If `xcostheta+ysintheta=z`, then what is the value of `(xsintheta-ycostheta)^(2)`?

Options

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

Explanation

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

**Given:**
\[ x \cos \theta + y \sin \theta = z \]

We need to find the value of:
\[ (x \sin \theta - y \cos \theta)^2 \]

### Step 1: Square both sides of the given equation
We start by squaring the equation \( z = x \cos \theta + y \sin \theta \):
\[ z^2 = (x \cos \theta + y \sin \theta)^2 \]

### Step 2: Expand the squared expression
Using the formula \( (a + b)^2 = a^2 + b^2 + 2ab \), we expand the right-hand side:
\[ z^2 = x^2 \cos^2 \theta + y^2 \sin^2 \theta + 2xy \sin \theta \cos \theta \]

### Step 3: Rearranging the equation
Now, we can rearrange this equation to express \( 2xy \sin \theta \cos \theta \):
\[ 2xy \sin \theta \cos \theta = z^2 - x^2 \cos^2 \theta - y^2 \sin^2 \theta \]

### Step 4: Define \( l \)
Let:
\[ l = (x \sin \theta - y \cos \theta)^2 \]
We will expand this expression:
\[ l = (x \sin \theta)^2 - 2(x \sin \theta)(y \cos \theta) + (y \cos \theta)^2 \]
\[ l = x^2 \sin^2 \theta - 2xy \sin \theta \cos \theta + y^2 \cos^2 \theta \]

### Step 5: Substitute for \( 2xy \sin \theta \cos \theta \)
Now we substitute the expression we found for \( 2xy \sin \theta \cos \theta \) into our equation for \( l \):
\[ l = x^2 \sin^2 \theta + y^2 \cos^2 \theta - (z^2 - x^2 \cos^2 \theta - y^2 \sin^2 \theta) \]

### Step 6: Simplify the expression
Now we simplify:
\[ l = x^2 \sin^2 \theta + y^2 \cos^2 \theta - z^2 + x^2 \cos^2 \theta + y^2 \sin^2 \theta \]
Grouping the terms:
\[ l = x^2 (\sin^2 \theta + \cos^2 \theta) + y^2 (\sin^2 \theta + \cos^2 \theta) - z^2 \]

### Step 7: Use the Pythagorean identity
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ l = x^2 \cdot 1 + y^2 \cdot 1 - z^2 \]
\[ l = x^2 + y^2 - z^2 \]

### Final Answer
Thus, the value of \( (x \sin \theta - y \cos \theta)^2 \) is:
\[ \boxed{x^2 + y^2 - z^2} \]