The value of $$(1+\cot^2 \theta)(1+ \cos \theta)(1 - \cos \theta) + (1 -\sin\theta)(1 + \sin \theta)(1 +\tan^2 \theta)$$​ is:

Correct option is B

Given:

​$$(1+\cot^2 \theta)(1+ \cos \theta)(1 - \cos \theta) + (1 -\sin\theta)(1 + \sin \theta)(1 +\tan^2 \theta)$$​​

Formula Used:

​$$\tan \theta = \frac{\sin \theta}{\cos\theta}$$​​

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

​$$\sin( \theta) = \frac{1}{\cosec \theta}$$​​

​$$(a-b)(a+b) = a^2 - b^2$$​​

​$$1+\cot^2 \theta = \cosec^2 \theta$$​​

​$$1+\tan^2 \theta = \sec^2 \theta$$​​

​$$\cosec \theta = \frac{1}{\sin \theta}$$​​

Solution:

​$$(1+\cot^2 \theta)(1+ \cos \theta)(1 - \cos \theta) + (1 -\sin\theta)(1 + \sin \theta)(1 +\tan^2 \theta)$$​​

$$=(\cosec^2 \theta)(1- \cos^2 \theta) + (1 -\sin^2 \theta)(\sec^2 \theta)$$​

$$=\frac{1}{\sin^2 \theta}(\sin^2 \theta) + \frac{1}{\sec^2 \theta}(\sec^2 \theta)$$​​

= 2