If \( \sqrt{2} \tan \alpha+1=0 \) and \( \cos \alpha>0 \), determine without the use of a calculator and with the of a diagram the value of: \( \begin{array}{ll}\text { (1) } \sqrt{3} \sin \alpha-\frac{\sqrt{2}}{\cos \alpha} & \text { (2) } \frac{1}{\cos ^{2} \alpha}+\tan ^{2} \alpha\end{array} \)

To solve the given equation \( \sqrt{2} \tan \alpha + 1 = 0 \), we isolate \(\tan \alpha\): \[ \tan \alpha = -\frac{1}{\sqrt{2}} \] Since \(\cos \alpha > 0\), \(\alpha\) must lie in the fourth quadrant (where \(\tan\) is negative). To find \(\sin \alpha\) and \(\cos \alpha\), we can use the identity \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\), leading to: \[ \sin \alpha = -\frac{1}{\sqrt{2}} \cos \alpha \] Recall the Pythagorean identity \(\sin^2 \alpha + \cos^2 \alpha = 1\). Substituting \(\sin \alpha\) yields: \[ \left(-\frac{1}{\sqrt{2}} \cos \alpha\right)^2 + \cos^2 \alpha = 1 \] \[ \frac{1}{2} \cos^2 \alpha + \cos^2 \alpha = 1 \] \[ \frac{3}{2} \cos^2 \alpha = 1 \implies \cos^2 \alpha = \frac{2}{3} \] Thus, \(\cos \alpha = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}\) (since \(\cos \alpha > 0\)). Now, substituting \(\cos \alpha\) back to find \(\sin \alpha\): \[ \sin \alpha = -\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{6}}{3} = -\frac{\sqrt{6}}{3\sqrt{2}} = -\frac{\sqrt{3}}{3} \] Next, we can compute the two requested expressions: 1) For \( \sqrt{3} \sin \alpha - \frac{\sqrt{2}}{\cos \alpha} \): \[ \sqrt{3} \sin \alpha = \sqrt{3}\left(-\frac{\sqrt{3}}{3}\right) = -1 \] \[ \frac{\sqrt{2}}{\cos \alpha} = \frac{\sqrt{2}}{\frac{\sqrt{6}}{3}} = \frac{3\sqrt{2}}{\sqrt{6}} = \frac{3\sqrt{2} \cdot \sqrt{6}}{6} = \frac{3\sqrt{12}}{6} = \frac{\sqrt{12}}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3} \] \[ \sqrt{3} \sin \alpha - \frac{\sqrt{2}}{\cos \alpha} = -1 - \sqrt{3} = - (1 + \sqrt{3}) \] 2) For \( \frac{1}{\cos^2 \alpha} + \tan^2 \alpha \): \[ \frac{1}{\cos^2 \alpha} = \frac{1}{\frac{2}{3}} = \frac{3}{2} \] \[ \tan^2 \alpha = \left(-\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \] \[ \frac{1}{\cos^2 \alpha} + \tan^2 \alpha = \frac{3}{2} + \frac{1}{2} = 2 \] Thus, the final values are: \[ \text{(1) } - (1 + \sqrt{3}), \quad \text{(2) } 2 \]