The greatest resultant of two vectors P and Q is n times their least resultant.Given magnitude of vector P is equal to magnitude of vector Q.When theta is the angle b/w 2 vectors the magnitude of resultant is half of the sum of magnitudes of two vectors.find cos theta

Rmax= P+Q
Rmin=P-Q
so, P+Q=n(P+Q)
so, (p+q)^2=n^2(p+q)^2
By solving this you get....
-(p^2+q^2)/2pq = (1-n^2)/(1+n^2)
Now,
rt(p^2+q^2+2pqcos(theta))=(p+q)/2
by solving the above above eq by squaring b.s.  and separating denominator
We get,
((3(p^2+q^2))/-8(pq))+(1/4)=cos(theta)
Now we substitute the value of (p^2+q^2)/-2pq
so we get,
(3(1+n^2)/4(1-n^2))+(1/4)=cos(theta)
Finally solving the above eq by taking LCM, we get,
cos(theta)=(n^2+2)/2(1-n^2)
that is your final answer...
hope it helps everyone....