Question

If \[\tan A=x\tan B\], prove that \[\dfrac{\sin \left( A-B \right)}{\sin \left( A+B\right)} = \dfrac{x-1}{x+1}\].

Answer 1

Hint: We will be using the concept of trigonometry to solve the problem. We will be apply componendo and dividendo in $\dfrac{\tan A}{\tan B}=x$, and then we will use trigonometric identity like $\tan A=\dfrac{\sin A}{\cos B}$ . Also we will use the trigonometric identities that $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ to further simplify the problem.

Complete step-by-step answer:

Now, we have been given that,

\[\tan A=x\tan B\]

We have to prove that,

\[\dfrac{\sin \left( A-B \right)}{\sin \left( A+B \right)}=\dfrac{x-1}{x+1}\]

Now, we have been given that,

\[\tan A=x\tan B\]

We will rearrange it as,

$\dfrac{\tan A}{\tan B}=x$

Now, we know that according to componendo and dividendo rule that,

$\dfrac{a}{b}=\dfrac{c}{d}$ is same as $\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}$. So, we have,

\[\dfrac{\tan A-\tan B}{\tan A+\tan B}=\dfrac{x-1}{x+1}\]

Now, we know a trigonometric identity that,

$\begin{align}

  & \tan A=\dfrac{\sin A}{\cos A} \\

 & \tan B=\dfrac{\sin B}{\cos B} \\
 & \dfrac{\dfrac{\sin A}{\cos A}-\dfrac{\sin B}{\cos B}}{\dfrac{\sin A}{\cos A}+\dfrac{\sin B}{\cos B}}=\dfrac{x-1}{x+1} \\

\end{align}$

Now, we will take cosAcosB as LCM. So, we have,

$\dfrac{\dfrac{\sin A\cos B-\sin B\cos A}{\cos A\cos B}}{\dfrac{\sin A\cos B+\sin B\cos A}{\cos A\cos B}}=\dfrac{x-1}{x+1}$

Now, we will cancel cosAcosB in numerator and denominator. Also, we will be using trigonometric identity that,

$\begin{align}

  & \sin \left( A-B \right)=\sin A\cos B-\cos A\sin B \\

 & \sin \left( A+B \right)=\sin A\cos B+\cos A\sin B \\

\end{align}$

So, we have,

\[\dfrac{\sin \left( A-B \right)}{\sin \left( A+B \right)}=\dfrac{x-1}{x+1}\]

Since, we have LHS = RHS. Hence, proved.

Note: To solve these type of questions it is important to note that we used componendo and dividendo to make the RHS of equation $\dfrac{\tan A}{\tan B}=\dfrac{x}{1}\ as\ \dfrac{x-1}{x+1}$. So, that the solution becomes simple and easy.