1. **State the problem:** Verify the trigonometric identities and solve the given trigonometric questions. 2. **Identity verification:** - For $\sec(\pi + \theta) = \sec \theta$, recall that $\sec x = \frac{1}{\cos x}$ and $\cos(\pi + \theta) = -\cos \theta$. Thus, $\sec(\pi + \theta) = \frac{1}{-\cos \theta} = -\sec \theta$, so the identity as stated is false unless considering absolute values. - For $\cos(\frac{3\pi}{2} + \theta) = \cos \theta$, use $\cos(a + b) = \cos a \cos b - \sin a \sin b$. Since $\cos(\frac{3\pi}{2})=0$ and $\sin(\frac{3\pi}{2})=-1$, we get $\cos(\frac{3\pi}{2} + \theta) = 0 \cdot \cos \theta - (-1) \cdot \sin \theta = \sin \theta$, which is not equal to $\cos \theta$. So this identity is false. - For $\tan(\frac{3\pi}{2} - \theta) = \cot \theta$, recall $\tan(\frac{3\pi}{2} - \theta) = \tan(-\frac{\pi}{2} + \theta) = -\cot \theta$, so the identity is false as stated. - For $\csc(\frac{3\pi}{2} - \theta) = \sec \theta$, since $\csc x = \frac{1}{\sin x}$ and $\sec x = \frac{1}{\cos x}$, and $\sin(\frac{3\pi}{2} - \theta) = -\cos \theta$, so $\csc(\frac{3\pi}{2} - \theta) = \frac{1}{-\cos \theta} = -\sec \theta$, not $\sec \theta$. 3. **Convert $\frac{7\pi}{90}$ radians to degrees:** Use $\text{degrees} = \frac{180}{\pi} \times \text{radians}$ $$\frac{180}{\pi} \times \frac{7\pi}{90} = \frac{180 \times 7}{90} = 14$$ So, $\frac{7\pi}{90}$ radians = 14°. 4. **Find angle for point $(-\frac{\sqrt{3}}{2}, \frac{1}{2})$:** Recall $\cos \theta = x$ and $\sin \theta = y$. Here, $\cos \theta = -\frac{\sqrt{3}}{2}$ and $\sin \theta = \frac{1}{2}$. This corresponds to $\theta = \frac{5\pi}{6}$ (150°) in the second quadrant. 5. **Exact values of trigonometric ratios:** (a) $\cos(-12\pi) = \cos(0) = 1$ since cosine is periodic with period $2\pi$. $\cos(-7\pi + (-5\pi)) = \cos(-12\pi) = 1$. $\cos(-\frac{1}{2} + \frac{1}{2}) = \cos(0) = 1$. (b) $\csc(\frac{11\pi}{3})$: Reduce angle: $\frac{11\pi}{3} = 2\pi + \frac{5\pi}{3}$, so $\csc(\frac{11\pi}{3}) = \csc(\frac{5\pi}{3})$. $\sin(\frac{5\pi}{3}) = -\frac{\sqrt{3}}{2}$, so $\csc(\frac{5\pi}{3}) = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$. $\sin(-\frac{2\pi}{\sqrt{3}})$ is not a standard angle; approximate or leave as is. 6. **Find $\sin(\frac{10\pi}{7})$ using identities:** (a) Using related acute angle: $\sin(\frac{21\pi}{14} - \frac{11\pi}{14}) = \sin(\frac{10\pi}{14}) = \sin(\frac{5\pi}{7})$. Since $\sin(\pi - x) = \sin x$, $\sin(\frac{10\pi}{7}) = -\sin(\frac{3\pi}{7}) = -c$. (b) Using cofunction identity: $\sin(\frac{\pi}{14} + \frac{6\pi}{14}) = \sin(\frac{7\pi}{14}) = \sin(\frac{\pi}{2}) = 1$. Also, $\sin(\frac{10\pi}{7}) = -\cos(\frac{4\pi}{7}) = -n$. 7. **Use angle formulas:** (a) $\sin(\frac{5\pi}{18}) \cos(\frac{10\pi}{9}) - \cos(\frac{5\pi}{18}) \sin(\frac{10\pi}{9}) = \sin(\frac{5\pi}{18} - \frac{10\pi}{9}) = \sin(-\frac{15\pi}{18}) = -\sin(\frac{5\pi}{6}) = -\frac{1}{2}$. (b) $\frac{2 \tan(\frac{\pi}{8})}{1 - \tan^2(\frac{\pi}{8})} = \tan(2 \times \frac{\pi}{8}) = \tan(\frac{\pi}{4}) = 1$. **Final answers:** - $\frac{7\pi}{90}$ radians = 14°. - Point $(-\frac{\sqrt{3}}{2}, \frac{1}{2})$ corresponds to $\frac{5\pi}{6}$. - $\cos(-12\pi) = 1$. - $\csc(\frac{11\pi}{3}) = -\frac{2\sqrt{3}}{3}$. - $\sin(\frac{10\pi}{7}) = -c = -\sin(\frac{3\pi}{7})$. - $\sin(\frac{5\pi}{18}) \cos(\frac{10\pi}{9}) - \cos(\frac{5\pi}{18}) \sin(\frac{10\pi}{9}) = -\frac{1}{2}$. - $\frac{2 \tan(\frac{\pi}{8})}{1 - \tan^2(\frac{\pi}{8})} = 1$.