Question

In \[\Delta ABC\], if \[{\sin ^2}\dfrac{A}{2}\], \[{\sin ^2}\dfrac{B}{2}\], \[{\sin ^2}\dfrac{C}{2}\] be in H.P, the find the progression in which a, b, c are connected.
A. A.P
B. G.P
C. H.P.
D. None of these

Answer 1

Hint: First we will apply the condition of HP on the given series. Then substitute the value \[{\sin ^2}\dfrac{A}{2}\], \[{\sin ^2}\dfrac{B}{2}\], \[{\sin ^2}\dfrac{C}{2}\] by using trigonometric functions of half angles of a triangle and simplify it to get the required solution.

Formula used:
Trigonometric functions of half angles
\[\sin \dfrac{A}{2} = \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{bc}}} \]
\[\sin \dfrac{B}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{ac}}} \]
\[\sin \dfrac{C}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{ab}}} \]
The condition of HP is: if a,b,c are in HP, then \[\dfrac{1}{a} + \dfrac{1}{c} = \dfrac{2}{b}\].

Complete step by step solution:
Given that, \[{\sin ^2}\dfrac{A}{2}\], \[{\sin ^2}\dfrac{B}{2}\], \[{\sin ^2}\dfrac{C}{2}\] are in H.P.
Apply the condition of HP:
\[\dfrac{1}{{{{\sin }^2}\dfrac{A}{2}}} + \dfrac{1}{{{{\sin }^2}\dfrac{C}{2}}} = \dfrac{2}{{{{\sin }^2}\dfrac{B}{2}}}\]
Now apply the formulas \[\sin \dfrac{A}{2} = \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{bc}}} \], \[\sin \dfrac{B}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{ac}}} \]\[\sin \dfrac{C}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{ab}}} \]:
\[ \Rightarrow \dfrac{1}{{{{\left( {\sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{bc}}} } \right)}^2}}} + \dfrac{1}{{{{\left( {\sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{ab}}} } \right)}^2}}} = \dfrac{2}{{{{\left( {\sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{ac}}} } \right)}^2}}}\]
\[ \Rightarrow \dfrac{{bc}}{{\left( {s - b} \right)\left( {s - c} \right)}} + \dfrac{{ab}}{{\left( {s - a} \right)\left( {s - b} \right)}} = \dfrac{{2ac}}{{\left( {s - a} \right)\left( {s - c} \right)}}\]
Now simplify the left side of the equation
\[ \Rightarrow \dfrac{{bc\left( {s - a} \right) + ab\left( {s - c} \right)}}{{\left( {s - b} \right)\left( {s - c} \right)\left( {s - a} \right)}} = \dfrac{{2ac}}{{\left( {s - a} \right)\left( {s - c} \right)}}\]
\[ \Rightarrow \dfrac{{sbc - abc + sab - abc}}{{\left( {s - b} \right)\left( {s - c} \right)\left( {s - a} \right)}} = \dfrac{{2ac}}{{\left( {s - a} \right)\left( {s - c} \right)}}\]
Cancel out \[\left( {s - a} \right)\left( {s - c} \right)\] from the denominator of both sides:
\[ \Rightarrow \dfrac{{sbc - abc + sab - abc}}{{\left( {s - b} \right)}} = \dfrac{{2ac}}{1}\]
Apply cross multiply
\[ \Rightarrow sbc - abc + sab - abc = 2ac\left( {s - b} \right)\]
\[ \Rightarrow sbc + sab - 2abc = 2sac - 2abc\]
Cancel out \[ - 2abc\] from both sides
\[ \Rightarrow sbc + sab = 2sac\]
Divide both sides by \[sabc\]
\[ \Rightarrow \dfrac{{sbc}}{{sabc}} + \dfrac{{sab}}{{sabc}} = \dfrac{{2sac}}{{sabc}}\]
\[ \Rightarrow \dfrac{1}{a} + \dfrac{1}{c} = \dfrac{2}{b}\]
It means a, b, and c are in H.P.
Hence option C is the correct option.

Note: Students often confused between Hp and AP. If a series a,b,c are in AP, then \[\dfrac{1}{a}\], \[\dfrac{1}{b}\], \[\dfrac{1}{c}\] are in HP. If a,b,c are in HP, then \[\dfrac{1}{a}\], \[\dfrac{1}{b}\], \[\dfrac{1}{c}\] are in AP.