# Math Help - A Sum and difference question

1. ## A Sum and difference question

$Tan(\frac{4\pi}{3}-\frac{\pi}{4}) = \frac{Tan(\frac{4\pi}{3})-Tan(\frac{\pi}{4})}{1+Tan(\frac{4\pi}{3}Tan(\frac{ \pi}{4})}=\frac{\sqrt{3}-1}{1+\sqrt{3}*1}= I do not know this step = 2-\sqrt{3}$

Could someone tell me what happens to take it from $\frac{\sqrt{3}-1}{1+\sqrt{3}*1} = 2-\sqrt{3}$

2. Originally Posted by Dannbr
$Tan(\frac{4\pi}{3}-\frac{\pi}{4}) = \frac{Tan(\frac{4\pi}{3})-Tan(\frac{\pi}{4})}{1+Tan(\frac{4\pi}{3}Tan(\frac{ \pi}{4})}=\frac{\sqrt{3}-1}{1+\sqrt{3}*1}= I do not know this step = 2-\sqrt{3}$

Could someone tell me what happens to take it from $\frac{\sqrt{3}-1}{1+\sqrt{3}*1} = 2-\sqrt{3}$
Multiply numerator and denominator by conjugate of denominator.

$\displaystyle \frac{\sqrt{3}-1}{1+\sqrt{3}}=\frac{(\sqrt{3}-1)(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}=\cdots=2-\sqrt{3}$