Two vector X and Y have equal magnitudes. If the magnitude of X+Y is n times the magnitude of X-Y, then angle between X and Y is

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$\sin^{- 1} (\frac{n - 1}{n + 1})$

$\cos^{- 1} (\frac{n - 1}{n + 1})$

$\cos^{- 1} (\frac{n^{2} - 1}{n^{2} + 1})$

$\sin^{- 1} (\frac{n^{2} - 1}{n^{2} + 1})$

answer is C.

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Detailed Solution

X=Y
Let angle between X and Y is $θ$
$| X + Y | = n | X - Y |$
On squaring both sides
$x^{2} + y^{2} + 2 x y \cos θ = n^{2} (x^{2} + y^{2} - 2 x y \cos θ)$
$x^{2} + x^{2} + 2 x^{2} \cos θ = n^{2} x^{2} + n^{2} x^{2} - 2 x^{2} n^{2} \cos θ$
$x^{2} [2 + 2 \cos θ] = x^{2} (2 n^{2} - 2 n^{2} \cos θ)$
$2 (1 + \cos θ) = 2 (n^{2} - n^{2} \cos θ)$
$1 + \cos θ = n^{2} - n^{2} \cos θ$
$n^{2} + 1 \cos θ = n^{2} - 1$
$\cos θ = \frac{n^{2} - 1}{n^{2} + 1}$