Let A = { theta : sin theta = tan theta } and | vedclass

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(B) Given sets are $A = \{ 	heta : \sin 	heta = 	an 	heta \}$ and $B = \{ 	heta : \cos 	heta = 1 \}$.For set $A$,$\sin 	heta = \frac{\sin 	het

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$A 
ot\subset B$

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(B) Given sets are $A = \{ 	heta : \sin 	heta = 	an 	heta \}$ and $B = \{ 	heta : \cos 	heta = 1 \}$.For set $A$,$\sin 	heta = \frac{\sin 	heta}{\cos 	heta}$,which implies $\sin 	heta (1 - \frac{1}{\cos 	heta}) = 0$.This gives $\sin 	heta = 0$ 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 $n \in \mathbb{Z}$.Thus,$A = \{ n\pi : n \in \mathbb{Z} \} = \{ 0, \pm \pi, \pm 2\pi, \dots \}$.For set $B$,$\cos 	heta = 1$,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,elements like $\pi$ are in $A$ but not in $B$,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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