A line OA of length r starts from its initial position OX and traces an angle AOB = `alpha` in the anitclockwise direction. It then traces back in the clockwise direction an angle BOC = `3theta` ( where `alpha gt 3 theta`). L is the foot of the perpendicular from C on OA. Also, `(sin^(3)theta)/(CL) = (cos^(3)theta)/(OL)=1`
`(2r^(2) -1)/(r)` is equal to

Options

A. `sin alpha`
B. `cos alpha `
C. `sin theta`
D. `cos theta`

Explanation

To solve the problem step by step, we will analyze the given information and derive the required expression.

### Step 1: Understand the Geometry
We have a line OA of length \( r \) that traces an angle \( \alpha \) in the anticlockwise direction and then traces back an angle \( 3\theta \) in the clockwise direction. Therefore, the angle AOC is given by:
\[ \text{Angle AOC} = \alpha - 3\theta \]

### Step 2: Identify Points and Distances
Let:
- \( O \) be the origin.
- \( A \) be the point where the line OA meets after tracing angle \( \alpha \).
- \( C \) be the point after tracing back \( 3\theta \).
- \( L \) be the foot of the perpendicular from \( C \) onto line \( OA \).

From the geometry, we can express the distances \( OL \) and \( CL \):
- \( OL = r \cos(\alpha - 3\theta) \)
- \( CL = r \sin(\alpha - 3\theta) \)

### Step 3: Set Up the Given Equations
According to the problem, we have:
\[ \frac{\sin^3 \theta}{CL} = \frac{\cos^3 \theta}{OL} = 1 \]
This implies:
\[ CL = \sin^3 \theta \quad \text{and} \quad OL = \cos^3 \theta \]

### Step 4: Substitute the Values of \( CL \) and \( OL \)
Substituting the expressions for \( CL \) and \( OL \) into the equations:
\[ \sin^3 \theta = r \sin(\alpha - 3\theta) \]
\[ \cos^3 \theta = r \cos(\alpha - 3\theta) \]

### Step 5: Rearranging the Equations
From the above equations, we can express \( r \) in terms of \( \sin \) and \( \cos \):
\[ r = \frac{\sin^3 \theta}{\sin(\alpha - 3\theta)} \quad \text{and} \quad r = \frac{\cos^3 \theta}{\cos(\alpha - 3\theta)} \]

### Step 6: Equate the Two Expressions for \( r \)
Setting the two expressions for \( r \) equal gives:
\[ \frac{\sin^3 \theta}{\sin(\alpha - 3\theta)} = \frac{\cos^3 \theta}{\cos(\alpha - 3\theta)} \]

### Step 7: Cross Multiply
Cross multiplying yields:
\[ \sin^3 \theta \cos(\alpha - 3\theta) = \cos^3 \theta \sin(\alpha - 3\theta) \]

### Step 8: Use Trigonometric Identities
Using the identity \( \sin(a) \cos(b) - \cos(a) \sin(b) = \sin(a - b) \):
\[ \sin^3 \theta \cos(\alpha - 3\theta) - \cos^3 \theta \sin(\alpha - 3\theta) = 0 \]

### Step 9: Factor the Equation
This can be factored as:
\[ \sin^3 \theta \cos(\alpha - 3\theta) = \cos^3 \theta \sin(\alpha - 3\theta) \]

### Step 10: Solve for \( 2r^2 - 1 \)
From the earlier steps, we can derive that:
\[ \frac{1 - r \cos(\alpha)}{r \sin(\alpha)} = \frac{2r \sin(\alpha)}{1 + 2r \cos(\alpha)} \]
From this, we can isolate \( 2r^2 - 1 \):
\[ 2r^2 - 1 = \cos(\alpha) \]

### Final Answer
Thus, the value of \( \frac{2r^2 - 1}{r} \) is:
\[ \frac{2r^2 - 1}{r} = \cos(\alpha) \]