Check equality of $\tan(\frac{7\pi}{4} + \frac{\pi}{6})$ and $\tan(\frac{\pi}{4} + \frac{5\pi}{3})$

Question

I'm trying to show that $\tan(\frac{7\pi}{4} + \frac{\pi}{6})$ is the same as $\tan(\frac{\pi}{4} + \frac{5\pi}{3})$.

Both of them should add up to $\tan(\frac{23\pi}{12})$.

When I calculate $\tan(\frac{\pi}{4} + \frac{5\pi}{3})$ I get $\frac{1-\sqrt{3}}{1+\sqrt{3}}$, which I know must be correct.

However I'm having an issue when I try to calculate $\tan(\frac{7\pi}{4} + \frac{\pi}{6})$.

By the angle sum identity: $\tan(\frac{7\pi}{4} + \frac{\pi}{6}) = \frac{\tan(\frac{7\pi}{4})+\tan(\frac{\pi}{6})}{1-\tan(\frac{7\pi}{4})\tan(\frac{\pi}{6})}$

I know from the common angles that $7\pi/4$ represents coordinates $(\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}})$ on the unit circle and $\tan$ is the slope so $\tan(\frac{7\pi}{4}) $ should equal -1.

Further I get that $\tan(\frac{7\pi}{4} + \frac{\pi}{6})$ is $\frac{-1+\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}$, which is not the same as $\frac{1-\sqrt{3}}{1+\sqrt{3}}$.

What the heck have I done wrong here, I just can't see it!

Answer 1

You can notice that $$\frac{7π}{4} + \frac{π}{6} = \frac{π}{4} + \left(\frac{6π}{4} + \frac{π}{6}\right) = \frac{π}{4} + \frac{18π + 2π}{12} = \frac{π}{4} + \frac{5π}{3}$$ so in fact the arguments are equal.

Answer 2

Hint:

You're almost there...with $\dfrac {-1 + \frac {1}{\sqrt 3}}{1+ \frac {1}{\sqrt 3}}$, you can multiply by a common value to clear fractions; your answer then follows.

ETA: Per Amaan M, that you can rationalize this fraction further to get an answer that does not contain fractions.