Question

If $b\sin \alpha =a\sin (\alpha +2\beta )$, then $\dfrac{a+b}{a-b}$ is equal to ?
A . $\dfrac{\tan \beta }{\tan (\alpha +\beta )}$
B . $\dfrac{\cot \beta }{\cot (\alpha -\beta )}$
C . $\dfrac{\cot \beta }{\cot (\alpha +\beta )}$
D . $\dfrac{\tan \beta }{\tan (\alpha -\beta )}$

Answer 1

Hint: In this question, we are given that $b\sin \alpha =a\sin (\alpha +2\beta )$ . First, we find the value of m from the given equation. Then we apply componendo and dividendo on the equation which is sometimes, in short, we called C and D rules 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 identities which are described 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 $b\sin \alpha =a\sin (\alpha +2\beta )$
Then $\dfrac{\sin \alpha }{\sin (\alpha +2\beta )}=\dfrac{a}{b}$
Now we apply componendo and dividendo in the above equation,
Componendo and dividendo is a theorem on proportions that allows for an easy way to perform calculations and helps in reducing the number of expansions needed.
We get
$\dfrac{\sin \alpha +\sin (\alpha +2\beta )}{\sin \alpha -\sin (\alpha +2\beta )}=\dfrac{a+b}{a-b}$……………………………… (1)
Now we use the trigonometric formulas 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 $\sin A-\sin B=2\cos \dfrac{A+B}{2}\sin \dfrac{A-B}{2}$
Put the above formulas in equation (1), we get
$\dfrac{2\sin \dfrac{\alpha +(\alpha +2\beta )}{2}\cos \dfrac{\alpha -(\alpha +2\beta )}{2}}{2\cos \dfrac{\alpha +(\alpha +2\beta )}{2}\sin \dfrac{\alpha -(\alpha +2\beta )}{2}}=\dfrac{a+b}{a-b}$
Solving further, we get
$\dfrac{\sin (\alpha +\beta )\cos (-\beta )}{\cos (\alpha +\beta )\sin (-\beta )}=\dfrac{a+b}{a-b}$
By simplifying the above equation, we get
$\tan\lgroup\alpha+\beta\rgroup~\cot\lgroup\beta\rgroup=\dfrac{a+b}{a-b}$
$\dfrac{\cot \beta }{\cot (\alpha +\beta )}=\dfrac{a+b}{a-b}$
Thus, Option ( C ) is the correct .

Note: In these types of questions, Students made mistakes in understanding the difference between componendo and dividendo. It is a mathematical rule that states that the product of two or more binomials can be divided by the sum of the individual terms within each binomial. In this, we add the numerator and denominator in the numerator and subtract both terms in the denominator. This allows for simplified algebraic equations and solutions. The dividendo part of the equation belongs to the first binomial, while the componendo belongs to all subsequent binomials.