Let A={θ:sinθ=tanθ} and B={θ:cosθ=1} Now, A={θ:sinθ=cosθsinθ} ={θ:sinθ(cosθ−1)=0} ={θ=0,π,2π,3π,……} For B:cosθ=1⇒θ=π,2π,4π,…… This shows that A is not contained in B. i.e. A⊂B. but B⊂A.
Let A={θ:sin(θ)=tan(θ)} and B=(θ:cos(θ)= 1} be two sets. Then:
Held on 25 Apr 2013 · Verified 6 Jul 2026.
A=B
A⊂B
B⊂A
A⊂B and B−A=ϕ
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