If `tantheta=(xsinphi)/(1-xcosphi)&tanphi=(ysintheta)/(1-ycostheta)`, then `(x)/(y)=?`

Options

A. `(sintheta)/(sinphi)`
B. `(sinphi)/(sintheta)`
C. `(sinphi)/(1-costheta)`
D. `(sintheta)/(1-cosphi)`

Answer

To solve the problem, we need to find the value of \( \frac{x}{y} \) given the equations:

\[ \tan \theta = \frac{x \sin \phi}{1 - x \cos \phi} \]
\[ \tan \phi = \frac{y \sin \theta}{1 - y \cos \theta} \]

### Step 1: Express \( x \) in terms of \( \tan \theta \) and \( \sin \phi \)

From the first equation, we can rearrange it to isolate \( x \):

\[ \tan \theta (1 - x \cos \phi) = x \sin \phi \]

Expanding this gives:

\[ \tan \theta - x \tan \theta \cos \phi = x \sin \phi \]

Now, we can collect all the \( x \) terms on one side:

\[ \tan \theta = x \sin \phi + x \tan \theta \cos \phi \]

Factoring out \( x \):

\[ \tan \theta = x (\sin \phi + \tan \theta \cos \phi) \]

Now, we can solve for \( x \):

\[ x = \frac{\tan \theta}{\sin \phi + \tan \theta \cos \phi} \]

### Step 2: Express \( y \) in terms of \( \tan \phi \) and \( \sin \theta \)

Using a similar approach with the second equation:

\[ \tan \phi (1 - y \cos \theta) = y \sin \theta \]

Expanding this gives:

\[ \tan \phi - y \tan \phi \cos \theta = y \sin \theta \]

Collecting \( y \) terms:

\[ \tan \phi = y \sin \theta + y \tan \phi \cos \theta \]

Factoring out \( y \):

\[ \tan \phi = y (\sin \theta + \tan \phi \cos \theta) \]

Now, we can solve for \( y \):

\[ y = \frac{\tan \phi}{\sin \theta + \tan \phi \cos \theta} \]

### Step 3: Find \( \frac{x}{y} \)

Now we can find \( \frac{x}{y} \):

\[ \frac{x}{y} = \frac{\frac{\tan \theta}{\sin \phi + \tan \theta \cos \phi}}{\frac{\tan \phi}{\sin \theta + \tan \phi \cos \theta}} \]

This simplifies to:

\[ \frac{x}{y} = \frac{\tan \theta}{\tan \phi} \cdot \frac{\sin \theta + \tan \phi \cos \theta}{\sin \phi + \tan \theta \cos \phi} \]

### Step 4: Substitute \( \tan \theta \) and \( \tan \phi \)

Recall that:

\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \quad \text{and} \quad \tan \phi = \frac{\sin \phi}{\cos \phi} \]

Substituting these into the equation gives:

\[ \frac{x}{y} = \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \phi}{\cos \phi}} \cdot \frac{\sin \theta + \frac{\sin \phi}{\cos \phi} \cos \theta}{\sin \phi + \frac{\sin \theta}{\cos \theta} \cos \phi} \]

This simplifies to:

\[ \frac{x}{y} = \frac{\sin \theta \cos \phi}{\sin \phi \cos \theta} \]

### Final Result

Thus, we have:

\[ \frac{x}{y} = \frac{\sin \theta}{\sin \phi} \]