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

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