The value of  $$2 - \frac{cos^2\theta}{1 + sin\theta} +\frac{1 + sin\theta}{cos \theta} - \frac{cos\theta}{1 - sin\theta}$$ is:

Correct option is D

Given: ​

​$$2 - \frac{cos^2\theta}{1 + sin\theta} +\frac{1 + sin\theta}{cos \theta} - \frac{cos\theta}{1 - sin\theta}$$ 

Formula Used: ​

​$$\sin^2 \theta + \cos^2 \theta = 1$$ 

Solution: ​

​$$2 - \frac{\cos^2\theta}{1 + \sin\theta} +\frac{1 + \sin\theta}{\cos \theta} - \frac{\cos\theta}{1 - \sin\theta} \\ \ \\ = 2 - \frac{1 -\sin^2\theta}{1 + \sin\theta} +\frac{(1 + \sin\theta)(1 - \sin\theta)-(\cos^2 \theta)}{(\cos \theta)(1 - \sin\theta)} \\ \ \\ = 2 - \frac{(1 -\sin\theta)(1+\sin \theta)}{1 + \sin\theta} +\frac{1 - \sin^2\theta - \cos^2 \theta}{\cos \theta(1 - \sin \theta)} \\ \ \\ = 2 - (1 -\sin\theta) +\frac{\cos^2\theta - \cos^2 \theta}{\cos \theta(1 - \sin \theta)} \\ \ \\ = 2 - (1 -\sin\theta) +\frac{0}{\cos \theta(1 - \sin \theta)} \\ \ \\ = 2 - 1+\sin \theta \\ \ \\ = 1 + \sin \theta$$​​