Let \(\mathrm{A}=\\{\theta: \tan \theta+\sec \theta=\sqrt{2} \sec \theta\\}\) and \(\mathrm{B}=\\{\theta: \sec \theta-\tan \theta=\sqrt{2} \tan \theta\\}\) be two sets then. (a) \(\mathrm{A}=\mathrm{B}\) (b) \(A \subset B\) (c) \(\mathrm{A} \neq \mathrm{B}\) (d) \(\mathrm{B} \subset \mathrm{A}\)

To solve \(\tan \theta + \sec \theta = \sqrt{2} \sec \theta\), first by isolating \(\tan \theta\) on the left side of the equation: \[ \tan \theta = \sqrt{2} \sec \theta - \sec \theta \] Now, factor out \(\sec \theta\) from the right side: \[ \tan \theta = \sec \theta (\sqrt{2} - 1) \] Remember that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and \(\sec \theta = \frac{1}{\cos \theta}\). We can substitute these expressions into our equation: \[ \frac{\sin \theta}{\cos \theta} = \frac{1}{\cos \theta} (\sqrt{2} - 1) \] Now multiply both sides by \(\cos \theta\) to get rid of the denominator: \[ \sin \theta = (\sqrt{2} - 1) \] Since this equation gives us a specific value for \(\sin \theta\), the set A contains the angles whose sine is equal to \(\sqrt{2} - 1\).

To solve \(\sec \theta - \tan \theta = \sqrt{2} \tan \theta\), first by isolating \(\tan \theta\) on the right side of the equation: \[ \sec \theta = \tan \theta(\sqrt{2} +1) \] Now, likewise, we'll substitute these expressions into our equation: \[ \frac{1}{\cos \theta} = \frac{\sin \theta}{\cos \theta}(\sqrt{2} + 1) \] Now multiply both sides by \(\cos \theta\) to get rid of the denominator: \[ 1 = (\sqrt{2} + 1)\sin \theta \] Divide both sides by \(\sqrt{2} + 1\): \[ \frac{1}{\sqrt{2} + 1} = \sin \theta \] Since this equation gives us a specific value for \(\sin \theta\), the set B contains the angles whose sine is equal to \(\frac{1}{\sqrt{2} + 1}\).

Now we can compare sets A and B: Set A has angles whose sine is equal to \(\sqrt{2} - 1\). Set B has angles whose sine is equal to \(\frac{1}{\sqrt{2} + 1}\). These sets have different angles (i.e., different values of \(\theta\)) since \(\sqrt{2} - 1 \neq \frac{1}{\sqrt{2} + 1}\). Therefore, the correct answer choice is: (c) \(\mathrm{A} \neq \mathrm{B}\)