{s\left( {s - a} \right)}}{{bc}}} \sqrt {\dfrac{{s\left( {s - b} \right)}}{{ac}}} }}\]
\[ = \dfrac{{c\sqrt {ab} \sqrt {\left( {s - a} \right)\left( {s - b} \right)} }}{{\sqrt {ab} \sqrt {s\left( {s - a} \right)} \sqrt {s\left( {s - b} \right)} }}\]
Cancel out \[\sqrt {ab} \sqrt {\left( {s - a} \right)\left( {s - b} \right)} \] from denominator and numerator
\[ = \dfrac{c}{s}\]
Putting \[s = \dfrac{{a + b + c}}{2}\]
\[ = \dfrac{c}{{\dfrac{{a + b + c}}{2}}}\]
\[ = \dfrac{{2c}}{{a + b + c}}\]
Hence option A is the correct option.

Note: Students often confuse with the value of s. They use s = a+b+c which is incorrect. S is semi perimeter. Thus \[s = \dfrac{{a + b + c}}{2}\].