A) \[\operatorname{A}=B\]
B) \[\operatorname{A}\not\subset B\]
C) \[\operatorname{B}\not\subset \operatorname{A}\]
D) \[A\subset B\]and\[B-A\ne \phi \]
Correct Answer: B
Solution :
Let \[A=\{\theta :\sin \theta =\tan \theta \}\]and\[B=\{\theta :\cos \theta =1\}\] Now,\[\left\{ \theta :\sin \theta =\frac{\sin \theta }{\cos \theta } \right\}\] \[=\{\theta :\sin \theta (\cos \theta -1)=0\}\] \[=\{\theta =0,\pi ,2\pi ,3\pi ,.....\}\] For\[B:\cos =\theta =1\Rightarrow \theta =\pi ,2\pi ,4\pi ,......\] This shows that A is not contained in B. i.e. \[A\not\subset B.\]but\[B\subset A.\]
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