Question

If $\dfrac{\sin (x+y)}{\sin (x-y)}=\dfrac{a+b}{a-b}$, show that$\dfrac{\tan x}{\tan y}=\dfrac{a}{b}$

Answer 1

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.