1. Simplify expression 7.1: $$\frac{\sin 210^\circ \cos 300^\circ \tan 240^\circ}{\cos 120^\circ \tan 150^\circ \sin 330^\circ}$$ Recall the exact values: - $\sin 210^\circ = -\frac{1}{2}$ - $\cos 300^\circ = \frac{1}{2}$ - $\tan 240^\circ = \sqrt{3}$ - $\cos 120^\circ = -\frac{1}{2}$ - $\tan 150^\circ = -\frac{1}{\sqrt{3}}$ - $\sin 330^\circ = -\frac{1}{2}$ Substitute: $$\frac{\left(-\frac{1}{2}\right) \left(\frac{1}{2}\right) \left(\sqrt{3}\right)}{\left(-\frac{1}{2}\right) \left(-\frac{1}{\sqrt{3}}\right) \left(-\frac{1}{2}\right)} = \frac{-\frac{\sqrt{3}}{4}}{-\frac{1}{4\sqrt{3}}}$$ Simplify numerator and denominator: $$\frac{-\frac{\sqrt{3}}{4}}{-\frac{1}{4\sqrt{3}}} = \frac{-\frac{\sqrt{3}}{4}}{-\frac{1}{4\sqrt{3}}} = \frac{-\frac{\sqrt{3}}{4}}{-\frac{1}{4\sqrt{3}}}$$ Cancel negatives and fractions: $$= \frac{\frac{\sqrt{3}}{4}}{\frac{1}{4\sqrt{3}}} = \frac{\sqrt{3}}{4} \times \frac{4\sqrt{3}}{1} = \sqrt{3} \times \sqrt{3} = 3$$ 2. Simplify expression 7.2: $$[\sin(-\theta) + \cos(360^\circ - \theta)] \times \left[\cos(90^\circ - \theta) + \frac{\sin \theta}{\tan \theta}\right]$$ Use identities: - $\sin(-\theta) = -\sin \theta$ - $\cos(360^\circ - \theta) = \cos \theta$ - $\cos(90^\circ - \theta) = \sin \theta$ - $\tan \theta = \frac{\sin \theta}{\cos \theta}$ Rewrite: $$(-\sin \theta + \cos \theta) \times \left(\sin \theta + \frac{\sin \theta}{\frac{\sin \theta}{\cos \theta}}\right) = (-\sin \theta + \cos \theta) \times \left(\sin \theta + \cos \theta\right)$$ Multiply: $$(-\sin \theta + \cos \theta)(\sin \theta + \cos \theta) = -\sin^2 \theta - \sin \theta \cos \theta + \cos \theta \sin \theta + \cos^2 \theta$$ Simplify middle terms: $$-\sin \theta \cos \theta + \cos \theta \sin \theta = 0$$ So expression reduces to: $$-\sin^2 \theta + \cos^2 \theta = \cos^2 \theta - \sin^2 \theta = \cos 2\theta$$ 3. Solve for $x$ in 7.3 given: $$\tan x = m + \frac{1}{m}, \quad 90^\circ \leq x \leq 270^\circ$$ $$m^2 + \frac{1}{m^2} = 1$$ Square $m + \frac{1}{m}$: $$\left(m + \frac{1}{m}\right)^2 = m^2 + 2 + \frac{1}{m^2} = 1 + 2 = 3$$ So: $$\tan^2 x = 3 \implies \tan x = \pm \sqrt{3}$$ Since $x$ is in $[90^\circ, 270^\circ]$ (2nd or 3rd quadrant), and tangent is negative in 2nd quadrant and positive in 3rd quadrant: - $\tan x = -\sqrt{3}$ in 2nd quadrant - $\tan x = \sqrt{3}$ in 3rd quadrant The angles with $\tan x = \sqrt{3}$ are $60^\circ$ and $240^\circ$. The angles with $\tan x = -\sqrt{3}$ are $120^\circ$ and $300^\circ$. Within $90^\circ \leq x \leq 270^\circ$, valid solutions are: $$x = 120^\circ \text{ or } 240^\circ$$ Final answers: - 7.1: $3$ - 7.2: $\cos 2\theta$ - 7.3: $x = 120^\circ$ or $240^\circ$