If a cos θ − b sin θ = c, then a sin θ + b cos θ =

Solution:

Given:

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

Squaring on both sides, we have

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

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

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

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

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

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

 

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

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

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

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

Hence, the correct option is (b).