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Step-by-Step Solution
Step 1: Express the magnitudes of (A + B) and (A − B)
Let each vector have magnitude $a$. If the angle between the vectors $ \overrightarrow{A} $ and $ \overrightarrow{B} $ is $ \theta $, then:
$ \bigl|\overrightarrow{A} + \overrightarrow{B}\bigr|
= \sqrt{\, a^2 + a^2 + 2 \, a \, a \cos\theta \,}
= \sqrt{\,2a^2(1 + \cos\theta)\,}
= 2a \cos\left(\frac{\theta}{2}\right). $
Similarly,
$ \bigl|\overrightarrow{A} - \overrightarrow{B}\bigr|
= \sqrt{\, a^2 + a^2 - 2 \, a \, a \cos\theta \,}
= \sqrt{\,2a^2(1 - \cos\theta)\,}
= 2a \sin\left(\frac{\theta}{2}\right). $
Step 2: Relate the magnitudes using the given condition
We are given that $ \bigl|\overrightarrow{A} + \overrightarrow{B}\bigr| $ is $n$ times $ \bigl|\overrightarrow{A} - \overrightarrow{B}\bigr| $. Mathematically:
$ \bigl|\overrightarrow{A} + \overrightarrow{B}\bigr|
= n \, \bigl|\overrightarrow{A} - \overrightarrow{B}\bigr|.
$
Substitute the expressions found in Step 1:
$ 2a \cos\left(\frac{\theta}{2}\right)
= n \times 2a \sin\left(\frac{\theta}{2}\right).
$
Simplifying,
$ \cos\left(\frac{\theta}{2}\right)
= n \, \sin\left(\frac{\theta}{2}\right)
\quad\Longrightarrow\quad
\tan\left(\frac{\theta}{2}\right) = n.
Step 3: Convert tangent to cosine form
If $ \tan\left(\frac{\theta}{2}\right) = n $, then let $ \frac{\theta}{2} = \alpha $. So $ \alpha = \tan^{-1}(n) $ and hence $ \theta = 2\alpha $.
We use the identity for cosine of a double angle:
$ \cos(\theta) = \cos(2\alpha)
= \frac{1 - \tan^2(\alpha)}{1 + \tan^2(\alpha)}.
$
Since $ \tan(\alpha) = n $, we have:
$ \cos(\theta)
= \frac{1 - n^2}{1 + n^2}.
$
Step 4: Determine the angle $\theta$
Hence,
$ \theta
= \cos^{-1} \Bigl[\frac{1 - n^2}{1 + n^2}\Bigr]
= \cos^{-1} \Bigl[\frac{n^2 - 1}{n^2 + 1}\Bigr] \quad (\text{by taking a negative sign inside carefully, we get an equivalent form}).
$
Typically, it is written as:
$ \theta = \cos^{-1}\left(\frac{n^2 - 1}{n^2 + 1}\right). $
Final Answer
The angle between the vectors is
$ \displaystyle \cos^{-1}\Bigl[\frac{n^2 - 1}{n^2 + 1}\Bigr].
$
