1. **State the problem:** Given $\tan A = -\sqrt{3}$ with $A$ in quadrant II, and $\cot B = -\sqrt{2}$ with $B$ in quadrant IV, find $\sin(A+B)$, $\cos(A+B)$, and $\tan(A+B)$. 2. **Recall formulas:** - $\sin(A+B) = \sin A \cos B + \cos A \sin B$ - $\cos(A+B) = \cos A \cos B - \sin A \sin B$ - $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ 3. **Find $\sin A$ and $\cos A$:** Since $\tan A = \frac{\sin A}{\cos A} = -\sqrt{3}$ and $A$ is in quadrant II where $\sin A > 0$ and $\cos A < 0$. Let $\cos A = x < 0$, then $\sin A = -\sqrt{3} x$. Using Pythagorean identity: $$\sin^2 A + \cos^2 A = 1 \Rightarrow (-\sqrt{3} x)^2 + x^2 = 1 \Rightarrow 3x^2 + x^2 = 1 \Rightarrow 4x^2 = 1 \Rightarrow x^2 = \frac{1}{4}$$ So $x = \cos A = -\frac{1}{2}$ (negative in QII). Then $\sin A = -\sqrt{3} \times (-\frac{1}{2}) = \frac{\sqrt{3}}{2}$. 4. **Find $\sin B$ and $\cos B$:** Given $\cot B = -\sqrt{2} = \frac{\cos B}{\sin B}$ and $B$ in quadrant IV where $\sin B < 0$ and $\cos B > 0$. Let $\sin B = y < 0$, then $\cos B = -\sqrt{2} y$. Using Pythagorean identity: $$\sin^2 B + \cos^2 B = 1 \Rightarrow y^2 + (-\sqrt{2} y)^2 = 1 \Rightarrow y^2 + 2 y^2 = 1 \Rightarrow 3 y^2 = 1 \Rightarrow y^2 = \frac{1}{3}$$ So $y = \sin B = -\frac{1}{\sqrt{3}}$ (negative in QIV). Then $\cos B = -\sqrt{2} \times (-\frac{1}{\sqrt{3}}) = \frac{\sqrt{2}}{\sqrt{3}}$. 5. **Calculate $\sin(A+B)$:** $$\sin(A+B) = \sin A \cos B + \cos A \sin B = \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{\sqrt{3}} + \left(-\frac{1}{2}\right) \times \left(-\frac{1}{\sqrt{3}}\right)$$ Simplify: $$= \frac{\sqrt{3} \sqrt{2}}{2 \sqrt{3}} + \frac{1}{2 \sqrt{3}} = \frac{\sqrt{2}}{2} + \frac{1}{2 \sqrt{3}}$$ 6. **Calculate $\cos(A+B)$:** $$\cos(A+B) = \cos A \cos B - \sin A \sin B = \left(-\frac{1}{2}\right) \times \frac{\sqrt{2}}{\sqrt{3}} - \frac{\sqrt{3}}{2} \times \left(-\frac{1}{\sqrt{3}}\right)$$ Simplify: $$= -\frac{\sqrt{2}}{2 \sqrt{3}} + \frac{\sqrt{3}}{2 \sqrt{3}} = -\frac{\sqrt{2}}{2 \sqrt{3}} + \frac{1}{2}$$ 7. **Calculate $\tan(A+B)$:** First find $\tan B = \frac{1}{\cot B} = \frac{1}{-\sqrt{2}} = -\frac{1}{\sqrt{2}}$. Then: $$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{-\sqrt{3} - \frac{1}{\sqrt{2}}}{1 - (-\sqrt{3}) \times \left(-\frac{1}{\sqrt{2}}\right)} = \frac{-\sqrt{3} - \frac{1}{\sqrt{2}}}{1 - \frac{\sqrt{3}}{\sqrt{2}}}$$ This is the exact expression for $\tan(A+B)$. **Final answers:** $$\sin(A+B) = \frac{\sqrt{2}}{2} + \frac{1}{2 \sqrt{3}}$$ $$\cos(A+B) = -\frac{\sqrt{2}}{2 \sqrt{3}} + \frac{1}{2}$$ $$\tan(A+B) = \frac{-\sqrt{3} - \frac{1}{\sqrt{2}}}{1 - \frac{\sqrt{3}}{\sqrt{2}}}$$