If $$\frac{1}{\sec \theta - \tan \theta} - \frac{1}{\cos \theta} = \sec \theta \times k, 0^\circ < \theta < 90^\circ$$, then k is equal to:

$$\frac{1}{\sec \theta - \tan \theta} - \frac{1}{\cos \theta} = \sec \theta \times k$$

$$\frac{1}{\frac{1}{\cos\ \theta}-\frac{\sin\ \theta}{\cos\ \theta}}-\frac{1}{\cos\theta}=\frac{1}{\cos\theta}\times k$$

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

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

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

We know that ($$\sin^2\ \theta + \cos^2\ \theta= 1$$).

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

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

$$\frac{\sin\ \theta-\sin^2\ \theta}{\left(1-\sin\ \theta\right)}=k$$
$$\frac{\sin\ \theta\left(1-\sin\ \theta\right)}{\left(1-\sin\ \theta\right)}=k$$

$$k = \sin\ \theta$$