If then show that sinα + cosα = √2 cos θ.

[Hint: Express tanθ = tan(α – π/2) θ = α – π/4]

Given,

To prove: sinα + cosα = √2 cos θ

∵

⇒

⇒ {∵ tan A = (sinA)/(cosA)}

⇒ {∵ tan π/4 = 1}

We know that: tan(x-y) =

∴

⇒ θ = α - π/4

⇒ α = θ + π/4 …(1)

As we have to prove - sinα + cosα = √2 cos θ

∵ LHS = sinα + cosα

⇒ LHS = sin(θ + π/4) + cos(θ + π/4) {using equation 1}

∵ sin(x + y) = sin x cos y + cos x sin y

And, cos(x + y) = cos x cos y – sin x sin y

∴ LHS = sin θ cos(π/4) + sin(π/4)cos θ + cos θ cos(π/4) - sin(π/4)sin θ

∵ sin(π/4)=cos(π/4) = 1/√2

⇒ LHS =

⇒ LHS =

⇒ LHS = √2 cos θ = RHS

Hence, sinα + cosα = √2 cos θ