The value of  $$2-\frac{sin^2⁡α}{1-cos⁡α }+\frac{1-cos⁡α}{sin⁡α} -\frac{sin⁡ α}{1+cos⁡α }$$​ is

Correct option is B

Given:​

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

Concept Used: 

$$\sin^2 \theta + \cos^2 \theta = 1 \\ \ \\ \implies 1 - \cos^2 \theta = \sin^2 \theta \\ \ \\ \implies (1-\cos \theta)(1+\cos \theta) = \sin^2 \theta$$ 

Solution: 

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