Two projectiles are fired from ground with same initial speeds from same point at angles $$(45^\circ + \alpha)$$ and $$(45^\circ - \alpha)$$ with horizontal direction. The ratio of their times of flights is

Let the common initial speed of the two projectiles be $$u$$ and let the acceleration due to gravity be $$g$$ (acting downward).

Case 1: Projection angle $$\theta_1 = 45^\circ + \alpha$$
Time of flight formula for a projectile launched from ground level is
$$T = \frac{2u \sin\theta}{g}$$
Thus, $$T_1 = \frac{2u \sin (45^\circ + \alpha)}{g}$$ $$-(1)$$

Case 2: Projection angle $$\theta_2 = 45^\circ - \alpha$$
Similarly,
$$T_2 = \frac{2u \sin (45^\circ - \alpha)}{g}$$ $$-(2)$$

We require the ratio $$\dfrac{T_1}{T_2}$$. Using $$(1)$$ and $$(2)$$, the common factors $$\frac{2u}{g}$$ cancel out:

$$\frac{T_1}{T_2} = \frac{\sin (45^\circ + \alpha)}{\sin (45^\circ - \alpha)}$$ $$-(3)$$

Apply the sine addition-subtraction identities:
$$\sin (45^\circ \pm \alpha) = \sin 45^\circ \cos\alpha \pm \cos 45^\circ \sin\alpha$$

Since $$\sin 45^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2}$$, we obtain
$$\sin (45^\circ + \alpha) = \frac{\sqrt{2}}{2}(\cos\alpha + \sin\alpha)$$
$$\sin (45^\circ - \alpha) = \frac{\sqrt{2}}{2}(\cos\alpha - \sin\alpha)$$

Substitute these into $$(3)$$:

$$\frac{T_1}{T_2} \;=\; \frac{\frac{\sqrt{2}}{2}(\cos\alpha + \sin\alpha)}{\frac{\sqrt{2}}{2}(\cos\alpha - \sin\alpha)}$$

The factor $$\frac{\sqrt{2}}{2}$$ cancels out, giving

$$\frac{T_1}{T_2} = \frac{\cos\alpha + \sin\alpha}{\cos\alpha - \sin\alpha}$$ $$-(4)$$

To convert the expression to a form involving $$\tan\alpha$$, divide numerator and denominator of $$(4)$$ by $$\cos\alpha$$:

$$\frac{T_1}{T_2} = \frac{1 + \tan\alpha}{1 - \tan\alpha}$$

Hence, the required ratio of times of flight is $$\dfrac{1 + \tan\alpha}{1 - \tan\alpha}$$, which corresponds to Option D.