Hint: $\sin 120^\circ = \sin(180^\circ - 60^\circ) = \frac{\sqrt{3}}{2}$ (Quadrant II, positive) and $\cos 150^\circ = \cos(180^\circ - 30^\circ) = -\frac{\sqrt{3}}{2}$ (Quadrant II, negative). Add them together.

Hint: $\tan 135^\circ = -1$ (Quadrant II) and $\sec 120^\circ = \frac{1}{\cos 120^\circ} = \frac{1}{-1/2} = -2$.

Hint: Both angles lie in Quadrant III where sine and cosine are negative. $\sin 210^\circ = \sin(180^\circ + 30^\circ) = -\frac{1}{2}$ and $\cos 240^\circ = \cos(180^\circ + 60^\circ) = -\frac{1}{2}$.

Hint: Angles lie in Quadrant III where tangent and cotangent are positive. $\tan 225^\circ = \tan(180^\circ + 45^\circ) = 1$ and $\cot 210^\circ = \cot(180^\circ + 30^\circ) = \sqrt{3}$.

Hint: Both angles lie in Quadrant IV. $\sin 300^\circ = \sin(360^\circ - 60^\circ) = -\frac{\sqrt{3}}{2}$ and $\cos 330^\circ = \cos(360^\circ - 30^\circ) = +\frac{\sqrt{3}}{2}$.

Hint: In Quadrant IV, $\tan$ is negative and $\sec$ is positive. $\tan 315^\circ = \tan(360^\circ - 45^\circ) = -1$ and $\sec 300^\circ = \frac{1}{\cos 300^\circ} = \frac{1}{1/2} = 2$.

Hint: Boundary angles: $\sin 270^\circ = -1$, $\cos 180^\circ = -1$, and $\tan 360^\circ = 0$.

Hint: In Quadrant III, both sine and cosine are negative, so their reciprocals are also negative. $\csc 210^\circ = \frac{1}{\sin 210^\circ} = -2$ and $\sec 225^\circ = \frac{1}{\cos 225^\circ} = -\sqrt{2}$.

Hint: $\tan 300^\circ = -\sqrt{3}$ (Quadrant IV) and $\cot 240^\circ = \frac{1}{\sqrt{3}}$ (Quadrant III). Multiply them directly.