$$\frac{2 + \tan^2 \theta + \cot^2 \theta}{\sec \theta \cosec \theta}$$ is equal to:

$$\frac{2 + \tan^2 \theta + \cot^2 \theta}{\sec \theta \cosec \theta}$$

= $$\frac{1 + \tan^2 \theta +1 + \cot^2 \theta}{\sec \theta \cosec \theta}$$

=$$\frac{ \sec^2 \theta + \cosec^2 \theta}{\sec \theta \cosec \theta}$$

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

= $$\frac{1}{sin\theta cos\theta}$$

= $$\sec\theta\cosec\theta$$

So , the answer would be option c)$$\sec\theta\cosec\theta$$.