1
You visited us
1
times! Enjoying our articles?
Unlock Full Access!
Byju's Answer
Standard XII
Mathematics
Properties Derived from Trigonometric Identities
The value of ...
Question
The value of
lim
|
x
|
→
∞
cos
(
tan
−
1
(
sin
(
tan
−
1
x
)
)
)
is equal to :
A
−
1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
√
2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
−
1
√
2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1
√
2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is
D
1
√
2
Given,
l
i
m
|
x
|
→
∞
cos
(
tan
−
1
(
sin
(
tan
−
1
(
x
)
)
)
)
=
cos
(
tan
−
1
(
sin
(
π
2
)
)
)
[
tan
π
2
=
∞
]
=
cos
(
tan
−
1
(
1
)
)
=
cos
(
π
4
)
=
1
√
2
Suggest Corrections
0
Similar questions
Q.
The value of
lim
|
x
|
→
∞
c
o
s
(
t
a
n
−
1
(
s
i
n
(
t
a
n
−
1
x
)
)
)
is equal to
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.
If
f
(
x
)
=
s
i
n
{
c
o
t
−
1
(
x
+
1
)
}
−
c
o
s
(
t
a
n
−
1
x
)
a
n
d
v
=
c
o
s
(
t
a
n
−
1
(
s
i
n
(
c
o
t
−
1
x
)
)
)
then the value of
v
2
for f(x) = 0 is equal to
Q.
The value of
sin
(
2
tan
−
1
1
3
)
+
cos
(
tan
−
1
2
√
2
)
is equal to
Q.
The value of x for which
s
i
n
(
c
o
t
−
1
(
1
+
x
)
)
=
c
o
s
(
t
a
n
−
1
x
)
is:
View More
Join BYJU'S Learning Program
Grade/Exam
1st Grade
2nd Grade
3rd Grade
4th Grade
5th Grade
6th grade
7th grade
8th Grade
9th Grade
10th Grade
11th Grade
12th Grade
Submit
Related Videos
Property 5
MATHEMATICS
Watch in App
Explore more
Properties Derived from Trigonometric Identities
Standard XII Mathematics
Join BYJU'S Learning Program
Grade/Exam
1st Grade
2nd Grade
3rd Grade
4th Grade
5th Grade
6th grade
7th grade
8th Grade
9th Grade
10th Grade
11th Grade
12th Grade
Submit
AI Tutor
Textbooks
Question Papers
Install app