If a cos θ + b sin θ = 4 and a sin θ − b sin θ = 3, then a2 + b2 =
Question:

If $a \cos \theta+b \sin \theta=4$ and $a \sin \theta-b \sin \theta=3$, then $a^{2}+b^{2}=$

(a) 7
(b) 12
(c) 25
(d) None of these

Solution:

Given:

$a \cos \theta+b \sin \theta=4$,

$a \sin \theta-b \cos \theta=3$

Squaring and then adding the above two equations, we have

$(a \cos \theta+b \sin \theta)^{2}+(a \sin \theta-b \cos \theta)^{2}=(4)^{2}+(3)^{2}$

$\Rightarrow\left(a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+2 \cdot a \cos \theta \cdot b \sin \theta\right)+\left(a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta-2 \cdot a \sin \theta \cdot b \cos \theta\right)=16+9$

$\Rightarrow a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+2 a b \sin \theta \cos \theta+a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta-2 a b \sin \theta \cos \theta=25$

$\Rightarrow a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta=25$

$\Rightarrow\left(a^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)+\left(b^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta\right)=25$

$\Rightarrow a^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)+b^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=25$

$\Rightarrow a^{2}(1)+b^{2}(1)=25$

$\Rightarrow a^{2}+b^{2}=25$

Hence, the correct option is (c).