wall cross

Equation of trajectory of the particle, $y = x \tan \theta (1-\frac {x}{R})$ .
$a = b \tan \theta (1- \frac {b}{R})$ . . . . . . . . . . . . . (1).
$b = a \tan \theta (1- \frac {a}{R})$ . . . . . . . . . . . . . . . .(2).
Hence, $\frac {a^{2}}{b^{2}} = \frac {R-b}{R-a}$ .
$R = \frac {a^{3}-b^{3}}{a^{2}-b^{2}}$ .
By substituting the value of R in any of the above equations, we will get the value of $\tan \theta = \boxed {\frac {a}{b} + \frac {b}{a} +1}$ .

Let the times the stone, with a projected velocity $v$ at an angle $\theta$ with the horizontal, clears the tops of the first and second walls be $t_1$ and $t_2$ respectively. Therefore, $t_1 = \cfrac{b}{u\cos{\theta}}, \quad t_2 = \cfrac{a}{u\cos{\theta}}$ and $a = u\sin{\theta}t_1-\cfrac{1}{2}gt_1^2, \quad b = u\sin{\theta}t_2-\cfrac{1}{2}gt_2^2$ Therefore, $a = u\sin{\theta}\cfrac{b}{u\cos{\theta}}-\cfrac{1}{2}g\left(\cfrac{b}{u\cos{\theta}}\right)^2$ $a = b\tan{\theta}-\cfrac{gb^2}{2u^2}\sec^2{\theta}$ $\div b^2$ throughout: $\frac{a}{b^2} = \cfrac{\tan{\theta}}{b}-\cfrac{g}{2u^2}\sec^2{\theta}$ Similarly, for the equation for $b$ , we have: $\frac{b}{a^2} = \cfrac{\tan{\theta}}{a}-\cfrac{g}{2u^2}\sec^2{\theta}$ Subtracting the above two equations: $\frac{a}{b^2}-\frac{b}{a^2} = \left( \cfrac { 1 }{ b } -\cfrac { 1 }{ a } \right) \tan{\theta}$ $\cfrac{a^3-b^3}{a^2b^2}=\cfrac{a-b}{ab}\tan{\theta}$ $\tan{\theta}=\cfrac {a^2+ab+b^2}{ab}=\boxed {1+\cfrac{a}{b}+\cfrac{b}{a}}$

Let the projectile travel a distance of 'b' when it's is at a height of 'a'. Let the initial velocity of projection be 'u' at an angle of 'x'. In X-axis b=ucos (x) t where 't' is the time the projectile takes to cover 'b'. therefore t=b/(ucos(x)) and In Y-axis a=usin(x) t-(g/2) (t^2)) putting the value of t in this equation, we obtain a=btan(x)-5 (b^2)/((ucos(x)^2))---------------------------(eqn-1)

Let the projectile travel a distance of 'a' when it's is at a height of 'b'. Let the initial velocity of projection be 'u' at an angle of 'x'. In X-axis a=ucos (x) t1 where 't1' is the time the projectile takes to cover 'b'. therefore t1=a/(ucos(x)) and In Y-axis b=usin(x) t-(g/2) (t1^2)) putting the value of t1 in this equation, we obtain b=atan(x)-5 (a^2)/((ucos(x)^2))------------------------------(eqn-2)

equate both the equations with the help of ((u*cos(x))^2) to obtain the answer.