SATHEE: Trigonometrical Equations 1 Question 12

Trigonometrical Equations 1 Question 12

12. If $P = θ : \sin θ - \cos θ = \sqrt{2} \cos θ$ and $Q = θ : \sin θ + \cos θ = \sqrt{2} \sin θ$ be two sets. Then, (2011)

(a) $P \subset Q$ and $Q - P \neq φ$

(b) $Q \notin P$

(d) $P = Q$

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Answer:

Correct Answer: 12. (d)

Solution:

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

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

$\Rightarrow \tan θ = \sqrt{2} + 1$

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

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

$\Rightarrow \tan θ = \frac{1}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = (\sqrt{2} + 1)$

$∴ P = Q$

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Trigonometrical Equations 1 Question 12

12. If P=θ:sin⁡θ−cos⁡θ=2cos⁡θ and Q=θ:sin⁡θ+cos⁡θ=2sin⁡θ be two sets. Then, (2011)

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12. If $P = θ : \sin θ - \cos θ = \sqrt{2} \cos θ$ and $Q = θ : \sin θ + \cos θ = \sqrt{2} \sin θ$ be two sets. Then, (2011)