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Multiplying both expressions together gives: $(\sin\alpha-\cos\beta)(\sin\beta-\cos\alpha)=\tan\gamma\cot\gamma=1$

The LHS:

$\quad(\sin\alpha-\cos\beta)(\sin\beta-\cos\alpha)$

$=\cos\alpha\cos\beta+\sin\alpha\sin\beta-\sin\alpha\cos\alpha-\sin\beta\cos\beta$

$=\cos(\alpha-\beta)-\frac12\sin2\alpha-\frac12\sin2\beta$ (Using compound angle formula for cosine and double angle formula for sine )

$=\cos(\alpha-\beta)-\frac12(2\sin(\alpha+\beta)\cos(\alpha-\beta))$ (Using a sum-to-product formula )

$=\cos(\alpha-\beta)-\sin(\alpha+\beta)\cos(\alpha-\beta)$

$=\cos(\alpha-\beta)(1-\sin(\alpha+\beta))$

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Since both $\alpha$ and $\beta$ are between $0^\circ$ and $90^\circ$ , $(\alpha-\beta)$ is between $-90^\circ$ and $90^\circ$ , so $\cos(\alpha-\beta)$ is between $0$ and $1$ .

Since both $\alpha$ and $\beta$ are between $0^\circ$ and $90^\circ$ , $(\alpha+\beta)$ is between $0^\circ$ and $180^\circ$ , so $\sin(\alpha+\beta)$ is between $0$ and $1$ , and $1-\sin(\alpha+\beta)$ is also between $0$ and $1$ .

Two numbers between $0$ and $1$ multiplied together gives $1$ , so they must be both $1$ .

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Therefore, $\cos(\alpha-\beta)=1$ and $\sin(\alpha+\beta)=0$ , which makes $\alpha$ and $\beta$ either both $0^\circ$ or both $90^\circ$ .

If both are $0^\circ$ , $\tan\gamma=\sin0^\circ-\cos0^\circ=-1$ , so it is rejected.

If both are $90^\circ$ , $\tan\gamma=\sin90^\circ-\cos90^\circ=1$ , so $\gamma=45^\circ$ .

$An\quad easy\quad solution\quad is\quad by\quad adding\quad LHS\quad and\quad RHS,We\quad get\\ sin(\alpha )+sin(\beta )-cos(\alpha )-cos(\beta )=tan(\gamma )+cot(\gamma )\\ A\quad simple\quad application\quad of\quad AM-GM\quad Ineq.\quad gives\\ sin(\alpha )=sin(\beta )=1\quad cos(\alpha )=cos(\beta )=0\quad and\quad tan(\gamma )=cot(\gamma )=1\\ So\quad the\quad answer\quad is\quad \boxed { 45 } .\\$

$\begin{cases} \sin{\alpha} - \cos{\beta} = \tan{\gamma} & \Rightarrow \sin{\alpha} \cos{\gamma} - \cos{\beta} \cos{\gamma} = \sin{\gamma} & ...(1) \\ \sin{\beta} - \cos{\alpha} = \cot{\gamma} & \Rightarrow \sin{\beta} \sin{\gamma} - \cos{\alpha} \sin{\gamma} = \cos{\gamma} & ...(2) \end{cases}$

$\begin{aligned} (1)+(2): \quad \sin{(\alpha - \gamma)} -\cos{(\beta + \gamma)} & = \sin{\gamma} + \cos{\gamma} \end{aligned}$

Equating the sine and cosine terms, we have:

$\begin{cases} \alpha - \gamma = \alpha & \Rightarrow \alpha = 2 \gamma \\ 180^\circ - \beta - \gamma = \gamma & \Rightarrow \beta = 180^\circ - 2 \gamma \end{cases}$

$\begin{aligned} \Rightarrow (1): \quad \sin{(2\gamma)} -\cos{(180^\circ - 2 \gamma)} & = \tan{\gamma} \\ \sin{(2\gamma)} + \cos{(2 \gamma)} & = \tan{\gamma} \\ \frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2} & = t \quad \quad \quad \small \color{#3D99F6} {[\text{Let } t = \tan{\gamma}]} \\ 2t + 1 - t^2 & = t + t^3 \\ t^3 + t^2 - t - 1 & = 0 \\ (t-1)(t+1)^2 & = 0 \\ \Rightarrow t = \tan{\gamma} & = \begin{cases} 1 & \Rightarrow \gamma = \boxed{45^\circ} \\ - 1 & \Rightarrow \color{#D61F06} {0^\circ > \gamma > 90^\circ \text{ rejected}} \end{cases} \end{aligned}$

We get the same result if we equate $\sin{(\alpha - \gamma)} = \cos{\gamma}$ and $-\cos{(\beta + \gamma)} = \sin{\gamma}$