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જો $A = \left\{ {\theta \,:\,\sin \,\left( \theta \right) = \tan \,\left( \theta \right)} \right\}$ અને $B = \left\{ {\theta \,:\,\cos \,\left( \theta \right) = 1} \right\}$ બે ગણ હોય તો ....
$A = B$
$A \not\subset B$
$B \not\subset A$
$A \subset B$ and $B - A \ne \phi $
Solution
Let $A = \left\{ {\theta \,:\,\sin \, \theta = \tan \,\theta } \right\}$
and $B = \left\{ {\theta \,:\,\cos \, \theta = 1} \right\}$
Now, $A\, = \,\left\{ {\theta \,\,:\,\,\sin \,\,\theta \, = \,\frac{{\sin \,\theta }}{{\cos \,\theta }}} \right\}$
$ = \,\{ \theta \,:\,\sin \,\theta \,\,(\cos \,\theta \,\, – \,1)\, = \,0\} $
$ = \,\{ \theta \, = \,0\,,\,\pi \,,\,2\pi \,,\,3\pi ,\,…..\} $
For $B{\mkern 1mu} \,:\,\cos \,\theta \, = \,1\,\, \Rightarrow \,\theta \, = \,\pi \,,\,2\pi \,,\,4\pi \,,…..{\mkern 1mu} $
This showns that $A$ is not contained in $B$ i.e.
$A\, \not\subset \,B$ but $B\, \subset \,A$.
