a projectile is fired vertically at an angle theta from ground with velocity v. speed of projectile when it is moving at an angle delta from horizontal is ​  

Dear student

when projectile fired vertically with an angle $\theta$ from the ground it's horizontal and vertical component of velocity are given as  $v_x=v\cos \theta and v_y=v\sin \theta$. Also there is no acceleration in the horizontal so the horizontal component of velocity does not change during the projectile motion. 
Let after a time t the projectile makes angle $\delta$ from the horizontal and let the velocity is u then the horizontal and vertical component of velocity at this time is given by 
$u_x=v_x=u\cos \delta =v\cos \theta$
since the acceleration in the vertical direction is g so the vertical velocity is given by 
$$\begin{gathered} u_y=v_y-gt \\ {u}_y=v\sin \theta -gt \end{gathered}$$
But since u makes the angle $\delta$ from the horizontal so 
$$\begin{gathered} u_x=u\cos \delta \\ {u}_y=u\sin \delta or \\ \tan \delta =\frac{u_y}{u_x}=\frac{v\sin \theta -gt}{v\cos \theta } \\ \Rightarrow v\cos \theta \tan \delta =v\sin \theta -gt \\ \Rightarrow gt=v\sin \theta -v\cos \theta \tan \delta \\ so \\ {u}^2={u_x}^2+{u_y}^2 \\ {u}^2={\left(v\cos \theta \right)}^2+{\left(v\sin \theta -gt\right)}^2 \\ {u}^2={\left(v\cos \theta \right)}^2+\left(v\sin \theta -v\sin \theta +v\cos \theta \tan \delta \right) \\ {u}^2={\left(v\cos \theta \right)}^2+{\left(v\cos \theta \tan \delta \right)}^2 \\ {u}^2=v^2{\cos }^2\theta +v^2{\cos }^2\theta {\tan }^2\delta \\ {u}^2=v^2{\cos }^2\theta \left(1+{\tan }^2\delta \right) \\ {u}^2=v^2{\cos }^2\theta ·se{c}^2\delta \\ u=v\cos \theta ·sec\delta \end{gathered}$$