A

Step 1: In projectile motion, the range R and the maximum height H are given by the formulas:
R = \frac{u^2 \sin(2\theta)}{g}
H = \frac{u^2 \sin^2(\theta)}{2g}

Step 2: Given that R = nH, substituting the formulas we get:
\frac{u^2 \sin(2\theta)}{g} = n \times \frac{u^2 \sin^2(\theta)}{2g}.

Step 3: Canceling out common terms and simplifying gives:
\sin(2\theta) = \frac{n}{2} \sin^2(\theta).

Step 4: Using the identity \sin(2\theta) = 2 \sin(\theta) \cos(\theta), the equation transforms into:
2 \sin(\theta) \cos(\theta) = \frac{n}{2} \sin^2(\theta).

Step 5: Rearranging, we have:
4 \cos(\theta) = n \sin(\theta).

Step 6: Dividing throughout by \cos(\theta), we find:
4 = n \tan(\theta).

Step 7: This leads us to the conclusion that:
\tan(\theta) = \frac{4}{n}.

Therefore, the angle of projection \theta is equal to \tan^{-1}(\frac{4}{n}).
Hence, the statement is confirmed.