If a cos θ + b sin θ = m and a sin θ − b cos θ = n, then a2 + b2 =

Given:

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

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

Squaring and adding these equations, we have

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

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

$\Rightarrow a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+2 a b \sin \theta \cos \theta+a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta-2 a b \sin \theta \cos \theta=m^{2}+n^{2}$

$\Rightarrow a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta=m^{2}+n^{2}$

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

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

$\Rightarrow a^{2}(1)+b^{2}(1)=m^{2}+n^{2}$

$\Rightarrow a^{2}+b^{2}=m^{2}+n^{2}$

Hence, the correct option is (d).