Question

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

विकल्प

•  7

• 12

• 25

• None of these

Answer 1

Given: 

`a cos θ+b sinθ=4` 

`a sin θ-b cosθ=3` 

Squaring and then adding the above two equations, we have

`(a cosθ+b sinθ)^2+(a sinθ-b cosθ)^2=(4)^2+(3)^2` 

`=(a^2cos^2θ+b^2 sin^2θ+2a cosθ.b.sinθ)+(a^2 sin^2θ+b^2 cos^2θ-2.a sinθ.b cosθ)=16+9`

`=a^2 cos^2θ+b^2 sin^2θ+ab sinθ cosθ+a^2 sin^2θ+b^2 cos^2θ-2ab sinθ cosθ=25`

`=a^2 cos^2θ+b^2 sin^2θ+a^2 sin^2θ+b^2 cos^2θ=25`

`=(a^2 cos^2θ+a^2sin^2θ)+(b^2 sin^2θ+b^2 cos^2θ)=25`

=`a^2(cos^2θ+sin^2θ)+b^2(sin^2θ+cos^2θ=25)`

`=a^2(1)+b^2(1)=25`

=`a^2+b^2=25``