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?

Notice, we have $$\cos \theta+\sin\theta=1$$
$$(\cos \theta+\sin\theta)^2=1$$$$\cos^2\theta+\sin^2\theta+2\sin \theta\cos \theta=1$$ $$1+2\sin \theta\cos \theta=1$$
$$\iff \sin\theta\cos \theta=0\tag 1$$
Now, we have
$$(\cos \theta-\sin\theta)^2=\cos^2\theta+\sin^2\theta-2\sin \theta\cos \theta$$
$$=(\cos^2\theta+\sin^2\theta+2\sin \theta\cos \theta)-4\sin \theta\cos \theta$$
$$=(\cos \theta+\sin\theta)^2-4\sin \theta\cos \theta$$
Substituting the corresponding values
$$=(1)^2-4(0)=1$$ $$\cos\theta-\sin\theta=\pm\sqrt 1=\pm 1$$
