If `tantheta=(x)/(y)`, then `(xsintheta+ycostheta)/(xsintheta-ycostheta)` is equal to

Options

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

Answer

To solve the problem, we start with the given equation:

\[ \tan \theta = \frac{x}{y} \]

We need to find the value of the expression:

\[ \frac{x \sin \theta + y \cos \theta}{x \sin \theta - y \cos \theta} \]

### Step 1: Express \(\sin \theta\) and \(\cos \theta\) in terms of \(x\) and \(y\)

From the definition of tangent, we can represent \(\sin \theta\) and \(\cos \theta\) using a right triangle where the opposite side (perpendicular) is \(x\) and the adjacent side (base) is \(y\).

Using the Pythagorean theorem, we find the hypotenuse \(h\):

\[ h = \sqrt{x^2 + y^2} \]

Thus, we can express \(\sin \theta\) and \(\cos \theta\) as follows:

\[ \sin \theta = \frac{x}{\sqrt{x^2 + y^2}} \]
\[ \cos \theta = \frac{y}{\sqrt{x^2 + y^2}} \]

### Step 2: Substitute \(\sin \theta\) and \(\cos \theta\) into the expression

Now, we substitute these values into the expression:

\[ x \sin \theta = x \cdot \frac{x}{\sqrt{x^2 + y^2}} = \frac{x^2}{\sqrt{x^2 + y^2}} \]
\[ y \cos \theta = y \cdot \frac{y}{\sqrt{x^2 + y^2}} = \frac{y^2}{\sqrt{x^2 + y^2}} \]

Now substituting these into the original expression:

\[ \frac{x \sin \theta + y \cos \theta}{x \sin \theta - y \cos \theta} = \frac{\frac{x^2}{\sqrt{x^2 + y^2}} + \frac{y^2}{\sqrt{x^2 + y^2}}}{\frac{x^2}{\sqrt{x^2 + y^2}} - \frac{y^2}{\sqrt{x^2 + y^2}}} \]

### Step 3: Simplify the expression

We can combine the fractions in the numerator and the denominator:

\[ = \frac{\frac{x^2 + y^2}{\sqrt{x^2 + y^2}}}{\frac{x^2 - y^2}{\sqrt{x^2 + y^2}}} \]

This simplifies to:

\[ = \frac{x^2 + y^2}{x^2 - y^2} \]

### Final Result

Thus, the final result is:

\[ \frac{x^2 + y^2}{x^2 - y^2} \]