
If $\dfrac{\sin (x+y)}{\sin (x-y)}=\dfrac{a+b}{a-b}$, show that$\dfrac{\tan x}{\tan y}=\dfrac{a}{b}$
Answer
162.3k+ views
Hint: In this question, we are given that $\dfrac{\sin (x+y)}{\sin (x-y)}=\dfrac{a+b}{a-b}$ and we have to show that$\dfrac{\tan x}{\tan y}=\dfrac{a}{b}$. First we apply componendo and dividendo on the given equation which is sometimes in short we called C And D rule and by using the formulas of $\sin A+\sin B$ and $\sin A-\sin B$and putting the formulas in the equation and simplifying it, we are able to get the desirable answer and choose the correct option.
Formula Used:
In this question, we use the trigonometric identities which is described as below :-
$\sin A+\sin B=2\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}$
And $\sin A-\sin B=2\cos \dfrac{A+B}{2}\sin \dfrac{A-B}{2}$
Complete step- by- step Solution:
Given equation is $\dfrac{\sin (x+y)}{\sin (x-y)}=\dfrac{a+b}{a-b}$
We have to show that $\dfrac{\tan x}{\tan y}=\dfrac{a}{b}$
Now we apply componendo and dividendo in the above equation,
Componendo and dividendo is a theorem on proportions which allows for an easy way to perform calculations and helps in reducing the number of expansions needed.
We get
$\dfrac{\sin (x+y)+\sin (x-y)}{\sin (x+y)-\sin (x-y)}=\dfrac{a+b+a-b}{a-b-(a-b)}$……………………………… (1)
Now we use the trigonometric identities according to the above equation, which is
We know $\sin A+\sin B=2\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}$
And $\sin A-\sin B=2\cos \dfrac{A+B}{2}\sin \dfrac{A-B}{2}$
On putting the above identities in equation (1), we get
$\dfrac{2\sin \dfrac{(x+y)+(x-y)}{2}\cos \dfrac{(x+y)-(x-y)}{2}}{2\cos \dfrac{(xy)+(x+y)}{2}\sin \dfrac{(x+y)-(x-y)}{2}}=\dfrac{a}{b}$
Solving further the above equation, we get
$\dfrac{2\sin x\cos y}{2\cos x\sin y}=\dfrac{a}{b}$
By simplifying the above equation, we get
$\dfrac{\tan x}{\tan y}=\dfrac{a}{b}$
Hence we proved that $\dfrac{\tan x}{\tan y}=\dfrac{a}{b}$
Note: Students can commit mistakes if they directly apply the sum and difference formula. This leads to a complex equation where student won’t be able to solve the question further.
Formula Used:
In this question, we use the trigonometric identities which is described as below :-
$\sin A+\sin B=2\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}$
And $\sin A-\sin B=2\cos \dfrac{A+B}{2}\sin \dfrac{A-B}{2}$
Complete step- by- step Solution:
Given equation is $\dfrac{\sin (x+y)}{\sin (x-y)}=\dfrac{a+b}{a-b}$
We have to show that $\dfrac{\tan x}{\tan y}=\dfrac{a}{b}$
Now we apply componendo and dividendo in the above equation,
Componendo and dividendo is a theorem on proportions which allows for an easy way to perform calculations and helps in reducing the number of expansions needed.
We get
$\dfrac{\sin (x+y)+\sin (x-y)}{\sin (x+y)-\sin (x-y)}=\dfrac{a+b+a-b}{a-b-(a-b)}$……………………………… (1)
Now we use the trigonometric identities according to the above equation, which is
We know $\sin A+\sin B=2\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}$
And $\sin A-\sin B=2\cos \dfrac{A+B}{2}\sin \dfrac{A-B}{2}$
On putting the above identities in equation (1), we get
$\dfrac{2\sin \dfrac{(x+y)+(x-y)}{2}\cos \dfrac{(x+y)-(x-y)}{2}}{2\cos \dfrac{(xy)+(x+y)}{2}\sin \dfrac{(x+y)-(x-y)}{2}}=\dfrac{a}{b}$
Solving further the above equation, we get
$\dfrac{2\sin x\cos y}{2\cos x\sin y}=\dfrac{a}{b}$
By simplifying the above equation, we get
$\dfrac{\tan x}{\tan y}=\dfrac{a}{b}$
Hence we proved that $\dfrac{\tan x}{\tan y}=\dfrac{a}{b}$
Note: Students can commit mistakes if they directly apply the sum and difference formula. This leads to a complex equation where student won’t be able to solve the question further.
Recently Updated Pages
Fluid Pressure - Important Concepts and Tips for JEE

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

Impulse Momentum Theorem Important Concepts and Tips for JEE

Graphical Methods of Vector Addition - Important Concepts for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

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

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

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

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

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

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

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

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

NEET 2025 – Every New Update You Need to Know
