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Byju's Answer
Standard XII
Mathematics
Basic Inverse Trigonometric Functions
Identify the ...
Question
Identify the pair of functions which are not identical
A
y
=
t
a
n
(
c
o
s
−
1
x
)
;
y
=
√
1
−
x
2
x
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B
y
=
t
a
n
(
c
o
t
−
1
x
)
;
y
=
1
x
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C
y
=
s
i
n
(
a
r
c
t
a
n
x
)
;
y
=
x
√
1
+
x
2
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D
y
=
c
o
s
(
t
a
n
−
1
x
)
;
y
=
s
i
n
(
t
a
n
−
1
x
)
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Solution
The correct option is
D
y
=
c
o
s
(
t
a
n
−
1
x
)
;
y
=
s
i
n
(
t
a
n
−
1
x
)
Conceptual, verify each of options
Suggest Corrections
2
Similar questions
Q.
Identify the pair(s) of functions which are identical.
i)
y
=
sin
(
a
r
c
tan
x
)
;
y
=
x
√
1
+
x
2
(ii)
y
=
cos
(
a
r
c
tan
x
)
;
y
=
sin
(
a
r
c
cot
x
)
.
(iii)
y
=
tan
(
cos
−
1
x
)
;
y
=
√
1
−
x
2
x
iv)
y
=
tan
(
cot
−
1
x
)
;
y
=
1
x
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
.
solve for x the following equations :
(a)
c
o
t
−
1
x
+
s
i
n
−
1
1
√
(
5
)
=
π
4
(b)
2
t
a
n
−
1
(
c
o
s
x
)
=
t
a
n
t
a
n
−
1
(
2
c
o
s
e
c
x
)
.
(c)
t
a
n
(
c
o
s
−
1
x
)
=
s
i
n
(
c
o
t
−
1
1
2
)
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
(
c
o
s
−
1
3
5
)
(b)
c
o
s
(
t
a
n
−
1
3
4
)
and
c
o
s
(
t
a
n
−
1
x
)
(c) If
s
i
n
(
c
o
t
−
1
(
1
+
x
)
)
=
c
o
s
(
t
a
n
−
1
x
)
then x is
(a)
1
/
2
(b)
1
(c)
0
(d)
−
1
/
2
(d)
s
i
n
(
c
o
t
−
1
x
)
(e)
s
i
n
(
2
s
i
n
−
1
0.8
)
Q.
Inverse circular functions,Principal values of
sin
−
1
x
,
cos
−
1
x
,
tan
−
1
x
.
tan
−
1
x
+
tan
−
1
y
=
tan
−
1
x
+
y
1
−
x
y
,
x
y
<
1
π
+
tan
−
1
x
+
y
1
−
x
y
,
x
y
>
1
.
Prove that
tan
−
1
1
−
x
1
+
x
tan
−
1
1
−
y
1
+
y
=
sin
−
1
y
−
x
√
1
+
x
2
√
1
+
y
2
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
.
If
0
<
x
<
1
, then
√
1
+
x
2
[
{
x
c
o
s
(
c
o
t
−
1
x
)
+
s
i
n
(
c
o
t
−
1
x
)
}
2
−
1
]
1
/
2
is equal to
(a)
x
√
1
+
x
2
(b)
x
(c)
x
√
1
+
x
2
(d)
√
1
+
x
2
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