JEE Exam > JEE Notes > Mathematics (Maths) for JEE Main & Advanced > JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions
JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
JEE Main Previous Year Questions
(2025): Inverse Trigonometric Functions
Question1: If for some ?? , ?? ; ?? = ?? , ?? + ?? = ?? and ?? ?? ?? ?? ? ( ?????? - ?? ? ?? ) + ? ?? ?? ?? ?? ?? ?? ( ?? ?? ?? - ?? ? ?? ) = ???? ,
then ?? ?? + ?? is _ _ _ _
JEE Main 2025 (Online) 24th January Morning Shift
Ans: 14
Solution:
If ( tan ? ( tan
- 1
? ( ?? ) ) + 1 ( c ot ? ( c ot
- 1
? ?? ) )
2
= 36
?? 2
+ ?? 2
= 34
???? = 15
?? = 3 , ?? = 5
? ?? 2
+ ?? = 9 + 5 = 14
Question2: Let ?? = { ?? : ?? ?? ?? - ?? ? ?? = ?? + ?? ???? - ?? ? ?? + ?? ???? - ?? ? [ ?? ?? + ?? ] }. Then ?
?? ? ?? ? ( ?? ?? - ?? )
?? is equal
to _ _ _ _ .
JEE Main 2025 (Online) 29th January Morning Shift
Solution:
c os
- 1
? ?? = ?? + sin
- 1
? ?? + sin
- 1
? ( 2 ?? + 1 )
2 c os
- 1
? ?? - sin
- 1
? ( 2 ?? + 1 ) =
3 ?? 2
2 ?? - ?? =
3 ?? 2
where c os
- 1
? ?? = ?? , sin
- 1
? ( 2 ?? + 1 ) = ??
2 ?? =
3 ?? 2
+ ??
c os ? 2 ?? = s i n ? ??
2 c os
2
? ?? - 1 = sin ? ??
2 ?? 2
- 1 = 2 ?? + 1
?? 2
- ?? - 1 = 0
? ?? =
1 - v 5
2
, { ?? =
1 + v 5
2
rejected }
? 4 ?? 2
- 4 ?? = 4
( 2 ?? - 1 )
2
= 5
Page 2
JEE Main Previous Year Questions
(2025): Inverse Trigonometric Functions
Question1: If for some ?? , ?? ; ?? = ?? , ?? + ?? = ?? and ?? ?? ?? ?? ? ( ?????? - ?? ? ?? ) + ? ?? ?? ?? ?? ?? ?? ( ?? ?? ?? - ?? ? ?? ) = ???? ,
then ?? ?? + ?? is _ _ _ _
JEE Main 2025 (Online) 24th January Morning Shift
Ans: 14
Solution:
If ( tan ? ( tan
- 1
? ( ?? ) ) + 1 ( c ot ? ( c ot
- 1
? ?? ) )
2
= 36
?? 2
+ ?? 2
= 34
???? = 15
?? = 3 , ?? = 5
? ?? 2
+ ?? = 9 + 5 = 14
Question2: Let ?? = { ?? : ?? ?? ?? - ?? ? ?? = ?? + ?? ???? - ?? ? ?? + ?? ???? - ?? ? [ ?? ?? + ?? ] }. Then ?
?? ? ?? ? ( ?? ?? - ?? )
?? is equal
to _ _ _ _ .
JEE Main 2025 (Online) 29th January Morning Shift
Solution:
c os
- 1
? ?? = ?? + sin
- 1
? ?? + sin
- 1
? ( 2 ?? + 1 )
2 c os
- 1
? ?? - sin
- 1
? ( 2 ?? + 1 ) =
3 ?? 2
2 ?? - ?? =
3 ?? 2
where c os
- 1
? ?? = ?? , sin
- 1
? ( 2 ?? + 1 ) = ??
2 ?? =
3 ?? 2
+ ??
c os ? 2 ?? = s i n ? ??
2 c os
2
? ?? - 1 = sin ? ??
2 ?? 2
- 1 = 2 ?? + 1
?? 2
- ?? - 1 = 0
? ?? =
1 - v 5
2
, { ?? =
1 + v 5
2
rejected }
? 4 ?? 2
- 4 ?? = 4
( 2 ?? - 1 )
2
= 5
Question3: If ?? = ?? ?? ?? ? (
?? ?? + ?? ?? ?? - ?? ?
?? ?? ) , then ( ?? - ?? )
?? + ?? ?? ?? is equal to
JEE Main 2025 (Online) 2nd April Evening Shift
Ans: 3
Solution:
?? = c os ? (
?? 3
+ c os
- 1
?
?? 2
)
= c os ? (
?? 3
) c os ? ( c os
- 1
? (
?? 2
) ) - sin ? (
?? 3 .
) sin ? ( c o s
- 1
? (
?? 2
) )
=
1
2
·
?? 2
-
v 3
2
·
v
1 -
?? 2
4
? 4 ?? = ?? - v 3
v
4 - ?? 2
? ( 4 ?? - ?? )
2
= 3 ( 4 - ?? 2
)
? 16 ?? 2
+ ?? 2
- 8 ???? = 12 - 3 ?? 2
?? 2
+ 4 ?? 2
- 2 ???? = 3
( ?? - ?? )
2
+ 3 ?? 2
= 3
Question4: Using the principal values of the inverse trigonometric functions, the sum of the
maximum and the minimum values of ???? ( ( ?? ?? ?? - ?? ? ?? )
?? + ( ?? ?? ?? ?? ?? - ?? ?? )
?? ) is :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 24 ?? 2
B. 18 ?? 2
C. 22 ?? 2
D. 31 ?? 2
Ans: C
Solution:
Let ?? ( ?? ) = 16 [ ( s e c
- 1
? ?? )
2
+ ( c os ec
- 1
?? )
2
].
We can express ?? ( ?? ) as:
?? ( ?? ) = 16 [ ( s ec
- 1
? ?? + c os ec
- 1
?? )
2
- 2 ( s ec
- 1
? ?? ) (
?? 2
- s ec
- 1
? ?? ) ]
This simplifies to:
?? ( ?? ) = 16 [
?? 2
4
- ?? s ec
- 1
? ?? + 2 ( s ec
- 1
? ?? )
2
] , ? where s ec
- 1
? ?? ? [ 0 , ?? ] - {
?? 2
}
Further simplification gives:
Page 3
JEE Main Previous Year Questions
(2025): Inverse Trigonometric Functions
Question1: If for some ?? , ?? ; ?? = ?? , ?? + ?? = ?? and ?? ?? ?? ?? ? ( ?????? - ?? ? ?? ) + ? ?? ?? ?? ?? ?? ?? ( ?? ?? ?? - ?? ? ?? ) = ???? ,
then ?? ?? + ?? is _ _ _ _
JEE Main 2025 (Online) 24th January Morning Shift
Ans: 14
Solution:
If ( tan ? ( tan
- 1
? ( ?? ) ) + 1 ( c ot ? ( c ot
- 1
? ?? ) )
2
= 36
?? 2
+ ?? 2
= 34
???? = 15
?? = 3 , ?? = 5
? ?? 2
+ ?? = 9 + 5 = 14
Question2: Let ?? = { ?? : ?? ?? ?? - ?? ? ?? = ?? + ?? ???? - ?? ? ?? + ?? ???? - ?? ? [ ?? ?? + ?? ] }. Then ?
?? ? ?? ? ( ?? ?? - ?? )
?? is equal
to _ _ _ _ .
JEE Main 2025 (Online) 29th January Morning Shift
Solution:
c os
- 1
? ?? = ?? + sin
- 1
? ?? + sin
- 1
? ( 2 ?? + 1 )
2 c os
- 1
? ?? - sin
- 1
? ( 2 ?? + 1 ) =
3 ?? 2
2 ?? - ?? =
3 ?? 2
where c os
- 1
? ?? = ?? , sin
- 1
? ( 2 ?? + 1 ) = ??
2 ?? =
3 ?? 2
+ ??
c os ? 2 ?? = s i n ? ??
2 c os
2
? ?? - 1 = sin ? ??
2 ?? 2
- 1 = 2 ?? + 1
?? 2
- ?? - 1 = 0
? ?? =
1 - v 5
2
, { ?? =
1 + v 5
2
rejected }
? 4 ?? 2
- 4 ?? = 4
( 2 ?? - 1 )
2
= 5
Question3: If ?? = ?? ?? ?? ? (
?? ?? + ?? ?? ?? - ?? ?
?? ?? ) , then ( ?? - ?? )
?? + ?? ?? ?? is equal to
JEE Main 2025 (Online) 2nd April Evening Shift
Ans: 3
Solution:
?? = c os ? (
?? 3
+ c os
- 1
?
?? 2
)
= c os ? (
?? 3
) c os ? ( c os
- 1
? (
?? 2
) ) - sin ? (
?? 3 .
) sin ? ( c o s
- 1
? (
?? 2
) )
=
1
2
·
?? 2
-
v 3
2
·
v
1 -
?? 2
4
? 4 ?? = ?? - v 3
v
4 - ?? 2
? ( 4 ?? - ?? )
2
= 3 ( 4 - ?? 2
)
? 16 ?? 2
+ ?? 2
- 8 ???? = 12 - 3 ?? 2
?? 2
+ 4 ?? 2
- 2 ???? = 3
( ?? - ?? )
2
+ 3 ?? 2
= 3
Question4: Using the principal values of the inverse trigonometric functions, the sum of the
maximum and the minimum values of ???? ( ( ?? ?? ?? - ?? ? ?? )
?? + ( ?? ?? ?? ?? ?? - ?? ?? )
?? ) is :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 24 ?? 2
B. 18 ?? 2
C. 22 ?? 2
D. 31 ?? 2
Ans: C
Solution:
Let ?? ( ?? ) = 16 [ ( s e c
- 1
? ?? )
2
+ ( c os ec
- 1
?? )
2
].
We can express ?? ( ?? ) as:
?? ( ?? ) = 16 [ ( s ec
- 1
? ?? + c os ec
- 1
?? )
2
- 2 ( s ec
- 1
? ?? ) (
?? 2
- s ec
- 1
? ?? ) ]
This simplifies to:
?? ( ?? ) = 16 [
?? 2
4
- ?? s ec
- 1
? ?? + 2 ( s ec
- 1
? ?? )
2
] , ? where s ec
- 1
? ?? ? [ 0 , ?? ] - {
?? 2
}
Further simplification gives:
?? ( ?? ) = 16 [ 2 ( s ec
- 1
? ?? -
?? 4
)
2
+
?? 2
4
-
?? 2
8
]
For the maximum value when s ec
- 1
? ?? = ?? :
m ax = 16 [ 2 ?? 2
- ?? 2
+
?? 2
4
] = 20 ?? 2
For the minimum value when s ec
- 1
? ?? =
?? 4
:
min = 16 [
2 × ?? 2
16
-
?? 2
4
+
?? 2
4
] = 2 ?? 2
Therefore, the sum of the maximum and minimum values is:
Sum = 22 ?? 2
Question5: If
?? ?? = ?? =
?? ?? ?? , then ?? ?? ?? - ?? ? (
????
????
?? ?? ?? ? ?? +
?? ????
?? ???? ? ?? ) is equal to:
JEE Main 2025 (Online) 23rd January Morning Shift
Options:
A. ?? + tan
- 1
?
5
12
B. ?? - tan
- 1
?
4
3
C. ?? + tan
- 1
?
4
5
D. ?? - tan
- 1
?
5
12
Ans: D
Solution:
?? 2
= ?? =
3 ?? 4
c os
- 1
? (
12
13
c os ? ?? +
5
12
sin ? ?? )
c os
- 1
? ( c o s ? ?? c os ? ?? + s i n ? ?? s i n ? ?? )
c os
- 1
? ( c o s ? ( ?? - ?? ) )
? ?? - ?? because ?? - ?? ? ( -
?? 2
,
?? 2
)
? ?? - tan
- 1
?
5
12
Question6: If ?? > ?? > ?? > ?? , then the expression ?? ?? ?? - ?? ? { ?? +
( ?? + ?? ?? )
( ?? - ?? )
} + ?? ?? ?? - ?? ? { ?? +
( ?? + ?? ?? )
( ?? - ?? )
} +
?? ?? ?? - ?? ? { ?? +
( ?? + ?? ?? )
( ?? - ?? )
} is equal to:
Page 4
JEE Main Previous Year Questions
(2025): Inverse Trigonometric Functions
Question1: If for some ?? , ?? ; ?? = ?? , ?? + ?? = ?? and ?? ?? ?? ?? ? ( ?????? - ?? ? ?? ) + ? ?? ?? ?? ?? ?? ?? ( ?? ?? ?? - ?? ? ?? ) = ???? ,
then ?? ?? + ?? is _ _ _ _
JEE Main 2025 (Online) 24th January Morning Shift
Ans: 14
Solution:
If ( tan ? ( tan
- 1
? ( ?? ) ) + 1 ( c ot ? ( c ot
- 1
? ?? ) )
2
= 36
?? 2
+ ?? 2
= 34
???? = 15
?? = 3 , ?? = 5
? ?? 2
+ ?? = 9 + 5 = 14
Question2: Let ?? = { ?? : ?? ?? ?? - ?? ? ?? = ?? + ?? ???? - ?? ? ?? + ?? ???? - ?? ? [ ?? ?? + ?? ] }. Then ?
?? ? ?? ? ( ?? ?? - ?? )
?? is equal
to _ _ _ _ .
JEE Main 2025 (Online) 29th January Morning Shift
Solution:
c os
- 1
? ?? = ?? + sin
- 1
? ?? + sin
- 1
? ( 2 ?? + 1 )
2 c os
- 1
? ?? - sin
- 1
? ( 2 ?? + 1 ) =
3 ?? 2
2 ?? - ?? =
3 ?? 2
where c os
- 1
? ?? = ?? , sin
- 1
? ( 2 ?? + 1 ) = ??
2 ?? =
3 ?? 2
+ ??
c os ? 2 ?? = s i n ? ??
2 c os
2
? ?? - 1 = sin ? ??
2 ?? 2
- 1 = 2 ?? + 1
?? 2
- ?? - 1 = 0
? ?? =
1 - v 5
2
, { ?? =
1 + v 5
2
rejected }
? 4 ?? 2
- 4 ?? = 4
( 2 ?? - 1 )
2
= 5
Question3: If ?? = ?? ?? ?? ? (
?? ?? + ?? ?? ?? - ?? ?
?? ?? ) , then ( ?? - ?? )
?? + ?? ?? ?? is equal to
JEE Main 2025 (Online) 2nd April Evening Shift
Ans: 3
Solution:
?? = c os ? (
?? 3
+ c os
- 1
?
?? 2
)
= c os ? (
?? 3
) c os ? ( c os
- 1
? (
?? 2
) ) - sin ? (
?? 3 .
) sin ? ( c o s
- 1
? (
?? 2
) )
=
1
2
·
?? 2
-
v 3
2
·
v
1 -
?? 2
4
? 4 ?? = ?? - v 3
v
4 - ?? 2
? ( 4 ?? - ?? )
2
= 3 ( 4 - ?? 2
)
? 16 ?? 2
+ ?? 2
- 8 ???? = 12 - 3 ?? 2
?? 2
+ 4 ?? 2
- 2 ???? = 3
( ?? - ?? )
2
+ 3 ?? 2
= 3
Question4: Using the principal values of the inverse trigonometric functions, the sum of the
maximum and the minimum values of ???? ( ( ?? ?? ?? - ?? ? ?? )
?? + ( ?? ?? ?? ?? ?? - ?? ?? )
?? ) is :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 24 ?? 2
B. 18 ?? 2
C. 22 ?? 2
D. 31 ?? 2
Ans: C
Solution:
Let ?? ( ?? ) = 16 [ ( s e c
- 1
? ?? )
2
+ ( c os ec
- 1
?? )
2
].
We can express ?? ( ?? ) as:
?? ( ?? ) = 16 [ ( s ec
- 1
? ?? + c os ec
- 1
?? )
2
- 2 ( s ec
- 1
? ?? ) (
?? 2
- s ec
- 1
? ?? ) ]
This simplifies to:
?? ( ?? ) = 16 [
?? 2
4
- ?? s ec
- 1
? ?? + 2 ( s ec
- 1
? ?? )
2
] , ? where s ec
- 1
? ?? ? [ 0 , ?? ] - {
?? 2
}
Further simplification gives:
?? ( ?? ) = 16 [ 2 ( s ec
- 1
? ?? -
?? 4
)
2
+
?? 2
4
-
?? 2
8
]
For the maximum value when s ec
- 1
? ?? = ?? :
m ax = 16 [ 2 ?? 2
- ?? 2
+
?? 2
4
] = 20 ?? 2
For the minimum value when s ec
- 1
? ?? =
?? 4
:
min = 16 [
2 × ?? 2
16
-
?? 2
4
+
?? 2
4
] = 2 ?? 2
Therefore, the sum of the maximum and minimum values is:
Sum = 22 ?? 2
Question5: If
?? ?? = ?? =
?? ?? ?? , then ?? ?? ?? - ?? ? (
????
????
?? ?? ?? ? ?? +
?? ????
?? ???? ? ?? ) is equal to:
JEE Main 2025 (Online) 23rd January Morning Shift
Options:
A. ?? + tan
- 1
?
5
12
B. ?? - tan
- 1
?
4
3
C. ?? + tan
- 1
?
4
5
D. ?? - tan
- 1
?
5
12
Ans: D
Solution:
?? 2
= ?? =
3 ?? 4
c os
- 1
? (
12
13
c os ? ?? +
5
12
sin ? ?? )
c os
- 1
? ( c o s ? ?? c os ? ?? + s i n ? ?? s i n ? ?? )
c os
- 1
? ( c o s ? ( ?? - ?? ) )
? ?? - ?? because ?? - ?? ? ( -
?? 2
,
?? 2
)
? ?? - tan
- 1
?
5
12
Question6: If ?? > ?? > ?? > ?? , then the expression ?? ?? ?? - ?? ? { ?? +
( ?? + ?? ?? )
( ?? - ?? )
} + ?? ?? ?? - ?? ? { ?? +
( ?? + ?? ?? )
( ?? - ?? )
} +
?? ?? ?? - ?? ? { ?? +
( ?? + ?? ?? )
( ?? - ?? )
} is equal to:
JEE Main 2025 (Online) 24th January Evening Shift
Options:
A. 3 ??
B.
?? 2
- ( ?? + ?? + ?? )
C. ??
D. 0
Ans: C
Solution:
? ? c ot
- 1
? (
???? + 1
?? - ?? ) + c ot
- 1
? (
???? + 1
?? - ?? ) + c ot
- 1
? (
???? + 1
?? - ?? )
? ? tan
- 1
? (
?? - ?? 1 + ????
) + tan
- 1
? (
?? - ?? 1 + ????
) + ?? + tan
- 1
? (
?? - ?? 1 + ????
)
? ? ( tan
- 1
? ?? - tan
- 1
? ?? ) + ( tan
- 1
? ?? - tan
- 1
? ?? ) + ( ?? + tan
- 1
? ?? - tan
- 1
? ?? )
? ? ??
Question7: ?? ?? ?? ? ( ?? ???? - ?? ?
?? ?? + ?? ???? - ?? ?
?? ????
+ ?? ???? - ?? ?
????
????
) is equal to:
JEE Main 2025 (Online) 28th January Morning Shift
Options:
A.
33
65
B. 1
C.
32
65
D. 0
Ans: D
Solution:
c os ? ( sin
- 1
?
3
5
+ sin
- 1
?
5
13
+ sin
- 1
?
33
65
)
c os ? ( tan
- 1
?
3
4
+ tan
- 1
?
5
12
+ tan
- 1
?
33
56
)
Page 5
JEE Main Previous Year Questions
(2025): Inverse Trigonometric Functions
Question1: If for some ?? , ?? ; ?? = ?? , ?? + ?? = ?? and ?? ?? ?? ?? ? ( ?????? - ?? ? ?? ) + ? ?? ?? ?? ?? ?? ?? ( ?? ?? ?? - ?? ? ?? ) = ???? ,
then ?? ?? + ?? is _ _ _ _
JEE Main 2025 (Online) 24th January Morning Shift
Ans: 14
Solution:
If ( tan ? ( tan
- 1
? ( ?? ) ) + 1 ( c ot ? ( c ot
- 1
? ?? ) )
2
= 36
?? 2
+ ?? 2
= 34
???? = 15
?? = 3 , ?? = 5
? ?? 2
+ ?? = 9 + 5 = 14
Question2: Let ?? = { ?? : ?? ?? ?? - ?? ? ?? = ?? + ?? ???? - ?? ? ?? + ?? ???? - ?? ? [ ?? ?? + ?? ] }. Then ?
?? ? ?? ? ( ?? ?? - ?? )
?? is equal
to _ _ _ _ .
JEE Main 2025 (Online) 29th January Morning Shift
Solution:
c os
- 1
? ?? = ?? + sin
- 1
? ?? + sin
- 1
? ( 2 ?? + 1 )
2 c os
- 1
? ?? - sin
- 1
? ( 2 ?? + 1 ) =
3 ?? 2
2 ?? - ?? =
3 ?? 2
where c os
- 1
? ?? = ?? , sin
- 1
? ( 2 ?? + 1 ) = ??
2 ?? =
3 ?? 2
+ ??
c os ? 2 ?? = s i n ? ??
2 c os
2
? ?? - 1 = sin ? ??
2 ?? 2
- 1 = 2 ?? + 1
?? 2
- ?? - 1 = 0
? ?? =
1 - v 5
2
, { ?? =
1 + v 5
2
rejected }
? 4 ?? 2
- 4 ?? = 4
( 2 ?? - 1 )
2
= 5
Question3: If ?? = ?? ?? ?? ? (
?? ?? + ?? ?? ?? - ?? ?
?? ?? ) , then ( ?? - ?? )
?? + ?? ?? ?? is equal to
JEE Main 2025 (Online) 2nd April Evening Shift
Ans: 3
Solution:
?? = c os ? (
?? 3
+ c os
- 1
?
?? 2
)
= c os ? (
?? 3
) c os ? ( c os
- 1
? (
?? 2
) ) - sin ? (
?? 3 .
) sin ? ( c o s
- 1
? (
?? 2
) )
=
1
2
·
?? 2
-
v 3
2
·
v
1 -
?? 2
4
? 4 ?? = ?? - v 3
v
4 - ?? 2
? ( 4 ?? - ?? )
2
= 3 ( 4 - ?? 2
)
? 16 ?? 2
+ ?? 2
- 8 ???? = 12 - 3 ?? 2
?? 2
+ 4 ?? 2
- 2 ???? = 3
( ?? - ?? )
2
+ 3 ?? 2
= 3
Question4: Using the principal values of the inverse trigonometric functions, the sum of the
maximum and the minimum values of ???? ( ( ?? ?? ?? - ?? ? ?? )
?? + ( ?? ?? ?? ?? ?? - ?? ?? )
?? ) is :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 24 ?? 2
B. 18 ?? 2
C. 22 ?? 2
D. 31 ?? 2
Ans: C
Solution:
Let ?? ( ?? ) = 16 [ ( s e c
- 1
? ?? )
2
+ ( c os ec
- 1
?? )
2
].
We can express ?? ( ?? ) as:
?? ( ?? ) = 16 [ ( s ec
- 1
? ?? + c os ec
- 1
?? )
2
- 2 ( s ec
- 1
? ?? ) (
?? 2
- s ec
- 1
? ?? ) ]
This simplifies to:
?? ( ?? ) = 16 [
?? 2
4
- ?? s ec
- 1
? ?? + 2 ( s ec
- 1
? ?? )
2
] , ? where s ec
- 1
? ?? ? [ 0 , ?? ] - {
?? 2
}
Further simplification gives:
?? ( ?? ) = 16 [ 2 ( s ec
- 1
? ?? -
?? 4
)
2
+
?? 2
4
-
?? 2
8
]
For the maximum value when s ec
- 1
? ?? = ?? :
m ax = 16 [ 2 ?? 2
- ?? 2
+
?? 2
4
] = 20 ?? 2
For the minimum value when s ec
- 1
? ?? =
?? 4
:
min = 16 [
2 × ?? 2
16
-
?? 2
4
+
?? 2
4
] = 2 ?? 2
Therefore, the sum of the maximum and minimum values is:
Sum = 22 ?? 2
Question5: If
?? ?? = ?? =
?? ?? ?? , then ?? ?? ?? - ?? ? (
????
????
?? ?? ?? ? ?? +
?? ????
?? ???? ? ?? ) is equal to:
JEE Main 2025 (Online) 23rd January Morning Shift
Options:
A. ?? + tan
- 1
?
5
12
B. ?? - tan
- 1
?
4
3
C. ?? + tan
- 1
?
4
5
D. ?? - tan
- 1
?
5
12
Ans: D
Solution:
?? 2
= ?? =
3 ?? 4
c os
- 1
? (
12
13
c os ? ?? +
5
12
sin ? ?? )
c os
- 1
? ( c o s ? ?? c os ? ?? + s i n ? ?? s i n ? ?? )
c os
- 1
? ( c o s ? ( ?? - ?? ) )
? ?? - ?? because ?? - ?? ? ( -
?? 2
,
?? 2
)
? ?? - tan
- 1
?
5
12
Question6: If ?? > ?? > ?? > ?? , then the expression ?? ?? ?? - ?? ? { ?? +
( ?? + ?? ?? )
( ?? - ?? )
} + ?? ?? ?? - ?? ? { ?? +
( ?? + ?? ?? )
( ?? - ?? )
} +
?? ?? ?? - ?? ? { ?? +
( ?? + ?? ?? )
( ?? - ?? )
} is equal to:
JEE Main 2025 (Online) 24th January Evening Shift
Options:
A. 3 ??
B.
?? 2
- ( ?? + ?? + ?? )
C. ??
D. 0
Ans: C
Solution:
? ? c ot
- 1
? (
???? + 1
?? - ?? ) + c ot
- 1
? (
???? + 1
?? - ?? ) + c ot
- 1
? (
???? + 1
?? - ?? )
? ? tan
- 1
? (
?? - ?? 1 + ????
) + tan
- 1
? (
?? - ?? 1 + ????
) + ?? + tan
- 1
? (
?? - ?? 1 + ????
)
? ? ( tan
- 1
? ?? - tan
- 1
? ?? ) + ( tan
- 1
? ?? - tan
- 1
? ?? ) + ( ?? + tan
- 1
? ?? - tan
- 1
? ?? )
? ? ??
Question7: ?? ?? ?? ? ( ?? ???? - ?? ?
?? ?? + ?? ???? - ?? ?
?? ????
+ ?? ???? - ?? ?
????
????
) is equal to:
JEE Main 2025 (Online) 28th January Morning Shift
Options:
A.
33
65
B. 1
C.
32
65
D. 0
Ans: D
Solution:
c os ? ( sin
- 1
?
3
5
+ sin
- 1
?
5
13
+ sin
- 1
?
33
65
)
c os ? ( tan
- 1
?
3
4
+ tan
- 1
?
5
12
+ tan
- 1
?
33
56
)
c os ? ( tan
- 1
? (
3
4
+
5
12
1 +
3
4
·
5
12
) + tan
- 1
?
33
56
)
c os ? ( tan
- 1
?
56
33
+ c ot
- 1
?
56
33
)
c os ? (
?? 2
) = 0
Question8: Let [ ?? ] denote the greatest integer less than or equal to ?? . Then the domain of
?? ( ?? ) = ?? ?? ?? - ?? ? ( ?? [ ?? ] + ?? ) is:
JEE Main 2025 (Online) 28th January Evening Shift
Options:
A. ( - 8 , 8 )
B. ( - 8 , 8 ) - { 0 }
C. ( - 8 , - 1 ] ? [ 0 , 8 )
D. ( - 8 , - 1 ] ? [ 1 , 8 )
Ans: A
Solution:
2 [ x ] + 1 = - 1 or 2 [ x ] + 1 = 1
? [ x ] = - 1 ? [ x ] = 0
? x ? ( - 8 , 0 ) ? x ? [ 0 , 8 )
? x ? ( - 8 , 8 )
Question9: Considering the principal values of the inverse trigonometric functions,
?? ???? - ?? ? (
v ?? ?? ?? +
?? ?? v ?? - ?? ?? ) , -
?? ?? < ?? <
?? v ?? , is equal to
JEE Main 2025 (Online) 4th April Morning Shift
Options:
A.
- 5 ?? 6
- s i n
- 1
? ??
B.
5 ?? 6
- sin
- 1
? ??
Read More
|
177 videos|596 docs|160 tests
|
FAQs on JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions - Mathematics (Maths) for JEE Main & Advanced
| 1. What are inverse trigonometric functions and why are they important in mathematics? |  |
Ans. Inverse trigonometric functions are the inverse functions of the basic trigonometric functions. They are important because they allow us to find angles when the values of the trigonometric functions are known. The six primary inverse trigonometric functions are arcsin, arccos, arctan, arccot, arcsec, and arccsc. These functions are used in various fields including physics, engineering, and computer science, particularly in solving triangles and modeling periodic phenomena.
| 2. How do you determine the range of inverse trigonometric functions? | |
Ans. The range of inverse trigonometric functions is determined by the output values that these functions can yield. For instance, the range of arcsin(x) is from -π/2 to π/2, the range of arccos(x) is from 0 to π, and the range of arctan(x) is from -π/2 to π/2. Knowing the range is crucial for solving equations involving inverse trigonometric functions and understanding their graphical representations.
| 3. What are the principal values of the inverse trigonometric functions? | |
Ans. The principal values of the inverse trigonometric functions refer to the specific values within their defined ranges that are typically used. For example, the principal value of arcsin(x) is within [-π/2, π/2], for arccos(x) it is [0, π], and for arctan(x) it is (-π/2, π/2). These principal values help in determining unique solutions for equations involving these functions.
| 4. Can you explain the relationship between inverse trigonometric functions and right triangles? | |
Ans. Inverse trigonometric functions have a direct relationship with right triangles. For example, if you know the lengths of two sides of a right triangle, you can use inverse trigonometric functions to find the angles. For instance, if you have the opposite side and the adjacent side, you can find the angle using arctan(opposite/adjacent). This application is essential in geometry, physics, and engineering for solving problems related to angles and distances.
| 5. How do you solve equations involving inverse trigonometric functions? | |
Ans. To solve equations involving inverse trigonometric functions, you can isolate the inverse function and then apply the corresponding trigonometric function to both sides of the equation. For example, if you have arcsin(x) = a, you can find x by taking sin(a). It’s important to consider the range and domain of the functions involved to ensure that the solutions are valid. Additionally, graphical methods can also be employed for visual understanding of the solutions.
About this Document
Oct 03, 2025
Last updated
Related Exams
Document Description: JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions for JEE 2025 is part of Mathematics (Maths) for JEE Main & Advanced preparation.
The notes and questions for JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions have been prepared according to the JEE exam syllabus. Information about JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions covers topics
like and JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions Example, for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions.
Introduction of JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions in English is available as part of our Mathematics (Maths) for JEE Main & Advanced
for JEE & JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions in Hindi for Mathematics (Maths) for JEE Main & Advanced course.
Download more important topics related with notes, lectures and mock test series for JEE
Exam by signing up for free. JEE: JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Description
Full syllabus notes, lecture & questions for JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced - JEE | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) for JEE Main & Advanced | Best notes, free PDF download
Information about JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions
In this doc you can find the meaning of JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions defined & explained in the simplest way possible. Besides explaining types of
JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions theory, EduRev gives you an ample number of questions to practice JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions tests, examples and also practice JEE
tests
|
177 videos|596 docs|160 tests
|
Download as PDF
Related Searches
Sample Paper
,practice quizzes
,JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
,study material
,shortcuts and tricks
,Viva Questions
,Free
,ppt
,Summary
,MCQs
,Semester Notes
,JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
,JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
,Extra Questions
,Previous Year Questions with Solutions
,video lectures
,Exam
,past year papers
,Objective type Questions
,Important questions
,mock tests for examination
;
Additional Information about JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions for JEE Preparation
JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions Free PDF Download
The JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions is an invaluable resource that delves deep into the core of the JEE exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions now and kickstart your journey towards success in the JEE exam.
Importance of JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions
The importance of JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions cannot be overstated, especially for JEE aspirants.
This document holds the key to success in the JEE exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions Notes
JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions Notes on EduRev are your ultimate resource for success.
JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions JEE
The "JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions JEE Questions" guide is a valuable resource for all aspiring students preparing for the
JEE exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions on the App
Students of JEE can study JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of JEE Main Previous Year Questions (2025): Inverse Trigonometric Functions is prepared as per the latest JEE syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup on EduRev and stay on top of your study goals
10M+ students crushing their study goals daily