$$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\frac{\frac{\sqrt3}3+(-2\sqrt2)}{1-\frac{\sqrt3}3\cdot(-2\sqrt2)}$$ The exact value of $\tan(\alpha+\beta)$ is shown below: $$\tan(\alpha+\beta)=\frac{8\sqrt2-9\sqrt3}5.$$
Hello. I am having troubles figuring out the steps involved in solving this problem. I can find the value for $\tan\alpha$ and $\tan\beta$ just fine, but I do not know what steps I am missing to solve this question correctly. I have tried many steps, such as multiplying the bottom by its conjugate (after having the fractions share the same denominator), but can't seem to get the answer shown. I have looked for examples online, but the examples are solved with simple multiplication of the conjugate/factoring with difference of squares.
If anyone could help/offer tips it would be much appreciated.
Firstly, multiply both numerator and denominator by $3$. We have $$\frac{\frac{\sqrt3}3-2\sqrt2}{1+\frac{\sqrt3}3\cdot2\sqrt2}=\frac{\sqrt3-6\sqrt2}{3+2\sqrt6}\cdot\color{red}{\frac{3-2\sqrt6}{3-2\sqrt6}}=\frac{3\sqrt3-2\sqrt{18}-18\sqrt2+12\sqrt{12}}{9-4\cdot6}$$ so...