Deriving Velocity and Normal Reaction in a Vertical Circular Motion

[jiboom]

2) a particle P of mass m moves in a vertical circle along the smooth inner surface of a fixed hollow sphere of internal radius a and centre O,the plane of the circle passing through O. The particle is projected from the lowest point of the sphere with a horizontal velocity u. when OP makes an angle T with the upward vertical,the velocity of the particle is v and the normal reaction between the particle and the sphere is R.Find expressions for v and R...done
v=[u^2-2ga(1+cos T)]^1/2 R=mu^2/a - mg(2+3cos T)
show if u^2,5ga the particle leaves the sphere where

cos T=(u^2-2ga)/3ga
if the particle leaves the sphere at a point A and its trajectory meets the sphere again at a point B such that AM is a diameter of the sphere show OA makes an angle of 45

from first part i have

cos T=(u^2-2ga)/3ga


v^2=[u^2-2ga(1+cos T)]

so when particle leaves sphere i get v^2=agcosT

now I am saying the angle of projection is equal to angle OP makes with upwards vertical,so from P to O we have

horizontal distance=asinT
vertical distance=acosT

for the projection

x=-tvcosT
y=tvsinT-gt^2/2

when x=-asinT => t=a(tanT)/v
so

-acosT=atanTsinT-g/2[(atanT)^2/(v^2)]
-acosT=atanTsinT-g/2[(atanT)^2/(agcosT)]
-2=2tan^2T-tan^2Tsec^2T
-2=tan^2T(2-sec^2T)
-2=tan^2T(2-1-tan^2T)
tan^4T-tan^2T-2=0
(tan^2T-2)(tan^2+1)=0

which is close as I am after tan T=1. is there an obvious error in my working?i hoping I am on right tracks but have dropped a sign somewhere?

 

[jiboom]

anyone able to spot my error here?