I'm trying to fill in the gaps in my knowledge of simplifying rational expressions using conjugates, but this one stumps me. Given $\tan(\frac{\pi}{4}-\frac{\pi}{6})$, I can work the formula down to:


Then I attempt to multiply top and bottom by the denominator's conjugate $9-3\sqrt{3}$, only to get something like:


which of course is wrong. The correct answer should be $2-\sqrt{3}$. Thanks in advance for any help. I simply can't figure out where my distribution is going wrong...


The arithmetic will be simpler if we immediately cancel a $3$. Even better is to cancel a $3\sqrt{3}$. If we do that, we get $$\frac{\sqrt{3}-1}{\sqrt{3}+1}.$$ Multiply top and bottom by $\sqrt{3}-1$. At the bottom we get $2$. At the top, we get $(\sqrt{3}-1)^2$, which is $3-2\sqrt{3}+1$, that is, $4-2\sqrt{3}$. finally, divide by $2$.

Remark: Your answer was not wrong. You got $\frac{108-54\sqrt{3}}{54}$. This is $\frac{108}{54}-\frac{54\sqrt{3}}{54}$, which is $2-\sqrt{3}$.

For the original problem, it would have been easier to use the formula $$\tan(x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y},$$ a mild variant of the more familiar $\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$.

  • $\begingroup$ Thank you for taking the time to explain this so thoroughly, and even point out that my answer was correct, albeit not simplified. Apparently my arithmetic in this area is weaker than I thought. I will give that formula for tangent a try next time! $\endgroup$ – Jason S. Apr 12 '14 at 21:10
  • $\begingroup$ You are welcome. $\endgroup$ – André Nicolas Apr 12 '14 at 21:41

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