The value of $$\sqrt{\sec^2 \theta + \cosec^2 \theta} \times \sqrt{\tan^2 \theta - \sin^2 \theta}$$ is equal to:

As per the question,

$$\sqrt{\sec^2 \theta + \cosec^2 \theta} \times \sqrt{\tan^2 \theta - \sin^2 \theta}$$

$$\Rightarrow \sqrt{\dfrac{1}{\cos^2 \theta} + \dfrac{1}{\sin^2 \theta}} \times \sqrt{\dfrac{\sin^2\theta}{\cos^2 \theta} - \sin^2 \theta}$$

$$\Rightarrow \sqrt{\dfrac{\sin^2 \theta + \cos^2 \theta }{ \cos^2 \theta \sin^2 \theta} }\times\sqrt{\dfrac{\sin^2\theta}{\cos^2 \theta} - \sin^2 \theta}$$

We know that,$$ \sin^2 \theta + \cos^2 \theta=1$$

$$\Rightarrow \sqrt{\dfrac{\sin^2\theta}{ \cos^2 \theta \sin^2 \theta (\cos^2 \theta)} -\dfrac{ \sin^2\theta}{ \cos^2 \theta \sin^2 \theta}}$$

$$\Rightarrow \sqrt{\dfrac{1}{ \cos^4 \theta} -\dfrac{ 1}{ \cos^2 \theta }}$$

$$\Rightarrow \sqrt{\dfrac{1-\cos^2\theta}{ \cos^4 \theta }}$$

$$\Rightarrow \sqrt{\dfrac{\sin^2\theta}{ \cos^4 \theta }}$$

$$\Rightarrow \sin \theta \sec^2 \theta $$