Two vectors `overset rarr (A)` "and" `overset rarr (B)` have equal in magnitudes. If magnitudes of `overset rarr (A)+ overset rarr (B)` is equal to n
times the magnitude of `overset rarr (A)- overset rarr (B)` , then the angle
between `overset rarr (A)"and" overset rarr (B)` is :

Options

A. `cos^(-1)((n+n^(2))/(n-n^(2)))`
B. `cos^(-1)((n^(2)+1)/(n^(2)-1))`
C. `cos^(-1)((n^(2)-1)/(n^(2)+1))`
D. `cos ^(-1)((2+n)/(2-n))`

Explanation

To solve the problem, we need to find the angle between two vectors \( \vec{A} \) and \( \vec{B} \) given that they have equal magnitudes and that the magnitude of \( \vec{A} + \vec{B} \) is equal to \( n \) times the magnitude of \( \vec{A} - \vec{B} \).

### Step-by-Step Solution:

1. **Define the Magnitudes of Vectors**:
Let the magnitude of both vectors \( \vec{A} \) and \( \vec{B} \) be \( A \) (i.e., \( |\vec{A}| = |\vec{B}| = A \)).

2. **Express the Magnitudes of the Sum and Difference**:
The magnitude of the sum of the vectors is given by:
\[ |\vec{A} + \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}||\vec{B}| \cos \theta} \]
Since \( |\vec{A}| = |\vec{B}| = A \), this simplifies to:
\[ |\vec{A} + \vec{B}| = \sqrt{A^2 + A^2 + 2A^2 \cos \theta} = \sqrt{2A^2(1 + \cos \theta)} = A \sqrt{2(1 + \cos \theta)} \]

The magnitude of the difference of the vectors is given by:
\[ |\vec{A} - \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 - 2 |\vec{A}||\vec{B}| \cos \theta} \]
This simplifies to:
\[ |\vec{A} - \vec{B}| = \sqrt{A^2 + A^2 - 2A^2 \cos \theta} = \sqrt{2A^2(1 - \cos \theta)} = A \sqrt{2(1 - \cos \theta)} \]

3. **Set Up the Given Relationship**:
According to the problem, we have:
\[ |\vec{A} + \vec{B}| = n |\vec{A} - \vec{B}| \]
Substituting the expressions we derived:
\[ A \sqrt{2(1 + \cos \theta)} = n A \sqrt{2(1 - \cos \theta)} \]

4. **Cancel Common Terms**:
We can cancel \( A \) (assuming \( A \neq 0 \)) and \( \sqrt{2} \) from both sides:
\[ \sqrt{1 + \cos \theta} = n \sqrt{1 - \cos \theta} \]

5. **Square Both Sides**:
Squaring both sides gives:
\[ 1 + \cos \theta = n^2 (1 - \cos \theta) \]

6. **Rearranging the Equation**:
Rearranging the equation leads to:
\[ 1 + \cos \theta = n^2 - n^2 \cos \theta \]
\[ \cos \theta + n^2 \cos \theta = n^2 - 1 \]
\[ \cos \theta (1 + n^2) = n^2 - 1 \]

7. **Solve for \( \cos \theta \)**:
Thus, we have:
\[ \cos \theta = \frac{n^2 - 1}{n^2 + 1} \]

8. **Find the Angle \( \theta \)**:
Finally, we can find the angle \( \theta \):
\[ \theta = \cos^{-1} \left( \frac{n^2 - 1}{n^2 + 1} \right) \]

### Final Answer:
The angle \( \theta \) between the vectors \( \vec{A} \) and \( \vec{B} \) is:
\[ \theta = \cos^{-1} \left( \frac{n^2 - 1}{n^2 + 1} \right) \]