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Byju's Answer
Standard XII
Mathematics
Properties Derived from Trigonometric Identities
Considering o...
Question
Considering only the principal values of inverse functions, the set
A
=
{
x
≥
0
:
tan
−
1
(
2
x
)
+
tan
−
1
(
3
x
)
=
π
4
}
A
is an empty set
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B
is a singleton
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C
contains two elements
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D
Contains more than two elements
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Solution
The correct option is
B
is a singleton
tan
−
1
2
x
+
tan
−
1
3
x
=
π
4
⇒
tan
−
1
2
x
+
3
x
1
−
2
x
.3
x
=
π
4
⇒
5
x
1
−
6
x
2
=
tan
π
4
5
x
=
1
−
6
x
2
⇒
6
x
2
+
5
x
−
1
=
0
⇒
(
x
+
1
)
(
x
−
1
6
)
=
0
As,
x
≥
0
⇒
x
=
1
6
Suggest Corrections
0
Similar questions
Q.
Considering only the principal values of inverse functions, the set
A
=
{
x
≥
0
:
tan
−
1
(
2
x
)
+
tan
−
1
(
3
x
)
=
π
4
}
Q.
Considering principal values, the number of solutions of
tan
−
1
2
x
+
tan
−
1
3
x
=
π
4
is
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
t
a
n
−
1
2
x
+
t
a
n
−
1
3
x
=
n
π
+
(
π
/
4
)
.
(b) Find all the positive integral solutions of
t
a
n
−
1
x
+
c
o
s
−
1
(
y
√
1
+
y
2
)
=
s
i
n
−
1
(
3
√
10
)
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
4
+
2
t
a
n
−
1
1
5
+
t
a
n
−
1
1
6
+
t
a
n
−
1
1
x
=
π
4
(b)
t
a
n
−
1
(
x
−
1
)
+
t
a
n
−
1
x
+
t
a
n
−
1
(
x
+
1
)
=
t
a
n
−
1
3
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
.
Prove that
t
a
n
−
1
(
1
2
t
a
n
2
A
)
+
t
a
n
−
1
(
c
o
t
A
)
+
t
a
n
−
1
(
c
o
t
3
A
)
0
if
π
/
4
<
A
<
π
/
2
and
=
π
if
0
<
A
<
π
/
4
.
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