Question

If \[\dfrac{\sin \left( x+y \right)}{\sin \left( x-y \right)}=\dfrac{a+b}{a-b}\], show that \[\dfrac{\tan x}{\tan y}=\dfrac{a}{b}\].

Answer 1

Hint: We will be using the concept of trigonometric identities to solve the problem. We will be first using the componendo and dividendo rule which states that $\dfrac{a}{b}=\dfrac{c}{d}$ is equivalent to $\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}$, then we will use trigonometric identity that $\sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ to further simplify the solution.

Complete step-by-step answer:

Now, we have been given that,

\[\dfrac{\sin \left( x+y \right)}{\sin \left( x-y \right)}=\dfrac{a+b}{a-b}\]

We have to prove that,

\[\dfrac{\tan x}{\tan y}=\dfrac{a}{b}\]

Now, we know that according to componendo and dividendo rule that, $\dfrac{a}{b}=\dfrac{c}{d}$ can be written as $\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}$.

Now, we will use this rule in (1). So, we have,

\[\dfrac{\sin \left( x+y \right)+\sin \left( x-y \right)}{\sin \left( x+y \right)-\sin \left( x-y \right)}=\dfrac{a+b+a-b}{a+b-a+b}\]

Now, we know a trigonometric identity that,

$\begin{align}

  & \sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) \\

 & \sin C-\sin D=2\sin \left( \dfrac{C-D}{2} \right)\cos \left( \dfrac{C+D}{2} \right) \\

\end{align}$

So, we have,

$\dfrac{2\sin \left( \dfrac{x+y+x-y}{2} \right)\cos \left( \dfrac{x+y-x+y}{2} \right)}{2\sin \left( \dfrac{x+y-x+y}{2} \right)\cos \left( \dfrac{x+y+x-y}{2} \right)}=\dfrac{2a}{2b}$

Now, we will further simplify it. So, we have,

$\dfrac{\sin \left( x \right)\cos \left( y \right)}{\sin \left( y \right)\cos \left( x \right)}=\dfrac{a}{b}$

Now, on rearranging terms we have,

$\dfrac{\dfrac{\sin \left( x \right)}{\cos \left( x \right)}}{\dfrac{\sin \left( y \right)}{\cos \left( y \right)}}=\dfrac{a}{b}$

Now, we know the trigonometric identity that,

\[\dfrac{\sin \left( x \right)}{\cos \left( x \right)}=\tan \left( x \right)\]

So, using this we have,

\[\dfrac{\tan \left( x \right)}{\tan y}=\dfrac{a}{b}\]

Note: To solve these type of questions it is important to note that we used componendo and dividendo in $\dfrac{\sin \left( x+y \right)}{\sin \left( x-y \right)}=\dfrac{a+b}{a-b}$. So, that we have $\dfrac{a}{b}$ left in RHS and then we further simplified RHS by using the formulae like $\sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$.