Explanation

(b) $v_{x} = v\cos(\alpha - \theta);v_{y} = v\sin(\alpha - \theta)$
$a_{x} = - g\sin\theta;a_{y} = - g\cos\theta$
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If $T$ is the time of flight, then
$0 = v\sin{(\alpha - \theta).T - \frac{1}{2}g\cos\theta}.T^{2}$
or $T = \frac{2v\sin(\alpha - \theta)}{g\cos\theta}$
$OB = v\cos\alpha \times T$
Now, $\cos\theta = \frac{OB}{OA}$ or $OA = \frac{OB}{\cos\theta}$
or $OA = \frac{v\sin{\alpha T}}{\cos\theta}$
or $OA = v\cos{\alpha \times \frac{2v\sin(\alpha - \theta)}{g\cos\theta} \times \frac{1}{\cos\theta}}$
or $OA = \frac{v^{2}}{g\cos^{2}\theta}\lbrack\sin(2\alpha - \theta)\cos\alpha\rbrack$
or $OA = \frac{v^{2}}{g\cos^{2}\theta}\lbrack sin(2\alpha - \theta) + sin( - \theta)\rbrack$
or $OA = \frac{v^{2}}{g\cos^{2}\theta}\lbrack\sin(2\alpha - \theta) - \sin\theta\rbrack$
Clearly, the range $R( = OA)$ will be maximum when sin $(2a - \theta)$ is maximum, $ie,1$.
This would mean
$2\ \alpha - \theta = \frac{\pi}{2}$ or $\alpha\ \frac{\theta}{2} + \frac{\pi}{4}$
Maximum range up the inclined plane,
$R_{\max} = \frac{v^{2}}{g\ cos^{2}\theta}\left( 1 - \sin\theta \right) = \frac{v^{2}(1 - \sin\theta)}{g(1 - \sin^{2}\theta)}$
$= \frac{v^{2}(1 - \sin\theta)}{g\left( 1 - \sin\theta \right)(1 + 1 - \sin\theta)} = \frac{v^{2}}{g(1 + \sin\theta)}$