Question

If the sides of a triangle are in A.P., then the cotangent of its half the angles will be in
A. H.P.
B. G.P.
C. A.P.
D. None of these

Answer 1

Hint: First we will assume that the cotangent of its half the angles are in A.P. Then check whether the progression is in A.P. or not using different laws of half angles of cosines and sine.

Formula used:
If a, b, c are in A.P., then a + c =2b.
Half angle formula for an oblique triangle
\[\sin \dfrac{A}{2} = \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{bc}}} \]
\[\sin \dfrac{B}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{ac}}} \]
\[\sin \dfrac{C}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{ab}}} \]
\[\cos \dfrac{A}{2} = \sqrt {\dfrac{{s\left( {s - a} \right)}}{{bc}}} \]
\[\cos \dfrac{B}{2} = \sqrt {\dfrac{{s\left( {s - b} \right)}}{{ac}}} \]
\[\cos \dfrac{C}{2} = \sqrt {\dfrac{{s\left( {s - c} \right)}}{{ab}}} \]

Complete step by step solution:
Assume that, \[\cot \dfrac{A}{2}\], \[\cot \dfrac{B}{2}\], \[\cot \dfrac{C}{2}\]are in A.P.
Thus, \[\cot \dfrac{A}{2} + \cot \dfrac{C}{2} = 2\cot \dfrac{B}{2}\]
L.H.S\[ = \cot \dfrac{A}{2} + \cot \dfrac{C}{2}\]
Apply the formula \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\]
\[ = \dfrac{{\cos \dfrac{A}{2}}}{{\sin \dfrac{A}{2}}} + \dfrac{{\cos \dfrac{C}{2}}}{{\sin \dfrac{C}{2}}}\]
Now we will apply the half angle formula
\[ = \dfrac{{\sqrt {\dfrac{{s\left( {s - a} \right)}}{{bc}}} }}{{\sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{bc}}} }} + \dfrac{{\sqrt {\dfrac{{s\left( {s - c} \right)}}{{ab}}} }}{{\sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{ab}}} }}\]
\[ = \dfrac{{\sqrt {s\left( {s - a} \right)} }}{{\sqrt {\left( {s - b} \right)\left( {s - c} \right)} }} + \dfrac{{\sqrt {s\left( {s - c} \right)} }}{{\sqrt {\left( {s - a} \right)\left( {s - b} \right)} }}\]
\[ = \dfrac{{\sqrt {s{{\left( {s - a} \right)}^2}} + \sqrt {s{{\left( {s - c} \right)}^2}} }}{{\sqrt {\left( {s - b} \right)\left( {s - c} \right)\left( {s - a} \right)} }}\]
\[ = \dfrac{{\sqrt s \left[ {\left( {s - a} \right) + \left( {s - c} \right)} \right]}}{{\sqrt {\left( {s - b} \right)\left( {s - c} \right)\left( {s - a} \right)} }}\]
\[ = \dfrac{{\sqrt s \left[ {2s - a - c} \right]}}{{\sqrt {\left( {s - b} \right)\left( {s - c} \right)\left( {s - a} \right)} }}\]
Given that, a,b,c are in AP, thus \[a + c = 2b\]
\[ = \dfrac{{\sqrt s \left( {2s - 2b} \right)}}{{\sqrt {\left( {s - b} \right)\left( {s - c} \right)\left( {s - a} \right)} }}\]
\[ = 2\dfrac{{\sqrt s \left( {s - b} \right)}}{{\sqrt {\left( {s - b} \right)\left( {s - c} \right)\left( {s - a} \right)} }}\]
\[ = 2\dfrac{{\sqrt {s{{\left( {s - b} \right)}^2}} }}{{\sqrt {\left( {s - b} \right)\left( {s - c} \right)\left( {s - a} \right)} }}\]
\[ = 2\dfrac{{\sqrt {s\left( {s - b} \right)} }}{{\sqrt {\left( {s - c} \right)\left( {s - a} \right)} }}\]
We know that, \[\sin \dfrac{B}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{ac}}} \] and \[\cos \dfrac{B}{2} = \sqrt {\dfrac{{s\left( {s - b} \right)}}{{ac}}} \] . So \[\cot \dfrac{B}{2} = \dfrac{{\sqrt {s\left( {s - b} \right)} }}{{\sqrt {\left( {s - c} \right)\left( {s - a} \right)} }}\]
\[ = 2\cot \dfrac{B}{2}\]
Thus, our assumption is correct.
Hence option C is the correct answer.

Note: Students often make the mistake of considering that the cotangent of its half the angles in G.P or H.P. First, we will check it for A.P. Then after that we will check other progressions.