$$\frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x}$$ is equal to:

$$\frac{\cos x}{1+\sin x}+\frac{1+\sin x}{\cos x}=\frac{\cos x}{1+\sin x}\times\frac{1-\sin x}{1-\sin x}+\frac{1+\sin x}{\cos\ x}$$

$$=\frac{\cos x\left(1-\sin x\right)}{\left(1-\sin^2x\right)}+\frac{1}{\cos x}+\frac{\sin x}{\cos\ x}$$

$$=\frac{\cos x\left(1-\sin x\right)}{\cos^2x}+\sec x+\tan x$$

$$=\frac{1-\sin x}{\cos x}+\sec x+\tan x$$

$$=\frac{1}{\cos x}-\frac{\sin x}{\cos x}+\sec x+\tan x$$

$$=\sec x-\tan x+\sec x+\tan x$$

$$=2\sec x$$

Hence, the correct answer is Option D