We have given two five digit numbers $x$ and $y$ consisting of all digits 0-9 satisfying $(1) \;\; \tan x - \tan y = 1 + \tan x \cdot \tan y$ ($x$ and $y$ are angles in degrees). Then the maximum of $x$ is 98730. Proof: Via the formula ${\textstyle \tan(x - y) = \frac{\tan x - \tan y}{1 + \tan x \cdot \tan y}}$ we find that equation (1) is equivalent to $\tan(x - y) = 1$, which solution is $(2) \;\; x - y = 45 + 180n \; (n \in \mathbb{Z})$. According to formula (2) we have $x - y \equiv 45 \pmod{180}$, a congruence which is satisfied (since $180 = 2^2 \cdot 3^2 \cdot 5$) iff $(3) \;\; x \equiv y \pmod{9}$, $(4) \;\; x \equiv y \pmod{5}$, $(5) \;\; x \equiv y+1 \pmod{4}$. Set $x = \overline{x_4x_3x_2x_1x_0}$ and $y = \overline{y_4y_3y_2y_1y_0}$. The fact that $x$ and $y$ consist of the 10 digits 0-9 implies $x + y \equiv \sum_{i=0}^4 (x_i + y_i) = \sum_{k=0}^9 k = 45 \equiv 0 \pmod{9}$, which combined with congruence (3) give us $(6) \;\; x \equiv y \equiv 0 \pmod{9}$. Futhermore by the congruences (4) and (5) we obtain $(7) \;\; x_0 \equiv y_0 \pmod{5}$ and $(8) \;\; 2x_1 + x_0 \equiv 2y_1 + y_0 + 1 \pmod{4}$. In order to maximize $x$, we choose $(x_4,x_3,x_2) = (9,8,7)$. According to congruence (6) we have $x \equiv \sum_{i=0}^0 x_i = 9 + 8 + 7 + x_1 + x_0 \equiv x_1 + x_0 - 3 \equiv 0 \pmod{9}$, i.e. $(9) \;\; x_1 + x_0 \equiv 3 \pmod{9}$. From the facts that $0 \leq x_0,x_1 \leq 6$ and $x_0 \neq x_1$ we obtain $0+1 \leq x_1 + x_0 \leq 5+6$, i.e. $1 \leq x_1 + x_0 \leq 11$, which combined with congruence (9) give us $(10) \;\; x_1 + x_0 = 3$. Hence in order to maximize $x$, we choose $(x_1,x_0) = (3,0)$, which according to congruence (7) implies $y_0 \equiv 0 \pmod{5}$, yielding $y_0 = 5$. Thus by congruence (8) we have $2y_1 + 5 + 1 \equiv 2 \cdot 3 + 0 \pmod{4}$, i.e. $y_1 \equiv 0 \pmod{2}$. In other words, $x = 98730$ and $y = 64125$ satisfies equation (2) ($98730 - 64125 = 34605 = 45 + 180 \cdot 192$). Conclusion: The maximal value of $x$ is 98730. q.e.d.

Then what is $\tan98730^\circ$ Solar Plexsus wrote: We have given two five digit numbers $x$ and $y$ consisting of all digits 0-9 satisfying $(1) \;\; \tan x - \tan y = 1 + \tan x \cdot \tan y$ ($x$ and $y$ are angles in degrees). Then the maximum of $x$ is 98730. Proof: Via the formula ${\textstyle \tan(x - y) = \frac{\tan x - \tan y}{1 + \tan x \cdot \tan y}}$ we find that equation (1) is equivalent to $\tan(x - y) = 1$, which solution is $(2) \;\; x - y = 45 + 180n \; (n \in \mathbb{Z})$. According to formula (2) we have $x - y \equiv 45 \pmod{180}$, a congruence which is satisfied (since $180 = 2^2 \cdot 3^2 \cdot 5$) iff $(3) \;\; x \equiv y \pmod{9}$, $(4) \;\; x \equiv y \pmod{5}$, $(5) \;\; x \equiv y+1 \pmod{4}$. Set $x = \overline{x_4x_3x_2x_1x_0}$ and $y = \overline{y_4y_3y_2y_1y_0}$. The fact that $x$ and $y$ consist of the 10 digits 0-9 implies $x + y \equiv \sum_{i=0}^4 (x_i + y_i) = \sum_{k=0}^9 k = 45 \equiv 0 \pmod{9}$, which combined with congruence (3) give us $(6) \;\; x \equiv y \equiv 0 \pmod{9}$. Futhermore by the congruences (4) and (5) we obtain $(7) \;\; x_0 \equiv y_0 \pmod{5}$ and $(8) \;\; 2x_1 + x_0 \equiv 2y_1 + y_0 + 1 \pmod{4}$. In order to maximize $x$, we choose $(x_4,x_3,x_2) = (9,8,7)$. According to congruence (6) we have $x \equiv \sum_{i=0}^0 x_i = 9 + 8 + 7 + x_1 + x_0 \equiv x_1 + x_0 - 3 \equiv 0 \pmod{9}$, i.e. $(9) \;\; x_1 + x_0 \equiv 3 \pmod{9}$. From the facts that $0 \leq x_0,x_1 \leq 6$ and $x_0 \neq x_1$ we obtain $0+1 \leq x_1 + x_0 \leq 5+6$, i.e. $1 \leq x_1 + x_0 \leq 11$, which combined with congruence (9) give us $(10) \;\; x_1 + x_0 = 3$. Hence in order to maximize $x$, we choose $(x_1,x_0) = (3,0)$, which according to congruence (7) implies $y_0 \equiv 0 \pmod{5}$, yielding $y_0 = 5$. Thus by congruence (8) we have $2y_1 + 5 + 1 \equiv 2 \cdot 3 + 0 \pmod{4}$, i.e. $y_1 \equiv 0 \pmod{2}$. In other words, $x = 98730$ and $y = 64125$ satisfies equation (2) ($98730 - 64125 = 34605 = 45 + 180 \cdot 192$). Conclusion: The maximal value of $x$ is 98730. q.e.d.