Let P = θ : sin θ - cos θ = 2 cos θ and Q = θ : sin θ + cos θ = 2 sin θ , be two sets. Then

P : sin θ - cos θ = 2 cos θ ⇒ sin θ = cos θ + 2 cos θ tan θ = 2 + 1 Q : sin θ + cos θ = 2 sin θ ⇒ 2 - 1 sin θ = cos θ = tan θ = 1 2 - 1 = 2 + 1

Mathematics - Trigonometric Equation Question with Solution | TestHub

MathematicsTrigonometric EquationTrigonometric EquationMedium2 minPYQ_2016

MathematicsMediumsingle choice

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

Answer:

Solution:

$P : \sin θ - \cos θ = \sqrt{2} \cos θ$

$\Rightarrow \sin θ = \cos θ + \sqrt{2} \cos θ$

$\tan θ = \sqrt{2} + 1$

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

$= \tan θ = \frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1$

Stream:JEESubject:MathematicsTopic:Trigonometric EquationSubtopic:Trigonometric Equation

⏱ 2mℹ️ Source: PYQ_2016

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