If $A=\left\{x \in[0,2 \pi] / \tan x-\tan ^2 x>0\right\}$ and $B=\left\{x \in[0,2 \pi] /|\sin x| <…
If $A=\left\{x \in[0,2 \pi] / \tan x-\tan ^2 x>0\right\}$ and $B=\left\{x \in[0,2 \pi] /|\sin x| < \frac{1}{2}\right\}$, then $A \cap B=$
- $\left(0, \frac{\pi}{6}\right) \cup\left(\pi, \frac{7 \pi}{6}\right)$
- $\left(0, \frac{\pi}{4}\right) \cup\left(\pi, \frac{7 \pi}{6}\right)$
- $\left(0, \frac{\pi}{6}\right) \cup\left(\frac{5 \pi}{6}, \frac{7 \pi}{6}\right)$
- $\left(\frac{\pi}{6}, \frac{7 \pi}{6}\right)$
Solution
$
\begin{aligned}
& \text { If } A=\left\{x \in[0,2 \pi] / \tan x-\tan ^2 x>0\right\} \\
& \therefore \quad \tan x-\tan ^2 x>0 \\
& \Rightarrow \tan x(1-\tan x)>0 \\
& \quad 0 < \tan x < 1
\end{aligned}
$
So, $\quad 0 < x < \frac{\pi}{4}$ and $\pi < x < \frac{5 \pi}{4}$
$
\Rightarrow \quad x \in\left(0, \frac{\pi}{4}\right) \cup\left(\pi, \frac{5 \pi}{4}\right)
$
Now, $B=\left\{x \in[0,2 \pi] /|\sin x| < \frac{1}{2}\right.$
$
\begin{aligned}
& \Rightarrow \quad-\frac{1}{2} < \sin x < \frac{1}{2} \\
& \quad \sin x < \frac{1}{2} \text { and } \sin x>\frac{-1}{2} \\
& \Rightarrow \quad 0 < x < \frac{\pi}{6} \text { and }\left(\frac{5 \pi}{6}, 2 \pi\right) \text { and } x \in\left(0, \frac{7 \pi}{6}\right) \\
& \Rightarrow \quad x \in\left(0, \frac{\pi}{6}\right) \cup\left(\frac{5 \pi}{6}, 2 \pi\right)
\end{aligned}
$
So, $A \cap B$ is defined in
$
\left(0, \frac{\pi}{6}\right) \cup\left(\pi, \frac{7 \pi}{6}\right)
$
Asked in: AP EAMCET 2018 (23 Apr Shift 1)
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