If $\sin\theta+\cos\theta=1$ prove that $\cos\theta-\sin\theta=\pm1$ So my work, 
Squaring both sides $$(\sin\theta+\cos\theta)^2=1$$
$$1+2\sin\theta\cos\theta=1\ \ \ \ \ \text{-------(i)}$$
$$\sin\theta\cos\theta=0 \ \ \ \ \ \text{------(ii)}$$
So reverting back to $(i)$, 
$$\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta-4\sin\theta\cos\theta=1-4\sin\theta\cos\theta$$
$$(\cos\theta-\sin\theta)^2=1-4\sin\theta\cos\theta$$
$$\cos\theta-\sin\theta=\pm1$$

But my teacher says that there is a shorter solution than that, so please can someone help me find that?

$\cos \theta + \sin \theta=1$ is easily solved for $\theta$ in a graphical way, since it describes the intersection between a line and the goniometric circle:
$$
\begin{cases}
X+Y=1\\
X^2+Y^2=1,
\end{cases}
$$
where $X=\cos\theta$, $Y=\sin\theta.$
So either $\cos\theta =0$, $\sin\theta=1$, or viceversa. Plugging this into $\cos\theta-\sin\theta$ you either get $+1$ or $-1$.
