Let $\mathrm{A}=\{\theta: \sin (\theta)=\tan (\theta)\}$ and $\mathrm{B}=(\theta: \cos (\theta)=$ 1\} be two…
Let $\mathrm{A}=\{\theta: \sin (\theta)=\tan (\theta)\}$ and $\mathrm{B}=(\theta: \cos (\theta)=$ 1\} be two sets. Then:
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$\mathrm{A}=\mathrm{B}$
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$A \not \subset B$
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$B \not \subset A$
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$A \subset B$ and $B-A \neq \phi$
Solution
Let $\mathrm{A}=\{\theta: \sin \theta=\tan \theta\}$ and $\mathrm{B}=\{\theta: \cos \theta=1\}$
Now, $A=\left\{\theta: \sin \theta=\frac{\sin \theta}{\cos \theta}\right\}$ $=\{\theta: \sin \theta(\cos \theta-1)=0\}$ $=\{\theta=0, \pi, 2 \pi, 3 \pi, \ldots \ldots\}$
For $\mathrm{B}: \cos \theta=1 \Rightarrow \theta=\pi, 2 \pi, 4 \pi, \ldots \ldots$
This shows that $\mathrm{A}$ is not contained in B. i.e. $\mathrm{A} \not \subset \mathrm{B}$. but $\mathrm{B} \subset \mathrm{A}$.
Asked in: JEE Main 2013 (25 Apr Online)
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