For a projectile projected with speed $u$ at angle $\theta$:

• Range, $R=\frac{u^2\sin 2\theta}{g}$

• Maximum height, $H=\frac{u^2\sin^2\theta}{2g}$

Given that range and maximum height are equal: $R=H$

So, $\frac{u^2\sin 2\theta}{g}=\frac{u^2\sin^2\theta}{2g}$

Cancel $\frac{u^2}{g}$ from both sides: $\sin 2\theta=\frac{\sin^2\theta}{2}$

Use $\sin 2\theta=2\sin\theta\cos\theta$: $2\sin\theta\cos\theta=\frac{\sin^2\theta}{2}$

Multiply both sides by $2$: $4\sin\theta\cos\theta=\sin^2\theta$

For non-zero projection angle, divide by $\sin\theta$: $4\cos\theta=\sin\theta$

Hence, $\tan\theta=4$

Therefore, the correct option is: $\boxed{4}$ यानी option $\boxed{D}$.

Comparison with stored answer:

Stored correct answer = $D$

Our derived answer = $D$

So they agree.