Question

If a cos θ − b sin θ = c, then a sin θ + b cos θ =

Options

• \[\pm \sqrt{a^2 + b^2 + c^2}\]

• \[\pm \sqrt{a^2 + b^2 - c^2}\]

• \[\pm \sqrt{c^2 - a^2 - b^2}\]

•  None of these

Answer 1

Given:

`a cosθ- b sinθ=c`

Squaring on both sides, we have

`(a cosθ-b sinθ ^2)=c^2`

`⇒ a^2 cos^2 θ+b^2 sin^2 θ-2. a cos θ. b sinθ=c^2`

``⇒a^2(1-sin ^2 θ)+b^2(1-cos^2θ)-2.a cosθ. b sin θ=c^2`

``⇒a^2-a^2 sin^2θ+b^2 cos^2 θ-2.acosθ. b sinθ=c^2`

``⇒-a^2 sin^2 θ-b^2 cos^2 θ-2 a cosθ. b sin θ=-a^2-b^2+c^2`

``⇒-(a^2 sin^2 θ+b^2 cos^2θ+2.a cosθ.b sin θ)=-(a^2+b^2-c^2)`

``⇒a^2 sin^2 θ+b^2 cos^2 θ+2.a sin θ.b cos θ=a^2+b^2-c^2`

``⇒(a sin θ+b cosθ)^2=a^2+b^2-c^2`

``⇒a sin θ+b cos θ=+- sqrt a^2+b^2-c^2`