Two vectors `vec A and vecB` have equal magnitudes.If magnitude of `(vecA+vecB)` is equal to n times of the magnitude of `(vecA-vecB)` then the angle between `vecA and vecB` is :-

Options

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

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. **Set Up the Problem:**
Let the magnitude of both vectors be \(A\) (i.e., \(|\vec{A}| = |\vec{B}| = A\)). We are given:
\[ |\vec{A} + \vec{B}| = n \cdot |\vec{A} - \vec{B}| \]

2. **Square Both Sides:**
To eliminate the square root, we square both sides:
\[ |\vec{A} + \vec{B}|^2 = n^2 \cdot |\vec{A} - \vec{B}|^2 \]

3. **Expand Both Sides:**
Using the formula for the square of the magnitude of a vector sum:
\[ |\vec{A} + \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2 \vec{A} \cdot \vec{B} \]
\[ |\vec{A} - \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2 \vec{A} \cdot \vec{B} \]
Since \(|\vec{A}| = |\vec{B}| = A\), we have:
\[ |\vec{A} + \vec{B}|^2 = A^2 + A^2 + 2 \vec{A} \cdot \vec{B} = 2A^2 + 2 \vec{A} \cdot \vec{B} \]
\[ |\vec{A} - \vec{B}|^2 = A^2 + A^2 - 2 \vec{A} \cdot \vec{B} = 2A^2 - 2 \vec{A} \cdot \vec{B} \]

4. **Substitute Back:**
Substitute these into the squared equation:
\[ 2A^2 + 2 \vec{A} \cdot \vec{B} = n^2 (2A^2 - 2 \vec{A} \cdot \vec{B}) \]

5. **Rearranging the Equation:**
Rearranging gives:
\[ 2A^2 + 2 \vec{A} \cdot \vec{B} = 2n^2 A^2 - 2n^2 \vec{A} \cdot \vec{B} \]
\[ 2 \vec{A} \cdot \vec{B} + 2n^2 \vec{A} \cdot \vec{B} = 2n^2 A^2 - 2A^2 \]
\[ (2 + 2n^2) \vec{A} \cdot \vec{B} = 2(n^2 - 1)A^2 \]

6. **Solving for \(\vec{A} \cdot \vec{B}\):**
Dividing both sides by 2:
\[ (1 + n^2) \vec{A} \cdot \vec{B} = (n^2 - 1)A^2 \]
\[ \vec{A} \cdot \vec{B} = \frac{(n^2 - 1)A^2}{(1 + n^2)} \]

7. **Using the Dot Product Relation:**
Recall that \(\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta = A^2 \cos \theta\):
\[ A^2 \cos \theta = \frac{(n^2 - 1)A^2}{(1 + n^2)} \]

8. **Solving for \(\cos \theta\):**
Dividing both sides by \(A^2\) (assuming \(A \neq 0\)):
\[ \cos \theta = \frac{n^2 - 1}{1 + n^2} \]

9. **Finding the Angle \(\theta\):**
Therefore, the angle \(\theta\) between the vectors is:
\[ \theta = \cos^{-1} \left( \frac{n^2 - 1}{n^2 + 1} \right) \]

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