1
You visited us
1
times! Enjoying our articles?
Unlock Full Access!
Byju's Answer
Standard XII
Mathematics
Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions
-1x+tan -13=π...
Question
cot
-1
(x) + tan
-1
(3) = π/2 , then x = ?
Open in App
Solution
The answer is x=3
Because
So if and only if both the angles are same the RHS is pie/2
So x must be equal to 3
Pls like if u r satisfied
Suggest Corrections
0
Similar questions
Q.
If
cot
−
1
x
+
tan
−
1
3
=
π
2
, then
x
=
Q.
Find x, if
tan
−
1
3
+
cot
−
1
x
=
π
2
.
Q.
Assertion :
cosec
−
1
(
3
/
2
)
+
cos
−
1
(
2
/
3
)
−
2
cot
−
1
(
1
/
7
)
−
cot
−
1
7
is equal to
cot
−
1
7
Reason:
sin
−
1
x
+
cos
−
1
x
=
π
/
2
,
tan
−
1
x
+
cot
−
1
x
=
π
/
2
,
cosec
−
1
x
=
sin
−
1
(
1
/
x
)
,
cot
−
1
(
1
/
x
)
=
tan
−
1
(
x
)
,
cot
−
1
(
x
)
=
tan
−
1
(
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
.
(a) Solve the equation :
c
o
s
−
1
(
√
6
x
)
+
c
o
s
−
1
(
3
√
3
x
2
)
=
π
2
(b) If
s
i
n
1
6
x
+
s
i
n
−
1
6
√
3
x
=
−
π
/
2
, then
x
=
.
.
.
.
.
.
.
(c)
s
i
n
−
1
x
+
s
i
n
−
1
2
x
=
π
/
3
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)
t
a
n
−
1
1
2
+
t
a
n
−
1
1
3
=
π
4
.
(b)
t
a
n
−
1
1
2
+
t
a
n
−
1
1
5
+
t
a
n
−
1
1
8
=
π
4
(c)
t
a
n
−
1
3
4
+
t
a
n
−
1
3
5
−
t
a
n
−
1
8
19
=
π
4
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 4
MATHEMATICS
Watch in App
Explore more
Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions
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