![SearchIcon](https://vmkt.vedantu.com/vmkt/PROD/png/bdcdbbd8-08a7-4688-98e6-4aa54e5e0800-1733305962725-4102606384256179.png)
$\cos \alpha .\sin (\beta -\gamma )+\cos \beta .\sin (\gamma -\alpha )+\cos \gamma .\sin (\alpha -\beta )$ is equal to
A . 0
B . $\dfrac{1}{2}$
C . 1
D . $\cos \alpha \cos \beta \cos \gamma $
Answer
124.8k+ views
Hint: In the given question we have to find the value of $\cos \alpha .\sin (\beta -\gamma )+\cos \beta .\sin (\gamma -\alpha )+\cos \gamma .\sin (\alpha -\beta )$. As we see an identity of sin is used in this question. So first we expand the equation by using the trigonometric identity of $\sin (x-y)$. Then by opening all the brackets and adding and subtracting the terms, we are able to get the desired answer and choose the correct option.
Formula Used:
In this question, we use the trigonometric identity which is described as below:-
$\sin (x-y)=\sin x\cos y-\cos x\sin y$
Complete step- by- step Solution:
Given that $\cos \alpha .\sin (\beta -\gamma )+\cos \beta .\sin (\gamma -\alpha )+\cos \gamma .\sin (\alpha -\beta )$……………………..(1)
We know the trigonometric identity
$\sin (x-y)=\sin x\cos y-\cos x\sin y$
then $\sin (\beta -\gamma )=(\sin \beta \cos \gamma -\sin \gamma \cos \beta )$
and $\sin (\gamma -\alpha )=(\sin \gamma \cos \alpha -\sin \alpha \cos \gamma )$
and $\sin (\alpha -\beta )=(\sin \alpha \cos \beta -\sin \beta \cos \alpha )$
Put the above identity in equation (1), we get
$\cos \alpha (\sin \beta \cos \gamma -\sin \gamma \cos \beta )+\cos \beta (sin\gamma \cos \alpha -\sin \alpha \cos \gamma )+\cos \gamma (\sin \alpha \cos \beta -\sin \beta \cos \alpha )$
Now by opening the brackets of the above equation, we get
$\cos \alpha \sin \beta \cos \gamma -\cos \alpha \sin \gamma \cos \beta +\cos \beta sin\gamma \cos \alpha -\cos \beta \sin \alpha \cos \gamma +\cos \gamma \sin \alpha \cos \beta -\cos \gamma \sin \beta \cos \alpha $
We see in the above equations, there are two similar terms with opposite signs. So they cancel each other.
Therefore, by cancelling all the similar terms with opposite signs, we get
$\cos \alpha .\sin (\beta -\gamma )+\cos \beta .\sin (\gamma -\alpha )+\cos \gamma .\sin (\alpha -\beta )$ = 0
Thus, Option (A) is correct.
Note: In these type of questions, students made mistake that they started solving the whole equation at one time. By solving the equation and putting the identities at one time makes us confused and we are not able to solve the question completely or we take extra time. By solving the equations in small parts and then combining them makes the question easy to solve and we solve it in lesser time.
Formula Used:
In this question, we use the trigonometric identity which is described as below:-
$\sin (x-y)=\sin x\cos y-\cos x\sin y$
Complete step- by- step Solution:
Given that $\cos \alpha .\sin (\beta -\gamma )+\cos \beta .\sin (\gamma -\alpha )+\cos \gamma .\sin (\alpha -\beta )$……………………..(1)
We know the trigonometric identity
$\sin (x-y)=\sin x\cos y-\cos x\sin y$
then $\sin (\beta -\gamma )=(\sin \beta \cos \gamma -\sin \gamma \cos \beta )$
and $\sin (\gamma -\alpha )=(\sin \gamma \cos \alpha -\sin \alpha \cos \gamma )$
and $\sin (\alpha -\beta )=(\sin \alpha \cos \beta -\sin \beta \cos \alpha )$
Put the above identity in equation (1), we get
$\cos \alpha (\sin \beta \cos \gamma -\sin \gamma \cos \beta )+\cos \beta (sin\gamma \cos \alpha -\sin \alpha \cos \gamma )+\cos \gamma (\sin \alpha \cos \beta -\sin \beta \cos \alpha )$
Now by opening the brackets of the above equation, we get
$\cos \alpha \sin \beta \cos \gamma -\cos \alpha \sin \gamma \cos \beta +\cos \beta sin\gamma \cos \alpha -\cos \beta \sin \alpha \cos \gamma +\cos \gamma \sin \alpha \cos \beta -\cos \gamma \sin \beta \cos \alpha $
We see in the above equations, there are two similar terms with opposite signs. So they cancel each other.
Therefore, by cancelling all the similar terms with opposite signs, we get
$\cos \alpha .\sin (\beta -\gamma )+\cos \beta .\sin (\gamma -\alpha )+\cos \gamma .\sin (\alpha -\beta )$ = 0
Thus, Option (A) is correct.
Note: In these type of questions, students made mistake that they started solving the whole equation at one time. By solving the equation and putting the identities at one time makes us confused and we are not able to solve the question completely or we take extra time. By solving the equations in small parts and then combining them makes the question easy to solve and we solve it in lesser time.
Recently Updated Pages
The real roots of the equation x23 + x13 2 0 are A class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Find the reminder when 798 is divided by 5 class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If there are 25 railway stations on a railway line class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Minimum area of the circle which touches the parabolas class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let A and B be two sets containing 2 elements and 4 class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A ray of light moving parallel to the xaxis gets reflected class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main Login 2045: Step-by-Step Instructions and Details
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
NCERT Solutions for Class 11 Maths Chapter 12 Limits and Derivatives
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
NCERT Solutions for Class 11 Maths Chapter 13 Statistics
![arrow-right](/cdn/images/seo-templates/arrow-right.png)