Question

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

विकल्प

• m² − n²

• m²n²

•  n² − m²

• m² + n²

Answer 1

Given: 

`a cosθ+b sinθ= m,` 

`a sinθ-b cos θ=n` 

Squaring and adding these equations, we have

`(a cos θ+bsin θ)^2+(a sinθ-b cosθ)^2=(m)^2+(n)^2`

`⇒ (a^2 cos^2θ+b^2sin^2θ+2.a cosθ.bsinθ)+(a^2 sin^2θ+b^2 cos^2θ-2.a sin θ.bcosθ)=m^2+n^2`

`⇒ a^2 cos^2θ+b^2 sin^2θ+2ab sin θ cosθ+a^2 sin^2θ+b^2 cos^2θ-2ab sinθ cos θ=m^2+n^2`

`⇒a^2 cos^2θ+b^2 sin^2θ+a^2 sin^2θ+b^2 cos^2=m^2+n^2` 

`⇒(a^2 cos^2θ+a^2 sin^2 θ)+(b^2 sin^2θ+b^2 cos^2θ)=m^2+n^2`

`⇒a^2 (cos^2θ+sin^2θ)+b^2(sin^2 θ+cos^2θ)=m^2+n^2`

`⇒ a^2(1)+b^2(1)=m^2+n^2`

`⇒ a^2+b^2=m^2+n^2`