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Problem
Problem
If
∣
1
−
∣
sin
x
∣
1
+
sin
x
∣
≥
2
3
\left| 1 - \frac{|\sin x|}{1 + \sin x} \right| \ge \frac{2}{3}
1
−
1
+
s
i
n
x
∣
s
i
n
x
∣
≥
3
2
, then
sin
x
\sin x
sin
x
lies in a.
(
−
∞
,
−
1
2
]
∪
[
1
2
,
∞
)
\left(-\infty, -\frac{1}{2}\right] \cup \left[\frac{1}{2}, \infty\right)
(
−
∞
,
−
2
1
]
∪
[
2
1
,
∞
)
b.
(
−
1
2
,
1
2
)
\left(-\frac{1}{2}, \frac{1}{2}\right)
(
−
2
1
,
2
1
)
c.
[
−
1
2
,
1
2
]
\left[-\frac{1}{2}, \frac{1}{2}\right]
[
−
2
1
,
2
1
]
d. None of the above
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