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Let $A=\left\{\theta \in R \mid \cos ^2(\sin \theta)+\sin ^2(\cos \theta)=1\right\}$ and $B=\{\theta \in R \mid \cos (\sin \theta) \sin (\cos \theta)=0\}$. Then, $A \cap B$
is the empty set
has exactly one clement
has more than one but finitely many elements
has infinitely many elements
Solution
(a)
We have,
$A=\left\{\theta \in R \mid \cos ^2(\sin \theta)+\sin ^2(\cos \theta)=1\right\}$
$\therefore \cos ^2(\sin \theta)+\sin ^2(\cos \theta)=1$
$\therefore \quad \sin \theta=\cos \theta \Rightarrow \tan \theta=1$
$\Rightarrow \quad \theta=n \pi+\frac{\pi}{4}$
$\quad B=\{\theta \in R \mid \cos (\sin \theta) \sin (\cos \theta)=0$
$\quad \cos (\sin \theta) \sin (\cos \theta)=0$
$\quad \cos (\sin \theta)=0 \text { or } \sin (\cos \theta)=0$
$\quad \sin \theta=(2 n+1) \frac{\pi}{2} \text { or } \cos \theta=2 n \pi$
$\therefore \text { Clearly } A \cap B=0$
$\therefore A \cap B \text { is an empty set. }$
