#### If a sinθ + b cosθ = c, then prove that a cosθ – b sinθ =$\sqrt{a^{2}+b^{2}-c^{2}}$

Solution.     Given:- asinθ + b cosθ = c
squaring both side we get
$\left ( a\sin \theta +b\cos \theta \right )^{2}= c^{2}$
$a^{2}\sin ^{2}\theta +b^{2}\cos ^{2}\theta+2ab\sin \theta \cos \theta = c^{2}$
$\left ( \because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab\right )$
$\Rightarrow 2ab\sin \theta \cos \theta = c^{2}-a^{2}\sin ^{2}\theta -b^{2}\cos ^{2}\theta \cdots \left ( 1 \right )$
To prove : acosθ – b sinθ =$\sqrt{a^{2}+b^{2}-c^{2}}$
Taking left hand side : a cosθ – b sinθ and square it we get
$\left ( a\cos \theta -b \sin \theta \right )^{2}$
$= a^{2}\cos ^{2}\theta +b^{2}\sin ^{2}\theta -2ab\cos \theta \sin \theta$
$\left [ \because \left ( a-b \right )^{2} = a^{2}+b^{2}-2ab\right ]$
$= a^{2}\cos ^{2}\theta +b^{2}\sin ^{2}\theta -\left ( c^{2}-a^{2}\sin ^{2}\theta -b^{2}\cos ^{2}\theta \right )$ $\text{[Using (1)]}$
$= a^{2}\cos ^{2}\theta +b^{2}\sin ^{2}\theta - c^{2}+a^{2}\sin ^{2}\theta +b^{2}\cos ^{2}\theta$
$= a^{2}\left ( \cos ^{2}\theta+\sin ^{2}\theta \right )+b^{2}\left ( \sin ^{2}\theta+ \cos ^{2}\right )-c^{2}$
$= a^{2}+b^{2} - c^{2}$      $\left ( \because \sin ^{2}\theta +\cos ^{2}\theta = 1 \right )$
Hence $\left ( a\cos \theta -b\sin \theta \right )^{2}= a^{2}+b^{2}-c^{2}$
$\Rightarrow \left ( a\cos \theta -b\sin \theta \right )= \sqrt{a^{2}+b^{2}-c^{2}}$
Hence proved.