1. Find the exact value of $\sin \frac{2\pi}{3} + \cos \frac{7\pi}{6} + \tan \frac{5\pi}{3}$. - $\sin \frac{2\pi}{3} = \sin 120^\circ = \frac{\sqrt{3}}{2}$ (since sine is positive in Q2). - $\cos \frac{7\pi}{6} = \cos 210^\circ = -\frac{\sqrt{3}}{2}$ (cosine is negative in Q3). - $\tan \frac{5\pi}{3} = \tan 300^\circ = -\sqrt{3}/3$ (tangent is negative in Q4). Sum: $$\frac{\sqrt{3}}{2} + \left(-\frac{\sqrt{3}}{2}\right) + \left(-\frac{\sqrt{3}}{3}\right) = 0 - \frac{\sqrt{3}}{3} = -\frac{\sqrt{3}}{3}$$ 2. Find $\sin \frac{2\pi}{3} \cos \frac{5\pi}{6} + \cos \frac{2\pi}{3} \sin \frac{5\pi}{6}$. Recall the sine addition formula: $$\sin(a+b) = \sin a \cos b + \cos a \sin b$$ Here, $a=\frac{2\pi}{3}$, $b=\frac{5\pi}{6}$. Calculate $a+b$: $$\frac{2\pi}{3} + \frac{5\pi}{6} = \frac{4\pi}{6} + \frac{5\pi}{6} = \frac{9\pi}{6} = \frac{3\pi}{2}$$ So, $$\sin \frac{3\pi}{2} = -1$$ 3. Find $\tan \frac{5\pi}{4} + \cot \frac{7\pi}{4} - \sec \frac{5\pi}{6}$. - $\tan \frac{5\pi}{4} = \tan 225^\circ = 1$ (tangent positive in Q3). - $\cot \frac{7\pi}{4} = \cot 315^\circ = 1$ (cotangent positive in Q4). - $\sec \frac{5\pi}{6} = \sec 150^\circ = -2$ (since $\cos 150^\circ = -\frac{\sqrt{3}}{2}$, so $\sec = \frac{1}{\cos} = -\frac{2}{\sqrt{3}}$ but exact value is $-2/\sqrt{3}$, rationalize denominator: $$-\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$$ So, $$\tan \frac{5\pi}{4} + \cot \frac{7\pi}{4} - \sec \frac{5\pi}{6} = 1 + 1 - \left(-\frac{2\sqrt{3}}{3}\right) = 2 + \frac{2\sqrt{3}}{3}$$ 4. Find $(\cos \frac{11\pi}{6} + \sin \frac{\pi}{3})(\tan \frac{\pi}{6} + \cot \frac{4\pi}{3})$. Calculate each term: - $\cos \frac{11\pi}{6} = \cos 330^\circ = \frac{\sqrt{3}}{2}$ - $\sin \frac{\pi}{3} = \sin 60^\circ = \frac{\sqrt{3}}{2}$ - $\tan \frac{\pi}{6} = \tan 30^\circ = \frac{1}{\sqrt{3}}$ - $\cot \frac{4\pi}{3} = \cot 240^\circ = \cot (180^\circ + 60^\circ) = \cot 60^\circ = \frac{1}{\sqrt{3}}$ (cotangent positive in Q3) Sum inside parentheses: $$\left(\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}\right) = \sqrt{3}$$ $$\left(\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}}\right) = \frac{2}{\sqrt{3}}$$ Multiply: $$\sqrt{3} \times \frac{2}{\sqrt{3}} = 2$$ --- Now, solve for $\theta$ in $[0,2\pi)$: 1. $\tan \theta = \frac{1}{\sqrt{3}}$ Recall $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$. Solutions in $[0,2\pi)$ where tangent is positive: $$\theta = \frac{\pi}{6}, \frac{7\pi}{6}$$ 2. $\tan \theta = -1$ Recall $\tan \frac{3\pi}{4} = -1$ and $\tan \frac{7\pi}{4} = -1$. Solutions: $$\theta = \frac{3\pi}{4}, \frac{7\pi}{4}$$ 3. $\sec \theta = 2$ Recall $\sec \theta = \frac{1}{\cos \theta}$, so $\cos \theta = \frac{1}{2}$. Solutions where cosine is positive: $$\theta = \frac{\pi}{3}, \frac{5\pi}{3}$$ 4. $\csc \theta = \frac{2}{\sqrt{3}}$ Recall $\csc \theta = \frac{1}{\sin \theta}$, so $\sin \theta = \frac{\sqrt{3}}{2}$. Solutions where sine is positive: $$\theta = \frac{\pi}{3}, \frac{2\pi}{3}$$