Question

If $\tan \left(\theta \right)=\frac{x\sin \left(\varphi \right)}{1-x\cos \left(\varphi \right)}$ and $\tan \left(\varphi \right)=\frac{y\sin \left(\theta \right)}{1-y\cos \left(\theta \right)}$, then $\frac{x}{y}=$

• A. $\frac{\sin \left(\varphi \right)}{\sin \left(\theta \right)}$
• B. $\frac{\sin \left(\theta \right)}{\sin \left(\varphi \right)}$
• C. $\frac{\sin \left(\theta \right)}{1-\cos \left(\theta \right)}$
• D. $\frac{\sin \left(\theta \right)}{1-\cos \left(\varphi \right)}$

Answer 1

The correct option is B

$\frac{\sin \left(\theta \right)}{\sin \left(\varphi \right)}$

Explanation for correct option

Step 1: Simplify in terms of $x$

Given: $\tan \left(\theta \right)=\frac{x\sin \left(\varphi \right)}{1-x\cos \left(\varphi \right)}$ and $\tan \left(\varphi \right)=\frac{y\sin \left(\theta \right)}{1-y\cos \left(\theta \right)}$

$$\begin{gathered} \tan \left(\theta \right)=\frac{x\sin \left(\varphi \right)}{1-x\cos \left(\varphi \right)} \\ \Rightarrow x\sin \left(\varphi \right)=\tan \left(\theta \right)\left(1-x\cos \left(\varphi \right)\right) \\ \Rightarrow x\sin \left(\varphi \right)=\tan \left(\theta \right)-x\tan \left(\theta \right)\cos \left(\varphi \right) \\ \Rightarrow \tan \left(\theta \right)=x\left[\tan \left(\theta \right)\cos \left(\varphi \right)+\sin \left(\varphi \right)\right] \\ \Rightarrow x=\frac{\tan \left(\theta \right)}{\left[\tan \left(\theta \right)\cos \left(\varphi \right)+\sin \left(\varphi \right)\right]} \end{gathered}$$

Step 2: Simplify in terms of $y$

$$\begin{gathered} \tan \left(\varphi \right)=\frac{y\sin \left(\theta \right)}{1-y\cos \left(\theta \right)} \\ \Rightarrow \tan \left(\varphi \right)\left(1-y\cos \left(\theta \right)\right)=y\sin \left(\theta \right) \\ \Rightarrow \tan \left(\varphi \right)-y\tan \left(\varphi \right)\cos \left(\theta \right)=y\sin \left(\theta \right) \\ \Rightarrow \tan \left(\varphi \right)=y\left(\tan \right(\varphi \left)\cos \right(\theta )+\sin (\theta \left)\right) \\ \Rightarrow y=\frac{\tan \left(\varphi \right)}{\left(\tan \right(\varphi \left)\cos \right(\theta )+\sin (\theta \left)\right)} \end{gathered}$$

Step 3: Solve for the value of $\frac{x}{y}$

$$\begin{gathered} \frac{x}{y}=\frac{{\displaystyle \frac{\tan \left(\theta \right)}{\left[\tan \left(\theta \right)\cos \left(\varphi \right)+\sin \left(\varphi \right)\right]}}}{{\displaystyle \frac{\tan \left(\varphi \right)}{\left[\tan \left(\varphi \right)\cos \left(\theta \right)+\sin \left(\theta \right)\right]}}} \\ \Rightarrow \frac{x}{y}=\frac{{\displaystyle \frac{{\displaystyle \frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}}}{\left[{\displaystyle \frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}}\cos \left(\varphi \right)+\sin \left(\varphi \right)\right]}}}{{\displaystyle \frac{{\displaystyle \frac{\sin \left(\varphi \right)}{\cos \left(\varphi \right)}}}{\left[{\displaystyle \frac{\sin \left(\varphi \right)}{\cos \left(\varphi \right)}}\cos \left(\theta \right)+\sin \left(\theta \right)\right]}}} \\ \Rightarrow \frac{x}{y}=\frac{{\displaystyle \frac{{\displaystyle \frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}}}{\left[{\displaystyle \frac{{\displaystyle \sin \left(\theta \right)\cos \left(\varphi \right)+\cos \left(\theta \right)\sin \left(\varphi \right)}}{\cos \left(\theta \right)}}\right]}}}{{\displaystyle \frac{{\displaystyle \frac{\sin \left(\varphi \right)}{\cos \left(\varphi \right)}}}{\left[{\displaystyle \frac{{\displaystyle \sin \left(\varphi \right)\cos \left(\theta \right)+\cos \left(\varphi \right)\sin \left(\theta \right)}}{\cos \left(\varphi \right)}}\right]}}} \\ \Rightarrow \frac{x}{y}=\frac{{\displaystyle \frac{\sin \left(\theta \right)}{\sin (\theta +\varphi )}}}{{\displaystyle \frac{\sin \left(\varphi \right)}{\sin (\theta +\varphi )}}}\left[\because \sin \left(x\right)\cos \left(y\right)+\cos \left(x\right)\sin \left(y\right)=\sin (x+y)\right] \\ \Rightarrow \frac{x}{y}=\frac{\sin \left(\theta \right)}{\sin \left(\varphi \right)} \end{gathered}$$

Hence, option B is correct.