If a \cos \\theta+b \sin \\theta=2 \& a \sin \\theta-b \cos \\theta=3 then a^{2}+b^{2}=

NIMCET 2021 Question Paper with solutions

Question : 12 of 120

Marks: +1, -0

a \cos \θ+b \sin \θ=2\ \&\ a \sin \θ-b \cos \θ=3

then

a^{2}+b^{2}=

Solution:

a \cos \θ+b \sin \θ=2

........(1)

a \sin \θ-b \cos \θ=3

........(2)
Squaring and adding equation (1) & (2)

a^{2} \cos ^{2} \θ+b^{2} \sin ^{2} \θ+2 a b \sin \θ\cos \θ

+a^{2} \sin ^{2} \θ+b^{2} \cos ^{2} \θ-2 a b

\sin \θ\cos \θ=2^{2}+3^{2}

a^{2}(\cos ^{2} \θ+\sin ^{2} \θ)+b^{2}(\sin ^{2} \θ+\cos ^{2} \θ)

=4+9

a^{2}+b^{2}=13

\;\;\;\; [\sin ^{2} \θ+\cos ^{2} \θ=1]

Go to Question:

Prev Question

Next Question