
In a \[\Delta ABC, 2a\sin (\dfrac{A-B+C}{2})\] is equal to [IIT Screening 2000]
A. ${{a}^{2}}+{{b}^{2}}-{{c}^{2}}$
B. ${{c}^{2}}+{{a}^{2}}-{{b}^{2}}$
C. ${{b}^{2}}-{{c}^{2}}-{{a}^{2}}$
D. ${{c}^{2}}-{{a}^{2}}-{{b}^{2}}$
Answer
242.7k+ views
Hint:
In this case, we will use the angle sum property of a triangle according to which the triangle's triangle's interior angles add up to $180^{o}$. Using this relation and putting it in the given scenario we will get the result.
Formula Used:
Three sides in the triangle $ABC$ are represented by the letters $AB, BC,$ and $CA$. The three vertices are $A, B$, and $C$, and the three interior angles are $\angle ABC, \angle BCA,$ and $\angle CAB$. Then according to the angle sum property of the triangle: $\angle ACB + \angle BAC + \angle CBA= 180^{o}$
Laws of cosine for triangle $ABC$ whose length is $a, b$, and $c$ respectively is given by;
$b^2 = a^2 +c^2 − 2ac.\cos B$.
Complete step-by-step solution:
We have, $2a\sin (\dfrac{A-B+C}{2})$
Since $ABC$ is a triangle, therefore we can say;
$A+B+C=180^{o}\\
A+C=180^{o}-B …(i)$
Now consider
$2a\sin (\dfrac{A-B+C}{2})$
Substitute value of $A+C$ using eq (i)
$\Rightarrow2a\sin \,\,\left( \dfrac{180^{o}-B-B}{2} \right)\\
\Rightarrow2a\sin(\dfrac{180^{o}-2B}{2})\\
\Rightarrow 2ac \sin(90^{o}-B)
\Rightarrow 2ac \cos B$
Using the cosine rule of the triangle we get;
$\therefore b^2 = a^2 +c^2 − 2ac.\cos B\\
\therefore \cos B=\dfrac{a^2+c^2-b^2}{2ac}\\
\Rightarrow 2ac \dfrac{a^2+c^2-b^2}{2ac}\\
\Rightarrow a^2+c^2-b^2$
$2a\sin \,\,\left( \dfrac{A-B+C}{2} \right)=a^2+c^2-b^2$.
Hence, $2a\sin \,\,\left( \dfrac{A-B+C}{2} \right)=a^2+c^2-b^2$. So, option B is correct .
Note:
Keep in mind that such type of question requires knowledge of the basic trigonometric conversation. Here we have uses the conversion of sine at $(90^{o}-B)$ which turns out to be $\cos B$ as it lies in the first quadrant so it is taken as positive. Using the law of cosine for angle B, students usually make mistakes in taking the wrong angles. Different angles use cosine rules in different ways.
In this case, we will use the angle sum property of a triangle according to which the triangle's triangle's interior angles add up to $180^{o}$. Using this relation and putting it in the given scenario we will get the result.
Formula Used:
Three sides in the triangle $ABC$ are represented by the letters $AB, BC,$ and $CA$. The three vertices are $A, B$, and $C$, and the three interior angles are $\angle ABC, \angle BCA,$ and $\angle CAB$. Then according to the angle sum property of the triangle: $\angle ACB + \angle BAC + \angle CBA= 180^{o}$
Laws of cosine for triangle $ABC$ whose length is $a, b$, and $c$ respectively is given by;
$b^2 = a^2 +c^2 − 2ac.\cos B$.
Complete step-by-step solution:
We have, $2a\sin (\dfrac{A-B+C}{2})$
Since $ABC$ is a triangle, therefore we can say;
$A+B+C=180^{o}\\
A+C=180^{o}-B …(i)$
Now consider
$2a\sin (\dfrac{A-B+C}{2})$
Substitute value of $A+C$ using eq (i)
$\Rightarrow2a\sin \,\,\left( \dfrac{180^{o}-B-B}{2} \right)\\
\Rightarrow2a\sin(\dfrac{180^{o}-2B}{2})\\
\Rightarrow 2ac \sin(90^{o}-B)
\Rightarrow 2ac \cos B$
Using the cosine rule of the triangle we get;
$\therefore b^2 = a^2 +c^2 − 2ac.\cos B\\
\therefore \cos B=\dfrac{a^2+c^2-b^2}{2ac}\\
\Rightarrow 2ac \dfrac{a^2+c^2-b^2}{2ac}\\
\Rightarrow a^2+c^2-b^2$
$2a\sin \,\,\left( \dfrac{A-B+C}{2} \right)=a^2+c^2-b^2$.
Hence, $2a\sin \,\,\left( \dfrac{A-B+C}{2} \right)=a^2+c^2-b^2$. So, option B is correct .
Note:
Keep in mind that such type of question requires knowledge of the basic trigonometric conversation. Here we have uses the conversion of sine at $(90^{o}-B)$ which turns out to be $\cos B$ as it lies in the first quadrant so it is taken as positive. Using the law of cosine for angle B, students usually make mistakes in taking the wrong angles. Different angles use cosine rules in different ways.
Recently Updated Pages
WBJEE 2026 Registration Started: Important Dates Eligibility Syllabus Exam Pattern

Geometry of Complex Numbers Explained

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE General Topics in Chemistry Important Concepts and Tips

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

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

Trending doubts
JEE Main 2026: Session 1 Results Out and Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

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

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

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

Understanding Average and RMS Value in Electrical Circuits

How to Convert a Galvanometer into an Ammeter or Voltmeter

Derivation of Equation of Trajectory Explained for Students

