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## Miscellaneous Exercise on Chapter 2 14.
sin
−
1
(
1
−
x
)
−
2
sin
−
1
x
=
π
2
\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}
sin
−
1
(
1
−
x
)
−
2
sin
−
1
x
=
2
π
, then
x
x
x
is equal to (A)
0
,
1
2
0, \frac{1}{2}
0
,
2
1
(B)
1
,
1
2
1, \frac{1}{2}
1
,
2
1
(C) 0 (D)
1
2
\frac{1}{2}
2
1
(A)
0
,
1
2
0, \frac{1}{2}
0
,
2
1
(B)
1
,
1
2
1, \frac{1}{2}
1
,
2
1
(C) 0 (D)
1
2
\frac{1}{2}
2
1
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