1. Formula for maximum height of a projectile

For a projectile fired with speed $u$ at angle $\theta$, the maximum height is

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

Since both projectiles have the same initial speed $u$, their maximum heights are proportional to $\sin^2\theta$.

2. Write the heights for the two angles

The two angles are:

$\theta_1=45^\circ-\alpha, \qquad \theta_2=45^\circ+\alpha$

So,

$H_1=\frac{u^2\sin^2(45^\circ-\alpha)}{2g}, \qquad H_2=\frac{u^2\sin^2(45^\circ+\alpha)}{2g}$

Hence,

$\frac{H_1}{H_2}=\frac{\sin^2(45^\circ-\alpha)}{\sin^2(45^\circ+\alpha)}$

3. Expand the sine terms

Using

$\sin(45^\circ\pm\alpha)=\frac{1}{\sqrt2}(\cos\alpha\pm\sin\alpha)$

we get

$\sin^2(45^\circ-\alpha)=\frac{1}{2}(\cos\alpha-\sin\alpha)^2$

$\sin^2(45^\circ+\alpha)=\frac{1}{2}(\cos\alpha+\sin\alpha)^2$

Therefore,

$\frac{H_1}{H_2}=\frac{(\cos\alpha-\sin\alpha)^2}{(\cos\alpha+\sin\alpha)^2}$

4. Simplify

Expand numerator and denominator:

$(\cos\alpha-\sin\alpha)^2=\cos^2\alpha+\sin^2\alpha-2\sin\alpha\cos\alpha=1-\sin2\alpha$

$(\cos\alpha+\sin\alpha)^2=\cos^2\alpha+\sin^2\alpha+2\sin\alpha\cos\alpha=1+\sin2\alpha$

Thus,

$\frac{H_1}{H_2}=\frac{1-\sin2\alpha}{1+\sin2\alpha}$

5. Match with the options

This matches Option D.

$\boxed{\frac{1-\sin2\alpha}{1+\sin2\alpha}}$