Let P - sin-cos-2cos- and Q - sin-cos-2sin- be two sets- ThenA- P - Q and Q - P -B- Q - PC- P - - QD- P - Q

The correct option is D P = QP = {θ : sin θ cos θ = √(2)cos θ} sin θ =(√(2)+1)cos θ, tan θ =√(2)+1 Q = {θ : sin θ + cos θ =√(2)sin θ} cos θ = (√(2) 1)sin ...

Let P =θ: sin...

Question

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

P \subset Q a n d Q - P \neq ϕ

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Q \subset̸ P

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P \subset̸ Q

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P = Q

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Solution

The correct option is D P = Q
$P = {θ : s i n θ - c o s θ = \sqrt{2} c o s θ} s i n θ = (\sqrt{2} + 1) c o s θ, t a n θ = \sqrt{2} + 1 Q = {θ : s i n θ + c o s θ = \sqrt{2} s i n θ} c o s θ = (\sqrt{2} - 1) s i n θ o r t a n θ = \sqrt{2} + 1 ∴ P = Q$

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Let P={θ:sinθ−cosθ=√2cosθ} and Q={θ:sinθ+cosθ=√2sinθ} be two sets. Then

The correct option is D P = QP={θ:sinθ−cosθ=√2cosθ}sin θ=(√2+1)cos θ, tanθ=√2+1Q={θ:sin θ+cos θ=√2sin θ}cos θ=(√2−1)sin θ or tan θ=√2+1∴ P=Q

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

The correct option is D P = Q
$P = {θ : s i n θ - c o s θ = \sqrt{2} c o s θ} s i n θ = (\sqrt{2} + 1) c o s θ, t a n θ = \sqrt{2} + 1 Q = {θ : s i n θ + c o s θ = \sqrt{2} s i n θ} c o s θ = (\sqrt{2} - 1) s i n θ o r t a n θ = \sqrt{2} + 1 ∴ P = Q$