If $a \sin \theta + b \cos \theta = c$,then prove that $a \cos \theta - b \sin \theta = \pm \sqrt{a^2 + b^2 - c^2}$,given $a^2 + b^2 \geq c^2$.

(N/A) Given: $a \sin \theta + b \cos \theta = c$ $(1)$
Let $x = a \cos \theta - b \sin \theta$ $(2)$
Squaring both equations $(1)$ and $(2)$ and adding them:
$(a \sin \theta + b \cos \theta)^2 + (a \cos \theta - b \sin \theta)^2 = c^2 + x^2$
$a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \sin \theta \cos \theta + a^2 \cos^2 \theta + b^2 \sin^2 \theta - 2ab \sin \theta \cos \theta = c^2 + x^2$
$a^2(\sin^2 \theta + \cos^2 \theta) + b^2(\cos^2 \theta + \sin^2 \theta) = c^2 + x^2$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have:
$a^2(1) + b^2(1) = c^2 + x^2$
$a^2 + b^2 = c^2 + x^2$
$x^2 = a^2 + b^2 - c^2$
$x = \pm \sqrt{a^2 + b^2 - c^2}$
Therefore,$a \cos \theta - b \sin \theta = \pm \sqrt{a^2 + b^2 - c^2}$.