If $a\cos \theta +b\sin \theta =1$ and $a\sin \theta -b\cos \theta =1$, then a correct statement is

$a\cos \theta +b\sin \theta =1$

Squaring on both sides

$a^2{\cos }^2\theta +b^2{\sin }^2\theta +2ab\cos \theta \sin \theta =1\cdots \left(1\right)$

$a\sin \theta -b\cos \theta =1$

Squaring on both sides

$a^2{\sin }^2\theta +b^2{\cos }^2\theta -2ab\sin \theta \cos \theta =1\cdots \left(2\right)$

By adding equation$\left(1\right)$ and equation$\left(2\right)$ 

$a^2{\cos }^2\theta +a^2{\sin }^2\theta +b^2{\sin }^2\theta +b^2{\cos }^2\theta +2ab\cos \theta \sin \theta -2ab\sin \theta \cos \theta =2$

$a^2({\cos }^2\theta +{\sin }^2\theta )+b^2({\sin }^2\theta +{\cos }^2\theta )=2$

$a^2+b^2=2$