AB is vertical tower. The point A is on the ground and C is the middle point of AB. The part CB subtend an angle `alpha` at a point P on the ground. If `AP = nAB`, then `tan alpha =`

Options

A. `n=(n^2+1) tan alpha`
B. `n=(2n^(2) -1) tan alpha`
C. `n^(2)=(2n^2 +1) tan alpha`
D. `n=(2n^(2)+1)tan alpha`

Answer

To solve the problem step by step, we will analyze the given information and apply trigonometric concepts.

### Step 1: Understand the Geometry
- Let \( AB \) be a vertical tower where \( A \) is at the ground level and \( B \) is at the top of the tower.
- The point \( C \) is the midpoint of \( AB \).
- The distance \( AP \) is given as \( n \cdot AB \).

### Step 2: Define Variables
- Let \( AB = h \) (the height of the tower).
- Therefore, \( C \) is at a height of \( \frac{h}{2} \) (since \( C \) is the midpoint).
- The distance \( AP = n \cdot h \).

### Step 3: Analyze Triangle \( ACP \)
- In triangle \( ACP \), we can denote the angle at \( C \) as \( \beta \).
- We can use the tangent function:
\[ \tan \beta = \frac{AC}{AP} \]
- Since \( AC = \frac{h}{2} \) and \( AP = n \cdot h \), we have:
\[ \tan \beta = \frac{\frac{h}{2}}{n \cdot h} = \frac{1}{2n} \]

### Step 4: Analyze Triangle \( ABP \)
- In triangle \( ABP \), the angle at \( A \) is \( \alpha + \beta \).
- We can use the tangent addition formula:
\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]
- Here, \( AB = h \) and \( AP = n \cdot h \), so:
\[ \tan(\alpha + \beta) = \frac{h}{n \cdot h} = \frac{1}{n} \]

### Step 5: Substitute Values
- Substitute \( \tan \beta = \frac{1}{2n} \) into the tangent addition formula:
\[ \frac{\tan \alpha + \frac{1}{2n}}{1 - \tan \alpha \cdot \frac{1}{2n}} = \frac{1}{n} \]

### Step 6: Cross Multiply
- Cross multiplying gives:
\[ \tan \alpha + \frac{1}{2n} = \frac{1}{n} \left( 1 - \tan \alpha \cdot \frac{1}{2n} \right) \]
- Rearranging leads to:
\[ n(\tan \alpha + \frac{1}{2n}) = 1 - \frac{\tan \alpha}{2n} \]

### Step 7: Simplify
- Multiply through by \( 2n \) to eliminate the fraction:
\[ 2n^2 \tan \alpha + 1 = 2n - \tan \alpha \]
- Rearranging gives:
\[ 2n^2 \tan \alpha + \tan \alpha = 2n - 1 \]
- Factor out \( \tan \alpha \):
\[ \tan \alpha (2n^2 + 1) = 2n - 1 \]

### Step 8: Solve for \( \tan \alpha \)
- Finally, we can express \( \tan \alpha \):
\[ \tan \alpha = \frac{2n - 1}{2n^2 + 1} \]

### Final Answer
Thus, the expression for \( \tan \alpha \) is:
\[ \tan \alpha = \frac{2n - 1}{2n^2 + 1} \]