Question

If $\sin \left( {\theta + \phi } \right) = n\sin \left( {\theta - \phi } \right)$ , $n \ne 1$, then the value of $\dfrac{{\tan \theta }}{{\tan \phi }}$ is:
A) $\dfrac{n}{{n - 1}}$
B) $\dfrac{{n + 1}}{{n - 1}}$
C) $\dfrac{n}{{1 - n}}$
D) $\dfrac{{1 + n}}{{1 - n}}$

Answer 1

Here we will use the formula of $\sin \left( {\text{A}} \right) + \sin \left( {\text{B}} \right) = 2\sin \left( {\dfrac{{{\text{A + B}}}}{{\text{2}}}} \right)\cos \left( {\dfrac{{{\text{A - B}}}}{2}} \right)$ and $\sin \left( {\text{A}} \right) - \sin \left( {\text{B}} \right) = 2\cos \left( {\dfrac{{{\text{A + B}}}}{{\text{2}}}} \right)\sin \left( {\dfrac{{{\text{A - B}}}}{2}} \right)$ where ${\text{A}}$ and ${\text{B}}$ are any variables.

Complete step-by-step solution:
Step 1: It is given that
$\sin \left( {\theta + \phi } \right) = n\sin \left( {\theta - \phi } \right)$ where
$n \ne 1$. We will apply the Componendo-Dividendo rule which states that if
$a$, $b$, $c$ and $d$ are four variables and
$\dfrac{{a + b}}{{a - b}} = \dfrac{{c + d}}{{c - d}}$ , then by applying this rule we will get the expression as
$\dfrac{{\left( {a + b} \right) + \left( {a - b} \right)}}{{\left( {a - b} \right) - \left( {a - b} \right)}} = \dfrac{{\left( {c + d} \right) + \left( {c - d} \right)}}{{\left( {c - d} \right) - \left( {c - d} \right)}}$. We can write the given expression as below:
$\dfrac{{\sin \left( {\theta + \phi } \right)}}{{\sin \left( {\theta - \phi } \right)}} = n$
By applying this rule in the given expression, we get:
$\Rightarrow \dfrac{{\sin \left( {\theta + \phi } \right) + \sin \left( {\theta - \phi } \right)}}{{\sin \left( {\theta - \phi } \right) - \sin \left( {\theta - \phi } \right)}} = \dfrac{{n + 1}}{{n - 1}}$ ………………………… (1)
Step 2: By applying the formula of
$\sin \left( {\text{A}} \right) + \sin \left( {\text{B}} \right) = 2\sin \left( {\dfrac{{{\text{A + B}}}}{{\text{2}}}} \right)\cos \left( {\dfrac{{{\text{A - B}}}}{2}} \right)$, in the above expression (1), we get:
$\Rightarrow \dfrac{{2\sin \left( {\dfrac{{\theta + \phi + \theta - \phi }}{2}} \right)\cos \left( {\dfrac{{\theta + \phi - \theta + \phi }}{2}} \right)}}{{2\cos \left( {\dfrac{{\theta + \phi + \theta - \phi }}{2}} \right)\sin \left( {\dfrac{{\theta + \phi - \theta + \phi }}{2}} \right)}} = \dfrac{{n + 1}}{{n - 1}}$
By simplifying the terms inside the brackets, we get:
$\Rightarrow \dfrac{{2\sin \left( {\dfrac{{2\theta }}{2}} \right)\cos \left( {\dfrac{{2\phi }}{2}} \right)}}{{2\cos \left( {\dfrac{{2\theta }}{2}} \right)\sin \left( {\dfrac{{2\phi }}{2}} \right)}} = \dfrac{{n + 1}}{{n - 1}}$
By simplifying the brackets of the above expression, we get:
$\Rightarrow \dfrac{{2\sin \left( \theta \right)\cos \left( \phi \right)}}{{2\cos \left( \theta \right)\sin \left( \phi \right)}} = \dfrac{{n + 1}}{{n - 1}}$
By eliminating
$2$from the LHS side of the above expression we get:
$\Rightarrow \dfrac{{\sin \left( \theta \right)\cos \left( \phi \right)}}{{\cos \left( \theta \right)\sin \left( \phi \right)}} = \dfrac{{n + 1}}{{n - 1}}$ …………………………. (2)
Step 3: Now, as we know that
$\dfrac{{\sin a}}{{\cos a}} = \tan a$, so by comparing it with the above expression (2), we get:
$\Rightarrow \dfrac{{\tan \theta }}{{\tan \phi }} = \dfrac{{n + 1}}{{n - 1}}$

$\because$ Option B is correct.

Note: Students need to remember some basic formulas ${\text{sinAcosB}}$ for solving these types of questions easily. Some of them are mentioned below:
$\sin \left( {{\text{A + B}}} \right) = \sin {\text{A cosB + cosA sinB}}$
$\sin \left( {{\text{A - B}}} \right) = \sin {\text{A cosB - cosA sinB}}$
$\sin \left( {\text{A}} \right) + \sin \left( {\text{B}} \right) = 2\sin \left( {\dfrac{{{\text{A + B}}}}{{\text{2}}}} \right)\cos \left( {\dfrac{{{\text{A - B}}}}{2}} \right)$
$\sin \left( {\text{A}} \right) - \sin \left( {\text{B}} \right) = 2\cos \left( {\dfrac{{{\text{A + B}}}}{{\text{2}}}} \right)\sin \left( {\dfrac{{{\text{A - B}}}}{2}} \right)$
Also, you should remember the Componendo-Dividendo rule which is a theorem that allows for a quick way to perform calculations and also it reduces the number of expansions needed.
It is used when dealing with equations that involve fractions or we can say rational fractions.
The theorem states that, if $a$, $b$, $c$ and $d$ are four numbers such that
$b$ and $d$ are non-zero and $\dfrac{a}{b} = \dfrac{c}{d}$, then the following holds as below:
Componendo: $\dfrac{{a + b}}{b} = \dfrac{{c + d}}{d}$
Dividendo: $\dfrac{{a - b}}{b} = \dfrac{{c - d}}{d}$
For $k \ne \dfrac{a}{b}$, $\dfrac{{a + kb}}{{a - kb}} = \dfrac{{c + kd}}{{c - kd}}$
For $k \ne \dfrac{{ - b}}{a}$, $\dfrac{a}{b} = \dfrac{{a + kc}}{{b + kd}}$