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Byju's Answer
Standard XII
Mathematics
Implicit Differentiation
The value of ...
Question
The value of
sin
−
1
⎡
⎢
⎣
cot
⎛
⎜
⎝
sin
−
1
⎷
(
2
−
√
3
4
)
+
cos
−
1
√
12
4
+
sec
−
1
√
2
⎞
⎟
⎠
⎤
⎥
⎦
is
A
0
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B
π
4
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C
π
6
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D
π
2
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Solution
The correct option is
A
0
We have
sin
−
1
⎡
⎢
⎣
cot
⎛
⎜
⎝
sin
−
1
⎷
(
2
−
√
3
4
)
+
cos
−
1
√
12
4
+
sec
−
1
√
2
⎞
⎟
⎠
⎤
⎥
⎦
=
sin
−
1
[
cot
(
sin
−
1
(
√
3
−
1
2
√
2
)
+
cos
−
1
√
3
2
+
cos
−
1
1
√
2
)
]
=
sin
−
1
[
cot
(
15
0
+
30
0
+
45
0
)
]
=
sin
−
1
[
cot
90
0
]
=
sin
−
1
0
=
0
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0
Similar questions
Q.
Inverse circular functions,Principal values of
s
i
n
−
1
x
,
c
o
s
−
1
x
,
t
a
n
−
1
x
.
t
a
n
−
1
x
+
t
a
n
−
1
y
=
t
a
n
−
1
x
+
y
1
−
x
y
,
x
y
<
1
π
+
t
a
n
−
1
x
+
y
1
−
x
y
,
x
y
>
1
.
(a)
s
i
n
−
1
(
3
x
/
5
)
+
s
i
n
−
1
(
4
x
/
5
)
=
s
i
n
−
1
x
(b)
c
o
s
−
1
x
+
s
i
n
−
1
(
1
2
x
)
=
π
6
(c) If
a
≤
t
a
n
−
1
(
1
−
x
1
+
x
)
≤
b
where
0
≤
x
≤
1
then
(
a
,
b
)
=
(a)
(
0
,
π
)
(b)
(
0
,
π
/
4
)
(c)
(
−
π
/
4
,
π
/
4
)
(d)
(
π
/
4
,
π
/
2
)
(d) If
a
≤
(
s
i
n
−
1
x
)
3
+
(
c
o
s
−
1
x
)
3
≤
b
then (a,b) is equal to
(
π
3
32
,
7
π
3
8
)
.
Q.
List I List II
A)
sin
−
1
(
1
2
)
+
cos
−
1
(
−
1
2
)
1)
π
2
B)
sin
−
1
(
−
1
2
)
+
cos
−
1
(
1
2
)
2)
π
6
C)
sin
−
1
(
1
2
)
+
cos
−
1
(
1
2
)
3)
5
π
6
D)
sin
−
1
(
1
)
+
cos
−
1
(
−
1
2
)
4)
7
π
6
Correct match from List I to List II is
Q.
Assertion :
s
i
n
−
1
[
x
−
x
2
2
+
x
3
4
.
.
.
.
]
=
π
/
2
−
c
o
s
−
1
[
x
2
−
x
4
2
+
x
6
4
.
.
.
.
]
for
0
<
|
x
|
<
√
2
has a unique solution. Reason:
t
a
n
−
1
√
x
(
x
+
1
)
+
s
i
n
−
1
√
x
2
+
x
+
1
=
π
/
2
has no solution for
−
√
2
<
x
<
0
Q.
For
0
≤
cos
−
1
x
≤
π
and
−
π
2
≤
sin
−
1
x
≤
π
2
, the value of
cos
(
sin
−
1
x
+
2
cos
−
1
x
)
at
x
=
1
5
is:
Q.
If tan(
π
cos x) = cot (
π
sin x), prove that
c
o
s
(
x
−
π
4
)
=
±
1
2
√
2
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