A particle is moving in XY-plane. It starts to move from origin O at an angle a with X-axis. It has a constant acceleration in negative Y-direction. After some time it passes through a point B in a direction making angle $\beta$ with the X-axis. If OB makes $\theta ={45}^{\circ }$ angle with the X-axis, the value of $\tan \alpha +\tan \beta$ is __________.

The X-coordinate of point B is

$x=\left(V_0\cos \alpha \right)t \text{ where, }{V}_0\text{ is initial velocity and }$

$y=\left(v_{0}t\sin \alpha -\frac{1}{2}{a}_0{t}^2\right)$.

where, a₀ is the acceleration

$\text{ Also, }\tan \theta =\frac{y}{x}$

Let the velocity of the particle is v at point B

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}v\cos \beta =v_0\cos \alpha \\ \text{ and }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}v\sin \beta =v_0\sin \alpha -a_{0}t\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\tan \beta =\frac{v_y}{v_x}=\frac{v_0\sin \alpha -a_{0}t}{v_0\cos \alpha }=\tan \alpha -\frac{a_{0}t}{v_0\cos \alpha } .....\left(\mathrm{i}\right)\end{array}$

$\mathrm{and} \tan \theta =\frac{y}{x}=\frac{v_{0}t\sin \alpha -\frac{1}{2}{a}_0{t}^2}{v_{0}t\cos \alpha }$

$\mathrm{or} 2\tan \theta =2\tan \alpha -\frac{a_{0}t}{v_0\cos \alpha } ......\left(\mathrm{ii}\right)$

From Eqs. (i) and (ii), we get

$\begin{array}{l}2\tan \theta -\tan \beta =2\tan \alpha -\tan \alpha \\ \mathrm{or} 2\tan \theta =\tan \alpha +\tan \beta \\ \therefore \tan \alpha +\tan \beta =2\tan {45}^{\circ }=2\end{array}$