​Solve the following.​

​$$1+\frac{1+cos\spaceθ}{sin\spaceθ}-\frac{sin^2\spaceθ}{1+cos\spaceθ}-\frac{sin\spaceθ}{1-cos\spaceθ}=$$ ?​

Correct option is A

Given:

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

Formula Used:
Trigonometric Identities:
​$$\sin^2 \theta + \cos^2 \theta = 1$$​​
Solution:
The third term is $$\frac{\sin^2 \theta}{1 + \cos \theta}$$​​

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

The fourth term is $$\frac{\sin \theta}{1 - \cos \theta}$$​​
Multiply numerator and denominator by (1 + c$$os \theta)$$​ (conjugate):
​$$\frac{\sin \theta (1 + \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)} = \frac{\sin \theta (1 + \cos \theta)}{1 - \cos^2 \theta} = \frac{\sin \theta (1 + \cos \theta)}{\sin^2 \theta} = \frac{1 + \cos \theta}{\sin \theta}$$​​
Now, the original expression becomes:

​$$1 + \frac{1 + \cos \theta}{\sin \theta} - (1 - \cos \theta) - \frac{1 + \cos \theta}{\sin \theta}$$​​

​$$= 1 - (1 - \cos \theta) = \cos \theta$$