Question

In \[\Delta ABC\], if \[\tan \dfrac{A}{2}\tan \dfrac{C}{2} = \dfrac{1}{2}\], then which of the following is true for a, b, c?
A. A.P.
B. G.P.
C. H.P.
D. None of these

Answer 1

Hint: First we will break \[\tan \] into \[\sin \] and \[\cos \]. Then we will apply the half angle formula to simplify the given equation. By this equation, we get the relation between a, b, c.

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

Complete step by step solution:
Given equation is \[\tan \dfrac{A}{2}\tan \dfrac{C}{2} = \dfrac{1}{2}\]
Apply the formula \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
\[\dfrac{{\sin \dfrac{A}{2}}}{{\cos \dfrac{A}{2}}} \cdot \dfrac{{\sin \dfrac{C}{2}}}{{\cos \dfrac{C}{2}}} = \dfrac{1}{2}\]
Now applying half angle formula
\[ \Rightarrow \dfrac{{\sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{bc}}} }}{{\sqrt {\dfrac{{s\left( {s - a} \right)}}{{bc}}} }} \cdot \dfrac{{\sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{ab}}} }}{{\sqrt {\dfrac{{s\left( {s - c} \right)}}{{ab}}} }} = \dfrac{1}{2}\]
Simplify the above equation:
\[ \Rightarrow \dfrac{{\sqrt {\left( {s - b} \right)\left( {s - c} \right)} }}{{\sqrt {s\left( {s - a} \right)} }} \cdot \dfrac{{\sqrt {\left( {s - a} \right)\left( {s - b} \right)} }}{{\sqrt {s\left( {s - c} \right)} }} = \dfrac{1}{2}\]
\[ \Rightarrow \dfrac{{s - b}}{s} = \dfrac{1}{2}\]
\[ \Rightarrow 2\left( {s - b} \right) = s\]
\[ \Rightarrow 2s - 2b = s\]
\[ \Rightarrow s - 2b = 0\]
Substitute \[s = \dfrac{{a + b + c}}{2}\]
\[ \Rightarrow \dfrac{{a + b + c}}{2} - 2b = 0\]
\[ \Rightarrow a + b + c - 4b = 0\]
\[ \Rightarrow a - 3b + c = 0\]
Thus it is not an AP series.
Hence option D is the correct option.

Note:Students are often confused when they get \[a - 3b + c = 0\] by solving the given equation. They mark option A as a correct option but the condition of A.P. is \[a - 2b + c = 0\].