Explanation

(a) We know that the range of projectile projected with velocity $u$, making an angle $\theta$ with the horizontal direction up the inclined plane, whose inclination with the horizontal direction is $\theta_{0},$ is
$R = \frac{u^{2}}{gcos^{2}\theta_{0}}\lbrack\sin\left( 2\theta - \theta_{0} \right) - \sin\theta_{0}\rbrack$
Here, $u = v,\theta = (90{^\circ} + \theta),\theta_{0} = \theta$
$\therefore R = \frac{v^{2}}{g\cos^{2}\theta_{0}}\{\sin{\left\lbrack 2(90{^\circ} + \theta) \right\rbrack - \theta}\} - \sin\theta\}$
$= \frac{v^{2}}{g\cos^{2}\theta_{0}}\lbrack\sin(180{^\circ} + \theta) - \sin\theta\rbrack$
$= - \frac{v^{2}}{g\cos^{2}\theta_{0}}2\sin\theta = - \frac{2u^{2}}{g}\tan\theta\sec\theta$
$= \frac{2v^{2}}{g}\tan\theta$ (in magnitude)