Inverse Trigonometric Functions: JEE Mains Previous Year Questions (2021-2024)

# Inverse Trigonometric Functions: JEE Mains Previous Year Questions (2021-2024) | Mathematics (Maths) for JEE Main & Advanced PDF Download

 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
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
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
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

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
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

1. Ans. (2)
2. Ans. (4)
3. Ans. (24)
4. Ans. (2)

Numerical Explanation
Ans. 1



## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests

## FAQs on Inverse Trigonometric Functions: JEE Mains Previous Year Questions (2021-2024) - Mathematics (Maths) for JEE Main & Advanced

 1. What are the basic properties of inverse trigonometric functions?
Ans. The basic properties of inverse trigonometric functions include the principal value branches, range, domain, and periodicity of functions such as sin$^{-1}x$, cos$^{-1}x$, and tan$^{-1}x$.
 2. How do we find the derivatives of inverse trigonometric functions?
Ans. To find the derivatives of inverse trigonometric functions, we can use the chain rule and the known derivatives of trigonometric functions. For example, the derivative of sin$^{-1}x$ is $\frac{1}{\sqrt{1-x^2}}$.
 3. What are some common identities involving inverse trigonometric functions?
Ans. Some common identities involving inverse trigonometric functions include sin(sin$^{-1}x) = x$, cos(cos$^{-1}x) = x$, and tan(tan$^{-1}x) = x$. These identities help simplify expressions involving inverse trigonometric functions.
 4. How do we solve equations involving inverse trigonometric functions?
Ans. To solve equations involving inverse trigonometric functions, we can apply the properties and identities of these functions. We can also use trigonometric identities to simplify the expressions and find the solutions.
 5. Can inverse trigonometric functions be used to find angles in a triangle?
Ans. Yes, inverse trigonometric functions can be used to find angles in a triangle. By using the properties of inverse trigonometric functions, we can determine unknown angles in a triangle based on the given sides or angles.

## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests

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