If `(sintheta)/(x)=(costheta)/(y)` then `sintheta-costheta` is equal to

Options

A. x-y
B. x+y
C. `(x-y)/(sqrt(x^(2)+y^(2)))`
D. `(y-x)/(sqrt(x^(2)+y^(2)))`

Answer

To solve the equation \(\frac{\sin \theta}{x} = \frac{\cos \theta}{y}\) and find the value of \(\sin \theta - \cos \theta\), we can follow these steps:

### Step 1: Rewrite the given equation
We start with the equation:
\[ \frac{\sin \theta}{x} = \frac{\cos \theta}{y} \]
Cross-multiplying gives us:
\[ \sin \theta \cdot y = \cos \theta \cdot x \]

### Step 2: Express \(\tan \theta\)
From the equation \(\sin \theta \cdot y = \cos \theta \cdot x\), we can divide both sides by \(\cos \theta\) (assuming \(\cos \theta \neq 0\)):
\[ \tan \theta = \frac{x}{y} \]

### Step 3: Construct a right triangle
In a right triangle, we can represent \(\tan \theta\) as the ratio of the opposite side (perpendicular) to the adjacent side (base):
- Let the opposite side be \(x\) and the adjacent side be \(y\).
- The hypotenuse \(h\) can be calculated using the Pythagorean theorem:
\[ h = \sqrt{x^2 + y^2} \]

### Step 4: Find \(\sin \theta\) and \(\cos \theta\)
Using the definitions of sine and cosine:
\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{x^2 + y^2}} \]
\[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{y}{\sqrt{x^2 + y^2}} \]

### Step 5: Calculate \(\sin \theta - \cos \theta\)
Now we can find \(\sin \theta - \cos \theta\):
\[ \sin \theta - \cos \theta = \frac{x}{\sqrt{x^2 + y^2}} - \frac{y}{\sqrt{x^2 + y^2}} \]
Combining the fractions:
\[ \sin \theta - \cos \theta = \frac{x - y}{\sqrt{x^2 + y^2}} \]

### Final Result
Thus, we have:
\[ \sin \theta - \cos \theta = \frac{x - y}{\sqrt{x^2 + y^2}} \]