The angle between the straight lines represented by `(x^(2)+y^(2))sin^(2)alpha=(x cos alpha-y sin alpha)^(2)` is

Options

A. `alpha`
B. `2alpha`
C. `pi-2alpha`
D. none of these

Answer

To find the angle between the straight lines represented by the equation

\[ (x^2 + y^2) \sin^2 \alpha = (x \cos \alpha - y \sin \alpha)^2, \]

we can follow these steps:

### Step 1: Expand the right-hand side
Start by expanding the right side of the equation:

\[ (x \cos \alpha - y \sin \alpha)^2 = x^2 \cos^2 \alpha - 2xy \sin \alpha \cos \alpha + y^2 \sin^2 \alpha. \]

### Step 2: Rewrite the equation
Substituting the expanded form back into the equation gives:

\[ (x^2 + y^2) \sin^2 \alpha = x^2 \cos^2 \alpha - 2xy \sin \alpha \cos \alpha + y^2 \sin^2 \alpha. \]

### Step 3: Rearranging the equation
Rearranging the equation leads to:

\[ x^2 \sin^2 \alpha + y^2 \sin^2 \alpha - x^2 \cos^2 \alpha + 2xy \sin \alpha \cos \alpha - y^2 \sin^2 \alpha = 0. \]

This simplifies to:

\[ x^2 (\sin^2 \alpha - \cos^2 \alpha) + 2xy \sin \alpha \cos \alpha = 0. \]

### Step 4: Identify coefficients
From the equation \(Ax^2 + Bxy + Cy^2 = 0\), we can identify:

- \(A = \sin^2 \alpha - \cos^2 \alpha\)
- \(B = 2 \sin \alpha \cos \alpha\)
- \(C = 0\)

### Step 5: Use the formula for the angle between two lines
The angle \(\theta\) between two lines represented by the equation \(Ax^2 + Bxy + Cy^2 = 0\) is given by:

\[ \tan \theta = \frac{B}{A - C}. \]

Substituting the values we have:

\[ \tan \theta = \frac{2 \sin \alpha \cos \alpha}{\sin^2 \alpha - \cos^2 \alpha}. \]

### Step 6: Simplify using trigonometric identities
Using the double angle identities, we can simplify:

\[ \tan \theta = \frac{\sin 2\alpha}{\cos 2\alpha} = \tan 2\alpha. \]

### Step 7: Find the angle
Thus, we find that:

\[ \theta = 2\alpha. \]

### Conclusion
The angle between the straight lines represented by the given equation is:

\[ \theta = 2\alpha. \]