Question

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⇒ No useless answers pls​

Answer 1

⚘ Option D is correct..

⚘ Explaination is in attachment..

⚘ Kindly, adjust with the handwriting.. ^•^

Answer 2

Basic Concept Used :-

Equal Sets :-

• Two sets A and B are said to be equal if every element of A is also an element of B and if every element of B is also an element of A.

$\large\underline{\sf{Solution-}}$

Given,

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

and

$\bf \: Q \: = \{ \theta : sin\theta + cos\theta = \sqrt{2}sin\theta \}$

Now,

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

$\rm :\longmapsto\:sin\theta - cos\theta = \sqrt{2} cos\theta$

$\rm :\longmapsto\:sin\theta = \sqrt{2}cos\theta + cos\theta$

$\rm :\longmapsto\:sin\theta = (\sqrt{2} + 1)cos\theta$

$\rm :\longmapsto\:sin\theta = (\sqrt{2} + 1) \: cos\theta \times \dfrac{ \sqrt{2} - 1}{ \sqrt{2} - 1 }$

$\rm :\longmapsto\:sin\theta = \: cos\theta \times \dfrac{ (\sqrt{2})^{2} - (1)^{2} }{ \sqrt{2} - 1 }$

$\: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \because \: \sf \: {x}^{2} - {y}^{2} = (x + y)(x - y)}$

$\rm :\longmapsto\:( \sqrt{2} - 1)sin\theta = cos\theta(2 - 1)$

$\rm :\longmapsto\:( \sqrt{2} - 1)sin\theta = cos\theta$

$\rm :\longmapsto\:\sqrt{2}sin\theta - sin\theta = cos\theta$

$\rm :\longmapsto\: \sqrt{2}sin\theta = cos\theta + sin\theta$

$\sf \: \therefore\: if \: sin\theta - cos\theta = \sqrt{2}cos\theta\ \:then\:sin\theta + cos\theta = \sqrt{2}cos\theta$

$\bf\implies \:P = Q$

$\boxed{ \bf \: Hence, \: Option \: (d) \: is \: correct}$

Additional Information :-

• Finite Set :- A set is said to be finite if its elements are countable. eg. Set of natural number less than 10

• Infinite Set :- A set is said to be infinite if its elements are not countable. eg :- Set of natural numbers.

• Empty Set or Void Set or Null Set :- A set is said to be null set if it doesn't contain any element and represented as { }.

• Subset :- A set A is subset of B if every element of A is contained in B and represented as A ⊂ B.