Advanced Functions, How to simplify $\tan \frac{5\pi}{12}$? I was asked to find the exact value of $\,\tan \left(\frac{5\pi}{12}\right)$, so I did up until this point where I got completely stuck.
I split the ratio into two, so $\;\dfrac{5\pi}{12} = \dfrac{\pi}4 + \dfrac{\pi}6\,.$
using the formula, $\tan(x+y) = \dfrac{\tan x+\tan y}{1-\tan x\tan y}$:
$$\tan(\pi/4 + \pi/6) = \frac{1+1/\sqrt 3 }{1-(1)(1/\sqrt 3)
}$$
...
$$\tan(\pi/4 + \pi/6) = \frac{3+\sqrt3}{ 3-\sqrt3 }$$
I'm stuck here, I looked up the answer online, and apparently, you have to times numerator and denominator both by $\,3+\sqrt3\,,\,$ but it's cheating, I don't understand why you have to do that.
 A: HINT
Another way to realize the solution:
\begin{align*}
\tan\left(\frac{5\pi}{12}\right) & = \tan\left(\frac{1}{2}\times\frac{5\pi}{6}\right)
\end{align*}
Now you can apply the corresponding half-angle formula.
Can you take it from here?
A: So, you've remembered (or re-derived) that $\tan(\frac{\pi}{4}) = 1$ and $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$, and plugging this into your sum-of-angles formula gives:
$$\tan(\frac{5\pi}{12}) = \frac{\tan(\frac{\pi}{4}) + \tan(\frac{\pi}{6})}{1 - \tan(\frac{\pi}{4})\tan(\frac{\pi}{6})} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - (1)(\frac{1}{\sqrt{3}})}$$
But fractions within fractions are just ugly, so let's multiply numerator and denominator by $\sqrt{3}$.
$$\tan(\frac{5\pi}{12}) = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$$
And this is a perfectly valid way of expressing the value (which numerically is approximately 3.732051).  But it's often considered good style to “rationalize the denominator”, i.e., have √ signs only on top and never on bottom.
Recall the FOIL method for multiplying two binomials, and that $$(a - b)(a + b) = a^2 + ab - ab - b^2 = a^2 - b^2$$
So, if you multiply a binomial by the same thing but with the opposite sign, then only squared terms occur in the product, so any √ signs go away.  So let's multiply the denominator by $\sqrt{3} + 1$.  Of course, we need to multiply the numerator by $\sqrt{3} + 1$ too so that the value of the expression doesn't change
$$\frac{\sqrt{3} + 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}$$
$$= \frac{3 + 2\sqrt{3} + 1}{3 - 1}$$
$$= \frac{4 + 2\sqrt{3}}{2}$$
$$= 2 + \sqrt{3}$$
A: $\tan\left(\dfrac{5\pi}{12}\right)=\tan\left(\dfrac\pi4+\dfrac\pi6\right)=\dfrac{3+\sqrt3}{3-\sqrt3}=\dfrac{3+\sqrt3}{3-\sqrt3}\cdot\dfrac{3+\sqrt3}{3+\sqrt3}=$
$=\dfrac{\left(3+\sqrt3\right)^2}{\left(3-\sqrt3\right)\left(3+\sqrt3\right)}=\dfrac{9+3+6\sqrt3}{9-3}=\dfrac{12+6\sqrt3}6=$
$=\dfrac{6\left(2+\sqrt3\right)}6=2+\sqrt3\;.$
You need to multiply both numerator and denominator by $\,3+\sqrt3\,$ in order to eliminate square roots in the denominator.
On the other hand, if you multiply both numerator and denominator by the same number, it does not change the value of the fraction.
A: 
I looked up the answer online, and apparently, you have to times numerator and denominator both by $3+\sqrt3$, but it's cheating, I don't understand why you have to do that.

The denominator $x = 3-\sqrt3 \;$ is an irrational number whose decimal expansion is non-terminating and non-repeating. So we do not know its exact value (to infinitely many decimal places) and hence are unable to compute a satisfying numerical value (decimal expansion) of the fraction $\dfrac{3+\sqrt3}{3-\sqrt3}\,$ using eg, long division.
However on making the denominator a rational number, aka "rationalizing the denominator" using the identity $(a+b)(a-b)=a^2-b^2 \;$ which removes the square roots, we get a nicer  $2+\sqrt{3}\,$ which can be computed to limited accuracy by hands/calculators or to much larger precision using computers.
