
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
164.1k+ views
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 \[a - 2b + c = 0\].
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 \[a - 2b + c = 0\].
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets
