A plane surface is inclined making an angle $\theta$ with the horizontal. From the bottom of this inclined plane, a bullet is fired with velocity $v$. The maximum possible range of the bullet on the inclined plane is

The correct option is B $\frac{v^2}{g(1+\sin \theta )}$
Resolving the motion parallel and perpendicular to the inclined plane as shown in the diagram,
$v_x=v\cos (\alpha -\theta )v_y=v\sin (\alpha -\theta )a_x=-g\sin \theta a_y=-g\cos \theta$
If $T$ is the time of flight, then
$0=v\sin (\alpha -\theta )T-\frac{1}{2}g\cos \theta T^2$
(Using second equation of motion along Y-axis)
$\begin{array}{}\Rightarrow T=\frac{2v\sin (\alpha -\theta )}{g\cos \theta }& \text{(1)}\end{array}$
If the bullet covers horizontal distance $OB$ in time $T$,
$OB=v\cos \alpha \times T$

Now, $\cos \theta =\frac{OB}{OA}$ (From the figure)
$\Rightarrow OA=\frac{OB}{\cos \theta }=\frac{v\cos \alpha \times T}{\cos \theta }$
Using $\left(1\right)$,
$\Rightarrow OA=\frac{v^2}{g{\cos }^2\theta }\left[2\sin \right(\alpha -\theta \left)\cos \alpha \right]$
Using the identity $2\sin A\cos B=\sin (A+B)+\sin (A-B)$

$OA=\frac{v^2}{g{\cos }^2\theta }\left[\sin \right(2\alpha -\theta )-\sin \theta ]$

Clearly, the range $R(=OA)$ will be maximum when $\sin (2\alpha -\theta )$ is maximum, i.e., 1.
This would mean $2\alpha -\theta =\frac{\pi }{2}\text{or}\alpha =\frac{\theta }{2}+\frac{\pi }{4}$

Maximum range on the inclined plane can be denoted by $R_{max}$.
$\therefore R_{max}=\frac{v^2}{g{\cos }^2\theta }(1-\sin \theta )=\frac{v^2(1-\sin \theta )}{g(1-{\sin }^2\theta )}=\frac{v^2(1-\sin \theta )}{g(1+\sin \theta )(1-\sin \theta )}=\frac{v^2}{g(1+\sin \theta )}$