Let P={θ:sin⁡θ−cos⁡θ=2cos⁡θ} ,Q={θ:sin⁡θ+cos⁡θ=2sin⁡θ}be two sets. Then

Let $P = {θ : \sin θ - \cos θ = \sqrt{2} \cos θ}$ , $Q = {θ : \sin θ + \cos θ = \sqrt{2} \sin θ}$ be two sets. Then

A
$P \subset Q and Q - P \neq ϕ$
B
$Q ⊄ P$
C
$P = Q$
D
$2 - \sqrt{3}$

Solution:

We have , $\sin θ - \cos θ = \sqrt{2} \cos θ$

$\Rightarrow \sin θ = (\sqrt{2} + 1) \cos θ = \frac{2 - 1}{\sqrt{2} - 1} \cos θ$

$\Rightarrow (\sqrt{2} - 1) \sin θ = \cos θ$ ..(i)

Also, $\sin θ + \cos θ = \sqrt{2} \sin θ \Rightarrow (\sqrt{2} - 1) \sin θ = \cos θ .$ …(ii)

Hence $, P = Q$ (From (i) & (ii))

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