A particle is thrown with a velocity u at an angle$\theta$ from the horizontal. Another particle is thrown with the same velocity at an angle $\theta$  from the vertical. The ratio of times of flight of the two particles will be

A particle  is projected with velocity = u

Angle of projection$=\theta$

Another particle B is projected with velocity = u

Angle of projection$=90-\theta$

Time of flight of A is $=\frac{2\text{u}\sin \theta }{g}$

Time of flight of B is$=\frac{\text{2u}\sin \left(90-\theta \right)}{g}$

then${\text{T}}_{\text{A}}:{\text{T}}_{\text{B}}=\frac{2\text{u}\sin \theta }{g}:\frac{\text{2u}\sin \left(90-\theta \right)}{g}$

${\text{T}}_{\text{A}}:{\text{T}}_{\text{B}}=\sin \theta :\cos \theta =\tan \theta :1$