Let A = {θ ∈ R | cos2 (sin θ) + sin2 (cosθ) = 1} and B = {θ ∈

1.	Let A = {θ ∈ R \| cos2 (sin θ) + sin2 (cosθ) = 1} and B = {θ ∈ R \| cos(sinθ) sin(cosθ) = 0}. Then A ∩ B (A) is the empty set (B) has exactly one element (C) has more than one but finitely many elements (D) has infinitely many elements
Answer» Correct Option :- (A) is the empty set Explanation :- for A ∩ B cos (sin θ) = 1 or –1 & sin(cos θ) = 0 which is not possible or cos (sin θ) = 0 & sin (cos θ) = 1 or –1 also not possible so A ∩ B is an empty set

Let A = {θ ∈ R | cos2 (sin θ) + sin2 (cosθ) = 1} and B = {θ ∈ R | cos(sinθ) sin(cosθ) = 0}. Then A ∩ B (A) is the empty set (B) has exactly one element (C) has more than one but finitely many elements (D) has infinitely many elements

Answer»

Correct Option :- (A) is the empty set

Explanation :-

for A ∩ B

cos (sin θ) = 1 or –1 & sin(cos θ) = 0

which is not possible or cos (sin θ) = 0 & sin (cos θ) = 1 or –1

also not possible

so A ∩ B is an empty set

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Let A = {θ ∈ R | cos2 (sin θ) + sin2 (cosθ) = 1} and B = {θ ∈ R | cos(sinθ) sin(cosθ) = 0}. Then A ∩ B (A) is the empty set (B) has exactly one element (C) has more than one but finitely many elements (D) has infinitely many elements

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