P = Q
P ⊆ Q
Q ⊆ P
P ⊆ Q and Q - P ≠ ∅
P = θ : sinθ - cosθ = √2 cosθ
sinθ - cosθ = √2 cosθ
Dividing by cosθ, we get
tanθ - 1 = √2
tanθ = √2 + 1
θ = arctan(√2 + 1)
Q = θ : sinθ + cosθ = √2 sinθ
sinθ + cosθ = √2 sinθ
cosθ = √2 sinθ - sinθ
cosθ = (√2 - 1) sinθ
cosθ / sinθ = √2 - 1
tanθ = 1 / (√2 - 1)
On rationalising, we get
tanθ = (√2 + 1) / (√2 - 1) * (√2 + 1) / (√2 + 1)
tanθ = (2 + 2√2 + 1) / (2 - 1)
tanθ = 3 + 2√2
This is incorrect. Let's try another approach.
cosθ = (√2 - 1) sinθ
cotθ = √2 - 1
tanθ = 1/(√2 - 1)
On rationalizing, we get
tanθ = √2 + 1
θ = arctan(√2 + 1)
Hence, sets P and Q both have the same values. So P = Q.