1. First, we show that $$\frac{1}{(1 + \csc \theta)(\sin \theta - \sin^2 \theta)} = \sec^2 \theta.$$ Step 1: Rewrite terms in sine and cosine. $$\csc \theta = \frac{1}{\sin \theta}$$ So, $$(1 + \csc \theta) = 1 + \frac{1}{\sin \theta} = \frac{\sin \theta + 1}{\sin \theta}.$$ Step 2: Simplify the other factor: $$\sin \theta - \sin^2 \theta = \sin \theta (1 - \sin \theta).$$ Step 3: Multiply both parts: $$(1 + \csc \theta)(\sin \theta - \sin^2 \theta) = \frac{\sin \theta + 1}{\sin \theta} \times \sin \theta (1 - \sin \theta) = (\sin \theta + 1)(1 - \sin \theta).$$ Step 4: Recognize this as a difference of squares: $$(\sin \theta + 1)(1 - \sin \theta) = 1 - \sin^2 \theta = \cos^2 \theta.$$ Step 5: Hence, $$\frac{1}{(1 + \csc \theta)(\sin \theta - \sin^2 \theta)} = \frac{1}{\cos^2 \theta} = \sec^2 \theta.$$ 2. Now solve $$(1 + \csc \theta)(\sin \theta - \sin^2 \theta) = \frac{3}{4}$$ Using above, $$\cos^2 \theta = \frac{3}{4}.$$ Step 1: Take square roots, $$\cos \theta = \pm \frac{\sqrt{3}}{2}.$$ Step 2: Find all angles $$\theta$$ in $$-180^\circ \leq \theta \leq 180^\circ$$ where $$\cos \theta = \pm \frac{\sqrt{3}}{2}$$. - $$\cos \theta = \frac{\sqrt{3}}{2}$$ at $$\theta = \pm 30^\circ$$ - $$\cos \theta = -\frac{\sqrt{3}}{2}$$ at $$\theta = \pm 150^\circ$$ Final answers: $$\theta = -150^\circ, -30^\circ, 30^\circ, 150^\circ.$$ 3. Solve $$3 \sec^2 (2\theta + \frac{\pi}{6}) = 4$$ for $$-\frac{\pi}{2} < \theta < \frac{\pi}{2}$$. Step 1: Divide both sides by 3: $$\sec^2 (2\theta + \frac{\pi}{6}) = \frac{4}{3}.$$ Step 2: Recall $$\sec^2 x = \frac{1}{\cos^2 x}$$, so $$\frac{1}{\cos^2 (2\theta + \frac{\pi}{6})} = \frac{4}{3} \Rightarrow \cos^2 (2\theta + \frac{\pi}{6}) = \frac{3}{4}.$$ Step 3: Taking square root, $$\cos (2\theta + \frac{\pi}{6}) = \pm \frac{\sqrt{3}}{2}.$$ Step 4: Solve for $$2\theta + \frac{\pi}{6}$$. Angles where $$\cos x = \frac{\sqrt{3}}{2}$$ are $$x = 2k\pi \pm \frac{\pi}{6}$$ but since $$\frac{\sqrt{3}}{2} = \cos \frac{\pi}{6}$$, actual solutions: $$2\theta + \frac{\pi}{6} = 2k\pi \pm \frac{\pi}{6}, \quad \text{or} \quad 2\theta + \frac{\pi}{6} = 2k\pi \pm \frac{5\pi}{6}$$ for integers $$k$$ because $$\cos x = -\frac{\sqrt{3}}{2}$$ at $$x = \pm \frac{5\pi}{6} + 2k\pi$$. Step 5: Solve each for $$\theta$$: - From $$\cos x = \frac{\sqrt{3}}{2}$$: $$2\theta + \frac{\pi}{6} = 2k\pi + \frac{\pi}{6} \Rightarrow 2\theta = 2k\pi \Rightarrow \theta = k\pi.$$ $$2\theta + \frac{\pi}{6} = 2k\pi - \frac{\pi}{6} \Rightarrow 2\theta = 2k\pi - \frac{\pi}{3} \Rightarrow \theta = k\pi - \frac{\pi}{6}.$$ - From $$\cos x = -\frac{\sqrt{3}}{2}$$: $$2\theta + \frac{\pi}{6} = 2k\pi + \frac{5\pi}{6} \Rightarrow 2\theta = 2k\pi + \frac{2\pi}{3} \Rightarrow \theta = k\pi + \frac{\pi}{3}.$$ $$2\theta + \frac{\pi}{6} = 2k\pi - \frac{5\pi}{6} \Rightarrow 2\theta = 2k\pi - \pi \Rightarrow \theta = k\pi - \frac{\pi}{2}.$$ Step 6: Check each solution within $$-\frac{\pi}{2} < \theta < \frac{\pi}{2}$$ for integer $$k$$. - $$\theta = k\pi$$ means $$k=0$$ (0), since for |k|≥1 it falls outside interval. - $$\theta = k\pi - \frac{\pi}{6}$$; for $$k=0$$, $$\theta = -\frac{\pi}{6}$$ (inside). For $$k=1$$, $$\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$ out of range. - $$\theta = k\pi + \frac{\pi}{3}$$; for $$k=0$$, $$\theta = \frac{\pi}{3}$$ in range. - $$\theta = k\pi - \frac{\pi}{2}$$; for $$k=0$$, $$\theta = -\frac{\pi}{2}$$ excluded since strict inequality. Final solutions inside interval: $$\theta = 0, -\frac{\pi}{6}, \frac{\pi}{3}.$$ Equivalent answers from the problem statement use a larger domain or different constants; given domain here yields above.