Shifting of origin (0,0) to (h,k)
`f(x,y)tof(x+h,y+k)`
Rotation of axes through an angle `theta`
By rotating the axes through an angle `theta` the equation `xy-y^(2)-3y+4=0` is transformed to the form which does not contain the term of xy then `sin theta=`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Substitute the new coordinates
We start with the equation \( xy - y^2 - 3y + 4 = 0 \). We will shift the origin and rotate the axes. The new coordinates after rotation by an angle \(\theta\) are given by:
\[ x = x' \cos \theta - y' \sin \theta \]
\[ y = x' \sin \theta + y' \cos \theta \]

### Step 2: Substitute into the equation
We substitute these expressions for \(x\) and \(y\) into the original equation:
\[ (x' \cos \theta - y' \sin \theta)(x' \sin \theta + y' \cos \theta) - (x' \sin \theta + y' \cos \theta)^2 - 3(x' \sin \theta + y' \cos \theta) + 4 = 0 \]

### Step 3: Expand the equation
Now we expand the left-hand side:
1. The term \( (x' \cos \theta - y' \sin \theta)(x' \sin \theta + y' \cos \theta) \):
\[ = x' \cos \theta \cdot x' \sin \theta + x' \cos \theta \cdot y' \cos \theta - y' \sin \theta \cdot x' \sin \theta - y' \sin \theta \cdot y' \cos \theta \]
\[ = x'^2 \cos \theta \sin \theta + x'y' \cos^2 \theta - x'y' \sin^2 \theta - y'^2 \sin \theta \cos \theta \]

2. The term \( -(x' \sin \theta + y' \cos \theta)^2 \):
\[ = - (x'^2 \sin^2 \theta + 2x'y' \sin \theta \cos \theta + y'^2 \cos^2 \theta) \]

3. The term \( -3(x' \sin \theta + y' \cos \theta) \):
\[ = -3x' \sin \theta - 3y' \cos \theta \]

### Step 4: Combine all terms
Combining all these terms, we get:
\[ x'^2 (\cos \theta \sin \theta - \sin^2 \theta) + y'^2 (\cos^2 \theta - \sin \theta \cos \theta) + x'y' (2\cos^2 \theta - 2\sin^2 \theta) - 3x' \sin \theta - 3y' \cos \theta + 4 = 0 \]

### Step 5: Collect coefficients of \(x'y'\)
We need to find the coefficient of \(x'y'\) and set it to zero:
\[ 2\cos^2 \theta - 2\sin^2 \theta = 0 \]
This simplifies to:
\[ \cos^2 \theta - \sin^2 \theta = 0 \]
Thus, we have:
\[ \cos^2 \theta = \sin^2 \theta \implies \tan^2 \theta = 1 \implies \tan \theta = 1 \implies \theta = \frac{\pi}{4} \]

### Step 6: Find \(\sin \theta\)
Now we can find \(\sin \theta\):
\[ \sin \theta = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \]

### Final Answer
Thus, the value of \(\sin \theta\) is:
\[ \sin \theta = \frac{\sqrt{2}}{2} \]