The greatest resultant of two vectors `vec P` and `vec Q` is (n) times their least reast resultant. Fiven
`| vec P | gt |vec Q|`. When `theta` is the angle between the two vectors, their resultant is half the sum of the two vectors. Show that,
`cos theta =- (n^(2) +2) // (n n^(2) -1)`.

Explanation

If `overset(rarr)R` is the resultant of `overset(rarr)A` and `overset(rarr)B` then `R^(2)=A^(2)+B^(2)+2AB cos theta`
we are given that
form equation (i) , (ii) and (iii)
`(nA)/(n+1)^(2)=A^(2)+(n-1)/(n+1)^(2)` or `(n^(2))/(n+1)^(2)=1+(n-1)^(2)/(n+1)^(2)+(2n-1)/(n+1)cos theta`
`cos theta =(n^(2)+2)/(2(n^(2)-1))`