<s>[INST] 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?
 [/INST] Let $x = \cos\theta$ and $y = \sin\theta$. Then $(x,y)$ is a point on
the unit circle. But the equation
$$\sin \theta + \cos \theta = 1$$
says that $x + y = 1$. So $(x,y)$ must be on the line given by $x + y = 1$,
that is, the line that intersects the unit circle at $(1,0)$ and $(0,1)$.
In fact, since $(x,y)$ is on that circle and on that line,
it must be one of those two points.
Case 1: $(x,y) = (0,1)$.
\begin{align}
\cos \theta &= 0 \\ \sin \theta &= 1 \\ \cos\theta - \sin\theta &= -1
\end{align}
Case 2: $(x,y) = (1,0)$
\begin{align}
\cos \theta &= 1 \\ \sin \theta &= 0 \\ \cos\theta - \sin\theta &= 1
\end{align}
And those are the only two possible cases that can occur.
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