Shifting or origin (0,0) to (b,k)
`f(x,y)impliesf(x+h,y+k)`
Rotation of axes through an angle `theta`
`f(x,y)impliesf(x cos theta-y sin theta,x sin theta+ycos theta)`
Now Answer the following questions:
(i) By rotating the axes through an angle `theta` the equation `xy-y^(2)-3y+4=0` is transformed to the form whcih does not contain the term of xy then `sin theta=`.........

Answer

To solve the problem of finding the value of \(\sin \theta\) such that the equation \(xy - y^2 - 3y + 4 = 0\) does not contain the term \(xy\) after rotating the axes through an angle \(\theta\), we will follow these steps:

### Step 1: Substitute the rotation formulas
We start by substituting the rotation formulas into the given equation. The transformations are:
\[ x' = x \cos \theta - y \sin \theta \]
\[ y' = x \sin \theta + y \cos \theta \]
We need to express the equation \(xy - y^2 - 3y + 4 = 0\) in terms of \(x'\) and \(y'\).

### Step 2: Rewrite the equation
Substituting \(x\) and \(y\) in terms of \(x'\) and \(y'\):
\[ xy = (x' \cos \theta + y' \sin \theta)(x' \sin \theta - y' \cos \theta) \]
Expanding this gives:
\[ xy = x' \cos \theta \cdot x' \sin \theta - x' \cos \theta \cdot y' \cos \theta + y' \sin \theta \cdot x' \sin \theta - y' \sin \theta \cdot y' \cos \theta \]
This simplifies to:
\[ xy = x' \sin \theta \cos \theta - y' \cos^2 \theta + y' \sin^2 \theta - y' \sin \theta \cdot y' \]

### Step 3: Collect terms
Now we need to collect all terms in the equation \(xy - y^2 - 3y + 4 = 0\) and set the coefficient of \(x'y'\) to zero (since we want to eliminate the \(xy\) term):
\[ \text{Coefficient of } x'y' = \cos^2 \theta - \sin^2 \theta - 2\sin \theta \cos \theta = 0 \]

### Step 4: Solve for \(\theta\)
Setting the coefficient of \(x'y'\) to zero gives us:
\[ \cos^2 \theta - \sin^2 \theta - 2\sin \theta \cos \theta = 0 \]
Using the identity \(\cos^2 \theta - \sin^2 \theta = \cos 2\theta\) and \(2\sin \theta \cos \theta = \sin 2\theta\), we can rewrite the equation as:
\[ \cos 2\theta - \sin 2\theta = 0 \]
This implies:
\[ \tan 2\theta = 1 \implies 2\theta = \frac{\pi}{4} + n\pi \implies \theta = \frac{\pi}{8} + \frac{n\pi}{2} \]

### Step 5: Calculate \(\sin \theta\)
Using \(\theta = \frac{\pi}{8}\):
\[ \sin \theta = \sin \left(\frac{\pi}{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2} \]

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