Let A = { theta : sin(theta) = tan(theta) } a | vedclass

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(B) For set $A$,we have $\sin(	heta) = 	an(	heta)$.This implies $\sin(	heta) = \frac{\sin(	heta)}{\cos(	heta)}$,which gives $\sin(	heta)(1 - \f

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$A 
ot\subset B$

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(B) For set $A$,we have $\sin(	heta) = 	an(	heta)$.This implies $\sin(	heta) = \frac{\sin(	heta)}{\cos(	heta)}$,which gives $\sin(	heta)(1 - \frac{1}{\cos(	heta)}) = 0$.Thus,$\sin(	heta) = 0$ or $\cos(	heta) = 1$.If $\sin(	heta) = 0$,then $	heta = n\pi$ for $n \in \mathbb{Z}$.If $\cos(	heta) = 1$,then $	heta = 2n\pi$ for $n \in \mathbb{Z}$.Combining these,$A = \{ n\pi : n \in \mathbb{Z} \} = \{ 0, \pm\pi, \pm 2\pi, \dots \}$.For set $B$,we have $\cos(	heta) = 1$,which implies $	heta = 2n\pi$ for $n \in \mathbb{Z}$.Thus,$B = \{ 2n\pi : n \in \mathbb{Z} \} = \{ 0, \pm 2\pi, \pm 4\pi, \dots \}$.Comparing the two sets,every element of $B$ is in $A$,so $B \subset A$.However,$\pi \in A$ but $\pi 
otin B$,so $A 
ot\subset B$.

Let $A = \{ \theta : \sin(\theta) = \tan(\theta) \}$ and $B = \{ \theta : \cos(\theta) = 1 \}$ be two sets. Then

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