The value of $$\left(\frac{1}{\sec^2 \theta - \cos^2 \theta} + \frac{1}{\cosec^2 \theta - \sin^2 \theta} \right) (\sin^2 \theta \cos^2 \theta)$$​

Correct option is C

Given:

​$$\left(\frac{1}{\sec^2 \theta - \cos^2 \theta} + \frac{1}{\cosec^2 \theta - \sin^2 \theta} \right) (\sin^2 \theta \cos^2 \theta)$$​​

Formula Used:

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

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

​$$(a^3 - b^3) =(a-b)(a^2 +b^2 +ab)$$​​

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

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

​$$\frac{1}{\cos \theta} = \sec\theta$$​​

Solution:

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

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

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