Page 1
JEE Mains Previous Year Questions
(2021-2024): Inverse Trigonometric
Functions
2024
Q1 - 2024 (27 Jan Shift 2)
Considering only the principal values of inverse trigonometric functions, the number of
positive real values of ?? satisfying t a n
- 1
? ( ?? ) + t a n
- 1
? ( 2 ?? ) =
?? 4
is :
(1) More than 2
(2) 1
(3) 2
(4) 0
Q2 - 2024 (29 Jan Shift 2)
Let x =
m
n
( m , n are co-prime natural numbers) be a solution of the equation
c o s ? ( 2 sin
- 1
? ?? ) =
1
9
and let ?? , ?? ( ?? > ?? ) be the roots of the equation mx
2
- nx - m + n = 0.
Then the point ( ?? , ?? ) lies on the line
(1) 3 ?? + 2 ?? = 2
(2) 5 ?? - 8 ?? = - 9
(3) 3 ?? - 2 ?? = - 2
(4) 5 ?? + 8 ?? = 9
Q3 - 2024 (31 Jan Shift 1)
For ?? , ?? , ?? ? 0. If sin
- 1
? ?? + sin
- 1
? ?? + sin
- 1
? ?? = ?? and ( ?? + ?? + ?? ) ( ?? - ?? + ?? ) = 3 ???? , then
?? equal to
(1)
v 3
2
(2)
1
v 2
Page 2
JEE Mains Previous Year Questions
(2021-2024): Inverse Trigonometric
Functions
2024
Q1 - 2024 (27 Jan Shift 2)
Considering only the principal values of inverse trigonometric functions, the number of
positive real values of ?? satisfying t a n
- 1
? ( ?? ) + t a n
- 1
? ( 2 ?? ) =
?? 4
is :
(1) More than 2
(2) 1
(3) 2
(4) 0
Q2 - 2024 (29 Jan Shift 2)
Let x =
m
n
( m , n are co-prime natural numbers) be a solution of the equation
c o s ? ( 2 sin
- 1
? ?? ) =
1
9
and let ?? , ?? ( ?? > ?? ) be the roots of the equation mx
2
- nx - m + n = 0.
Then the point ( ?? , ?? ) lies on the line
(1) 3 ?? + 2 ?? = 2
(2) 5 ?? - 8 ?? = - 9
(3) 3 ?? - 2 ?? = - 2
(4) 5 ?? + 8 ?? = 9
Q3 - 2024 (31 Jan Shift 1)
For ?? , ?? , ?? ? 0. If sin
- 1
? ?? + sin
- 1
? ?? + sin
- 1
? ?? = ?? and ( ?? + ?? + ?? ) ( ?? - ?? + ?? ) = 3 ???? , then
?? equal to
(1)
v 3
2
(2)
1
v 2
(3)
v 3 - 1
2 v 2
(4) v 3
Q4 - 2024 (31 Jan Shift 2)
If ?? = sin
- 1
? ( s i n ? ( 5 ) ) and ?? = c o s
- 1
? ( c o s ? ( 5 ) ), then ?? 2
+ ?? 2
is equal to
(1) 4 ?? 2
+ 25
(2) 8 ?? 2
- 40 ?? + 50
(3) 4 ?? 2
- 20 ?? + 50
(4) 25
Answer Key
Q1 (2)
Q2 (4)
Q3 (1)
Q4 (2)
Solutions
Q1
t a n
- 1
? ?? + t a n
- 1
? 2 ?? =
?? 4
; ?? > 0
? t a n
- 1
? 2x =
?? 4
- t a n
- 1
? x
Taking tan both sides
? 2 ?? =
1 - ?? 1 + ??
? 2 ?? 2
+ 3 ?? - 1 = 0
?? =
- 3 ± v 9 + 8
8
=
- 3 ± v 17
8
Only possible ?? =
- 3 + v 17
8
Q2
Page 3
JEE Mains Previous Year Questions
(2021-2024): Inverse Trigonometric
Functions
2024
Q1 - 2024 (27 Jan Shift 2)
Considering only the principal values of inverse trigonometric functions, the number of
positive real values of ?? satisfying t a n
- 1
? ( ?? ) + t a n
- 1
? ( 2 ?? ) =
?? 4
is :
(1) More than 2
(2) 1
(3) 2
(4) 0
Q2 - 2024 (29 Jan Shift 2)
Let x =
m
n
( m , n are co-prime natural numbers) be a solution of the equation
c o s ? ( 2 sin
- 1
? ?? ) =
1
9
and let ?? , ?? ( ?? > ?? ) be the roots of the equation mx
2
- nx - m + n = 0.
Then the point ( ?? , ?? ) lies on the line
(1) 3 ?? + 2 ?? = 2
(2) 5 ?? - 8 ?? = - 9
(3) 3 ?? - 2 ?? = - 2
(4) 5 ?? + 8 ?? = 9
Q3 - 2024 (31 Jan Shift 1)
For ?? , ?? , ?? ? 0. If sin
- 1
? ?? + sin
- 1
? ?? + sin
- 1
? ?? = ?? and ( ?? + ?? + ?? ) ( ?? - ?? + ?? ) = 3 ???? , then
?? equal to
(1)
v 3
2
(2)
1
v 2
(3)
v 3 - 1
2 v 2
(4) v 3
Q4 - 2024 (31 Jan Shift 2)
If ?? = sin
- 1
? ( s i n ? ( 5 ) ) and ?? = c o s
- 1
? ( c o s ? ( 5 ) ), then ?? 2
+ ?? 2
is equal to
(1) 4 ?? 2
+ 25
(2) 8 ?? 2
- 40 ?? + 50
(3) 4 ?? 2
- 20 ?? + 50
(4) 25
Answer Key
Q1 (2)
Q2 (4)
Q3 (1)
Q4 (2)
Solutions
Q1
t a n
- 1
? ?? + t a n
- 1
? 2 ?? =
?? 4
; ?? > 0
? t a n
- 1
? 2x =
?? 4
- t a n
- 1
? x
Taking tan both sides
? 2 ?? =
1 - ?? 1 + ??
? 2 ?? 2
+ 3 ?? - 1 = 0
?? =
- 3 ± v 9 + 8
8
=
- 3 ± v 17
8
Only possible ?? =
- 3 + v 17
8
Q2
Assume sin
- 1
? ?? = ??
c o s ? ( 2 ?? ) =
1
9
sin ? ?? = ±
2
3
as m and n are co-prime natural numbers,
?? =
2
3
i.e. ?? = 2 , ?? = 3
So, the quadratic equation becomes 2 ?? 2
- 3 ?? + 1 = 0 whose roots are ?? = 1 , ?? =
1
2
( 1 ,
1
2
)
lies on 5 ?? + 8 ?? = 9
Q3
Let sin
- 1
? ?? = A , sin
- 1
? ?? = B , sin
- 1
? ?? = C
A + B + C = ??
( ?? + ?? )
2
- ?? 2
= 3 ????
?? 2
+ ?? 2
- ?? 2
= ????
?? 2
+ ?? 2
- ?? 2
2 ????
=
1
2
? c o s ? C =
1
2
sin ? C = ??
c o s ? C = v 1 - ?? 2
=
1
2
?? =
v 3
2
Q4
?? = sin
- 1
? ( sin ? 5 ) = 5 - 2 ??
and ?? = c o s
- 1
? ( c o s ? 5 ) = 2 ?? - 5
? ?? 2
+ ?? 2
= ( 5 - 2 ?? )
2
+ ( 2 ?? - 5 )
2
= 8 ?? 2
- 40 ?? + 50
Page 4
JEE Mains Previous Year Questions
(2021-2024): Inverse Trigonometric
Functions
2024
Q1 - 2024 (27 Jan Shift 2)
Considering only the principal values of inverse trigonometric functions, the number of
positive real values of ?? satisfying t a n
- 1
? ( ?? ) + t a n
- 1
? ( 2 ?? ) =
?? 4
is :
(1) More than 2
(2) 1
(3) 2
(4) 0
Q2 - 2024 (29 Jan Shift 2)
Let x =
m
n
( m , n are co-prime natural numbers) be a solution of the equation
c o s ? ( 2 sin
- 1
? ?? ) =
1
9
and let ?? , ?? ( ?? > ?? ) be the roots of the equation mx
2
- nx - m + n = 0.
Then the point ( ?? , ?? ) lies on the line
(1) 3 ?? + 2 ?? = 2
(2) 5 ?? - 8 ?? = - 9
(3) 3 ?? - 2 ?? = - 2
(4) 5 ?? + 8 ?? = 9
Q3 - 2024 (31 Jan Shift 1)
For ?? , ?? , ?? ? 0. If sin
- 1
? ?? + sin
- 1
? ?? + sin
- 1
? ?? = ?? and ( ?? + ?? + ?? ) ( ?? - ?? + ?? ) = 3 ???? , then
?? equal to
(1)
v 3
2
(2)
1
v 2
(3)
v 3 - 1
2 v 2
(4) v 3
Q4 - 2024 (31 Jan Shift 2)
If ?? = sin
- 1
? ( s i n ? ( 5 ) ) and ?? = c o s
- 1
? ( c o s ? ( 5 ) ), then ?? 2
+ ?? 2
is equal to
(1) 4 ?? 2
+ 25
(2) 8 ?? 2
- 40 ?? + 50
(3) 4 ?? 2
- 20 ?? + 50
(4) 25
Answer Key
Q1 (2)
Q2 (4)
Q3 (1)
Q4 (2)
Solutions
Q1
t a n
- 1
? ?? + t a n
- 1
? 2 ?? =
?? 4
; ?? > 0
? t a n
- 1
? 2x =
?? 4
- t a n
- 1
? x
Taking tan both sides
? 2 ?? =
1 - ?? 1 + ??
? 2 ?? 2
+ 3 ?? - 1 = 0
?? =
- 3 ± v 9 + 8
8
=
- 3 ± v 17
8
Only possible ?? =
- 3 + v 17
8
Q2
Assume sin
- 1
? ?? = ??
c o s ? ( 2 ?? ) =
1
9
sin ? ?? = ±
2
3
as m and n are co-prime natural numbers,
?? =
2
3
i.e. ?? = 2 , ?? = 3
So, the quadratic equation becomes 2 ?? 2
- 3 ?? + 1 = 0 whose roots are ?? = 1 , ?? =
1
2
( 1 ,
1
2
)
lies on 5 ?? + 8 ?? = 9
Q3
Let sin
- 1
? ?? = A , sin
- 1
? ?? = B , sin
- 1
? ?? = C
A + B + C = ??
( ?? + ?? )
2
- ?? 2
= 3 ????
?? 2
+ ?? 2
- ?? 2
= ????
?? 2
+ ?? 2
- ?? 2
2 ????
=
1
2
? c o s ? C =
1
2
sin ? C = ??
c o s ? C = v 1 - ?? 2
=
1
2
?? =
v 3
2
Q4
?? = sin
- 1
? ( sin ? 5 ) = 5 - 2 ??
and ?? = c o s
- 1
? ( c o s ? 5 ) = 2 ?? - 5
? ?? 2
+ ?? 2
= ( 5 - 2 ?? )
2
+ ( 2 ?? - 5 )
2
= 8 ?? 2
- 40 ?? + 50
Numerical 2023
Question:1
JEE Main 2023 (Online) 13th April Evening Shift
Question:2
JEE Main 2023 (Online) 13th April Morning Shift
Question:3
JEE Main 2023 (Online) 10th April Evening Shift
Question:4
JEE Main 2023 (Online) 25th January Morning Shift
Numerical Answer Key
1. Ans. (2)
2. Ans. (4)
3. Ans. (24)
4. Ans. (2)
Page 5
JEE Mains Previous Year Questions
(2021-2024): Inverse Trigonometric
Functions
2024
Q1 - 2024 (27 Jan Shift 2)
Considering only the principal values of inverse trigonometric functions, the number of
positive real values of ?? satisfying t a n
- 1
? ( ?? ) + t a n
- 1
? ( 2 ?? ) =
?? 4
is :
(1) More than 2
(2) 1
(3) 2
(4) 0
Q2 - 2024 (29 Jan Shift 2)
Let x =
m
n
( m , n are co-prime natural numbers) be a solution of the equation
c o s ? ( 2 sin
- 1
? ?? ) =
1
9
and let ?? , ?? ( ?? > ?? ) be the roots of the equation mx
2
- nx - m + n = 0.
Then the point ( ?? , ?? ) lies on the line
(1) 3 ?? + 2 ?? = 2
(2) 5 ?? - 8 ?? = - 9
(3) 3 ?? - 2 ?? = - 2
(4) 5 ?? + 8 ?? = 9
Q3 - 2024 (31 Jan Shift 1)
For ?? , ?? , ?? ? 0. If sin
- 1
? ?? + sin
- 1
? ?? + sin
- 1
? ?? = ?? and ( ?? + ?? + ?? ) ( ?? - ?? + ?? ) = 3 ???? , then
?? equal to
(1)
v 3
2
(2)
1
v 2
(3)
v 3 - 1
2 v 2
(4) v 3
Q4 - 2024 (31 Jan Shift 2)
If ?? = sin
- 1
? ( s i n ? ( 5 ) ) and ?? = c o s
- 1
? ( c o s ? ( 5 ) ), then ?? 2
+ ?? 2
is equal to
(1) 4 ?? 2
+ 25
(2) 8 ?? 2
- 40 ?? + 50
(3) 4 ?? 2
- 20 ?? + 50
(4) 25
Answer Key
Q1 (2)
Q2 (4)
Q3 (1)
Q4 (2)
Solutions
Q1
t a n
- 1
? ?? + t a n
- 1
? 2 ?? =
?? 4
; ?? > 0
? t a n
- 1
? 2x =
?? 4
- t a n
- 1
? x
Taking tan both sides
? 2 ?? =
1 - ?? 1 + ??
? 2 ?? 2
+ 3 ?? - 1 = 0
?? =
- 3 ± v 9 + 8
8
=
- 3 ± v 17
8
Only possible ?? =
- 3 + v 17
8
Q2
Assume sin
- 1
? ?? = ??
c o s ? ( 2 ?? ) =
1
9
sin ? ?? = ±
2
3
as m and n are co-prime natural numbers,
?? =
2
3
i.e. ?? = 2 , ?? = 3
So, the quadratic equation becomes 2 ?? 2
- 3 ?? + 1 = 0 whose roots are ?? = 1 , ?? =
1
2
( 1 ,
1
2
)
lies on 5 ?? + 8 ?? = 9
Q3
Let sin
- 1
? ?? = A , sin
- 1
? ?? = B , sin
- 1
? ?? = C
A + B + C = ??
( ?? + ?? )
2
- ?? 2
= 3 ????
?? 2
+ ?? 2
- ?? 2
= ????
?? 2
+ ?? 2
- ?? 2
2 ????
=
1
2
? c o s ? C =
1
2
sin ? C = ??
c o s ? C = v 1 - ?? 2
=
1
2
?? =
v 3
2
Q4
?? = sin
- 1
? ( sin ? 5 ) = 5 - 2 ??
and ?? = c o s
- 1
? ( c o s ? 5 ) = 2 ?? - 5
? ?? 2
+ ?? 2
= ( 5 - 2 ?? )
2
+ ( 2 ?? - 5 )
2
= 8 ?? 2
- 40 ?? + 50
Numerical 2023
Question:1
JEE Main 2023 (Online) 13th April Evening Shift
Question:2
JEE Main 2023 (Online) 13th April Morning Shift
Question:3
JEE Main 2023 (Online) 10th April Evening Shift
Question:4
JEE Main 2023 (Online) 25th January Morning Shift
Numerical Answer Key
1. Ans. (2)
2. Ans. (4)
3. Ans. (24)
4. Ans. (2)
Numerical Explanation
Ans. 1
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