Let $0^{\circ}\leq\alpha,\beta,\gamma\leq90^{\circ}$ be angles such that \[\sin\alpha-\cos\beta=\tan\gamma\] \[\sin\beta-\cos\alpha=\cot\gamma\]
Compute the sum of all possible values of $\gamma$ in degrees.

[i]Proposed by Michael Ren[/i]
1. Given the equations:
   \[
   \sin\alpha - \cos\beta = \tan\gamma
   \]
   \[
   \sin\beta - \cos\alpha = \cot\gamma
   \]
   we need to find the sum of all possible values of $\gamma$ in degrees.

2. Note that $\tan\gamma \cdot \cot\gamma = 1$ because $\tan\gamma = \frac{\sin\gamma}{\cos\gamma}$ and $\cot\gamma = \frac{\cos\gamma}{\sin\gamma}$, so:
   \[
   \tan\gamma \cdot \cot\gamma = \left(\frac{\sin\gamma}{\cos\gamma}\right) \left(\frac{\cos\gamma}{\sin\gamma}\right) = 1
   \]

3. Since $\sin\alpha, \cos\alpha, \sin\beta, \cos\beta$ are all between $0$ and $1$, the expressions $\sin\alpha - \cos\beta$ and $\sin\beta - \cos\alpha$ must also lie between $-1$ and $1$.

4. For $\tan\gamma$ and $\cot\gamma$ to both be between $-1$ and $1$, and given that $\tan\gamma \cdot \cot\gamma = 1$, the only possible values for $\tan\gamma$ and $\cot\gamma$ are $1$ or $-1$.

5. Since $0^\circ \leq \gamma \leq 90^\circ$, $\tan\gamma = \cot\gamma = -1$ is not possible because $\tan\gamma$ and $\cot\gamma$ are positive in this range. Therefore, we must have:
   \[
   \tan\gamma = \cot\gamma = 1
   \]

6. The condition $\tan\gamma = 1$ implies:
   \[
   \gamma = 45^\circ
   \]

7. Therefore, the only possible value for $\gamma$ is $45^\circ$.

The final answer is $\boxed{45^\circ}$.