Prove without using a calculator: \( \quad \frac{\cos 315^{\circ}+1}{\sin 315^{\circ}-1}=-1 \) Evaluate without using a calculator: \( \sqrt{4^{\sin 150^{\circ}} .2^{3 \tan 225^{\circ}}} \) Show that \( \theta=30^{\circ} \) is a solution to \( (\sin \theta)^{\sin \theta}=\frac{1}{\sqrt{2}} \)

To prove \( \frac{\cos 315^{\circ}+1}{\sin 315^{\circ}-1}=-1 \), we start by determining the values of \( \cos 315^{\circ} \) and \( \sin 315^{\circ} \). Since \( 315^{\circ} \) is in the fourth quadrant, we have: \[ \cos 315^{\circ} = \cos(360^{\circ} - 45^{\circ}) = \cos 45^{\circ} = \frac{\sqrt{2}}{2}, \] \[ \sin 315^{\circ} = \sin(360^{\circ} - 45^{\circ}) = -\sin 45^{\circ} = -\frac{\sqrt{2}}{2}. \] Now substituting these values into the original expression: \[ \frac{\frac{\sqrt{2}}{2} + 1}{-\frac{\sqrt{2}}{2} - 1} = \frac{\frac{\sqrt{2}}{2} + \frac{2}{2}}{-\frac{\sqrt{2}}{2} - \frac{2}{2}} = \frac{\frac{\sqrt{2} + 2}{2}}{-\frac{\sqrt{2} + 2}{2}} = -1. \] Thus, the equation holds. Now, let's evaluate \( \sqrt{4^{\sin 150^{\circ}} \cdot 2^{3 \tan 225^{\circ}}} \). We know: \[ \sin 150^{\circ} = \sin(180^{\circ} - 30^{\circ}) = \sin 30^{\circ} = \frac{1}{2}, \] \[ \tan 225^{\circ} = \tan(180^{\circ} + 45^{\circ}) = \tan 45^{\circ} = 1. \] Substituting the trigonometric values: \[ 4^{\sin 150^{\circ}} = 4^{\frac{1}{2}} = 2, \] \[ 2^{3 \tan 225^{\circ}} = 2^{3 \cdot 1} = 2^{3} = 8. \] Now we compute the expression: \[ \sqrt{4^{\sin 150^{\circ}} \cdot 2^{3 \tan 225^{\circ}}} = \sqrt{2 \cdot 8} = \sqrt{16} = 4. \] Finally, we'll show that \( \theta = 30^{\circ} \) is a solution to the equation \( (\sin \theta)^{\sin \theta} = \frac{1}{\sqrt{2}} \): Calculating \( \sin 30^{\circ} \): \[ \sin 30^{\circ} = \frac{1}{2}. \] Now substituting this into the equation: \[ \left(\frac{1}{2}\right)^{\frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}. \] Since both sides of the equation are equal, \( \theta = 30^{\circ} \) is indeed a solution to the equation.