Can't... Resist... Bashing...

By rearranging the expression, we get

$a_{n+2} = \frac{a_{n} + a_{n+1}}{1 - a_{n}a_{n+1}}$

This looks like the formula

$\textrm{tan (A + B)} = \frac{\textrm{tan A + tan B}}{\textrm{1- tan A tanB}}$

Since $a_{1} = \sqrt{3}$ , which is $\textrm{tan } 60^{\circ}$

and $a_{2} = 1$ , which is $\textrm{tan } 45^{\circ}$

Therefore, $a_{3} = \textrm{tan } (60^{\circ} + 45^{\circ}) = \textrm{tan } 105^{\circ}$

Similarly, $a_{4} = \textrm{tan } (45^{\circ} + 105^{\circ}) = \textrm{tan } 150^{\circ}$

$a_{5} = \textrm{tan } (105^{\circ} + 150^{\circ}) = \textrm{tan } 255^{\circ}$

$a_{6} = \textrm{tan } (150^{\circ} + 255^{\circ}) = \textrm{tan } 405^{\circ}$

$a_{7} = \textrm{tan } (255^{\circ} + 405^{\circ}) = \textrm{tan } 660^{\circ}$

$a_{8} = \textrm{tan } (405^{\circ} + 660^{\circ}) = \textrm{tan } 1065^{\circ}$

$a_{9} = \textrm{tan } (660^{\circ} + 1065^{\circ}) = \textrm{tan } 1725^{\circ}$

$a_{10} = \textrm{tan } (1065^{\circ} + 1725^{\circ}) = \textrm{tan } 2790^{\circ}$

However, $\textrm{tan}$ of odd multiples of $90^{\circ}$ is not defined, and $2790^{\circ}$ is an odd multiple of $90^{\circ}$ , so $\{a_{n}\}^{k}_{n=1}$ is defined only for $1\leq n \leq 9$

Therefore, $\boxed{k = 9}$

This one is tricky. Rearranging the recurrence, I found a computable form: $a_{n+2} = \frac{a_n + a_{n+1}}{1 - a_na_{n+1}}$ so clearly we're looking for $k$ such that $a_{k-1}a_{k-2} = 1$ . I wasn't sure how to proceed with this problem (since there doesn't seem to be a nice way to get the closed form solution of this sequence, so I just computed out a couple of sequences by hand. After a while, I started searching for common functions that produces $\sqrt{3}$ , and a quick google search took me to the trigonometric section. I tried the inverse $\sin$ function to no avail, but when I tried the inverse $\tan$ function: Bingo, all of the terms were some rational multiple of $\pi$ !

This got me excited, and a quick search on trigonometric identities gave the following: $\tan(b_n + b_{n+1}) = \frac{\tan(b_n) + \tan(b_{n+1})}{1 - \tan(b_n)\tan(b_{n+1})}$ which has the same structure as our recurrence! This means that the sequence $a_n$ can be thought of as the injection of some sequence $b_n$ via the $\tan$ function! That is, let $a_n = \tan(b_n)$ , then $a_{n+2} = \tan(b_{n+2}) = \frac{\tan(b_n) + \tan(b_{n+1})}{1 - \tan(b_n)\tan(b_{n+1})} = \tan(b_n + b_{n+1})$ then, it's easy to see that letting $b_1 = \arctan(\sqrt{3}) = \frac{\pi}{3}$ and $b_2 = \arctan(1) = \frac{\pi}{4}$ , the sequence $b_{n+2} = b_{n+1} + b_n$ can be transformed into $a_n$ via $a_n = \tan(b_n)$ .

But then $b_n$ is just the fibonacci sequence, so we can expression $b_n = f_{n-2}b_1 + f_{n-1}b_2$ (it would be n-1 and n, but we started our index at one instead of zero). Now, this sequence terminates immediately $\frac{\pi}{2}\mid b_n$ , which occurs at $k+1 = 10$ because $b_{10} = \frac{31\pi}{2}$ , so $\boxed{k = 9}$ .

Rearranging the given equation gives $a_{n+2} = \dfrac{a_n + a_{n+1}}{1 - a_n a_{n+1}}$ . This is exactly the recurrence satisfied by $\tan(x)$ . We know $a_1 = \tan(60)$ , $a_2 = \tan(45)$ , so $\begin{aligned} a_3 &=& \tan(60 + 45) = \tan(105) \\ a_4 &=& \tan(45 + 105) = \tan(150) \\ a_5 &=& \tan(105 + 150) = \tan(255) = \tan(75) \\ a_6 &=& \tan(150 + 75) = \tan(225) = \tan(45) \\ a_7 &=& \tan(75 + 45) = \tan(120) \\ a_8 &=& \tan(45 + 120) = \tan(165) \\ a_9 &=& \tan(120 + 165) = \tan(105) \end{aligned}$ However, if $a_n$ is defined for $n=10$ , we would have $a_{10} = \tan(165 + 105) = \tan(270)$ which is undefined. Therefore, $a_n$ can only defined up until $k=9$ .