Given that. $\quad a \sin \theta+b \cos \theta=c$

On squaring both sides.

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

$\Rightarrow$ $a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta+2 a b \sin \theta \cdot \cos \theta=c^{2} \quad\left[\because(x+y)^{2}=x^{2}+2 x y+y^{2}\right)$

$\Rightarrow \quad a^{2}\left(1-\cos ^{2} \theta\right)+b^{2}\left(1-\sin ^{2} \theta\right)+2 a b \sin \theta \cdot \cos \theta=c^{2}$

$\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$

$\Rightarrow \quad a^{2}-a^{2} \cos ^{2} \theta+b^{2}-b^{2} \sin ^{2} \theta+2 a b \sin \theta \cdot \cos \theta=c^{2}$

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

$\Rightarrow$ $\left(a^{2}+b^{2}-c^{2}\right)=(a \cos \theta-b \sin \theta)^{2}$

$\left[\because a^{2}+b^{2}-2 a b=(a-b)^{2}\right]$

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

$\Rightarrow$ $a \cos \theta-b \sin \theta=\sqrt{a^{2}+b^{2}-c^{2}} \quad$ Hence proved.