A particle is projected at an angle ‘a’ .After t seconds it appears to have an angle of elevation of ‘b’ as se

A particle is projected at an angle ‘a’ .After t seconds it appears to have an angle of elevation of ‘b’ as seen from the point of projection.
Prove that initial velocity is gt cos b/ {2sin (a-b)}

tan b = vertical displacement at time t / horizontal displacement at that instant.
tan b = (u sin a)*t – gt²/2) / (u cos a) t
tan b = tan a – (gt / [2u cos a])
tan a - tan b = (gt / [2u cos a])
u = gt / [2cos a* (tan a - tan b)]
Multiplying both numerator and denominator by cos b
u = gt cos b / [2(sin a cos b - cos a sin b)]
u = gt cos b / {2sin (a-b)}
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Let the initial velocity be "V" at angle "a". The parametric form of the trajectory is

x = (V cos(a)) t
y = (V sin(a))) t - (1/2)gt^2

so the angle "b" that it appears at time "t" is y/x:

tan(b) = tan(a) - gt/(2 V cos(a))

V = (gt/2) / (cos(a)(tan(a) - tan(b)))

= (gt/2) / (sec(b) sin(a - b))

= (gt/2) cos(b) / sin(a - b)