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JEE Main Previous Year Questions (2025): Trigonometric Ratios, Functions & Equations

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JEE Main Previous Year Questions (2025): 
Trigonometric Ratios, Functions & 
Equations 
Q1: The value of ( ?? ???? ? ????
°
) ( ?? ?? ?? ? ????
°
?? ?? ?? ? ????
°
- ?? ) is 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 0 
B. 2 / 3 
C. 1 
D. 3 / 2 
Ans: C 
Solution: 
sin ? 70
°
( c o t ? 10
°
c ot ? 70
°
- 1 ) 
?
c os ? ( 80
°
)
sin ? 10
= 1 
 
Q2: Let the range of the function 
?? ( ?? ) = ?? + ???? ?? ?? ?? ? ?? · ?? ?? ?? ? (
?? ?? - ?? ) · ?? ?? ?? ? (
?? ?? + ?? ) · ?? ???? ? ?? ?? · ?? ?? ?? ? ?? ?? , ?? ? ?? be [ ?? , ?? ]. Then the 
distance of the point ( ?? , ?? ) from the line ?? ?? + ?? ?? + ???? = ?? is : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. 11 
B. 10 
C. 8 
D. 9 
Ans: A 
Solution: 
?? ( ?? ) ? = 6 + 16 (
1
4
c os ? 3 ?? ) s i n ? 3 ?? · c os ? 6 ?? ? = 6 + 4c os ? 3 ?? sin ? 3 ?? c os ? 6 ?? ? = 6 + sin ? 12 ?? 
Range of ?? ( ?? ) is [5,7] 
( ?? , ?? ) = ( 5 , 7 ) 
distance = |
15 + 28 + 12
5
| = 11 
Page 2


JEE Main Previous Year Questions (2025): 
Trigonometric Ratios, Functions & 
Equations 
Q1: The value of ( ?? ???? ? ????
°
) ( ?? ?? ?? ? ????
°
?? ?? ?? ? ????
°
- ?? ) is 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 0 
B. 2 / 3 
C. 1 
D. 3 / 2 
Ans: C 
Solution: 
sin ? 70
°
( c o t ? 10
°
c ot ? 70
°
- 1 ) 
?
c os ? ( 80
°
)
sin ? 10
= 1 
 
Q2: Let the range of the function 
?? ( ?? ) = ?? + ???? ?? ?? ?? ? ?? · ?? ?? ?? ? (
?? ?? - ?? ) · ?? ?? ?? ? (
?? ?? + ?? ) · ?? ???? ? ?? ?? · ?? ?? ?? ? ?? ?? , ?? ? ?? be [ ?? , ?? ]. Then the 
distance of the point ( ?? , ?? ) from the line ?? ?? + ?? ?? + ???? = ?? is : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. 11 
B. 10 
C. 8 
D. 9 
Ans: A 
Solution: 
?? ( ?? ) ? = 6 + 16 (
1
4
c os ? 3 ?? ) s i n ? 3 ?? · c os ? 6 ?? ? = 6 + 4c os ? 3 ?? sin ? 3 ?? c os ? 6 ?? ? = 6 + sin ? 12 ?? 
Range of ?? ( ?? ) is [5,7] 
( ?? , ?? ) = ( 5 , 7 ) 
distance = |
15 + 28 + 12
5
| = 11 
Q3: If ?
?? = ?? ????
? {
?? ?? ?? ?? ? (
?? ?? + ( ?? - ?? )
?? ?? ) ?? ?? ?? ? (
?? ?? +
????
?? )
} = ?? v ?? + ?? , ?? , ?? ? ?? , then ?? ?? + ?? ?? is equal to : 
JEE Main 2025 (Online) 28th January Evening Shift 
Options: 
A. 10 
B. 4 
C. 8 
D. 2 
 
Ans: C 
Solution: 
1
sin ?
?? 6
?
?? = 1
13
?
sin ? [ (
?? 4
+
r ?? 6
) - (
?? 4
) - ( r - 1 )
?? 6
]
sin ? (
?? 4
+ ( r - 1 )
?? 6
) sin ? (
?? 4
+
r ?? 6
)
 
1
sin ?
?? 6
?
r = 1
13
? ( c o t ? (
?? 4
+ ( r - 1 )
?? 6
) - c ot ? (
?? 4
+
r ?? 6
) ) 
= 2 v 3 - 2 = ?? v 3 + b 
So a
2
+ b
2
= 8 
 
Q4: If ?? ???? ? ?? + ?? ???? ?? ? ?? = ?? , ?? ? ( ?? ,
?? ?? ), then 
( ?? ?? ?? ????
? ?? + ?????? ????
? ?? ) + ?? ( ?? ?? ?? ????
? ?? + ?????? ????
? ?? + ?? ?? ?? ?? ? ?? + ?????? ?? ? ?? ) + ( ?? ?? ?? ?? ? ?? + ?????? ?? ? ?? ) is equal to: 
JEE Main 2025 (Online) 29th January Evening Shift 
Options: 
A. 3 
B. 4 
C. 2 
D. 1 
 
Ans: C 
Solution: 
sin ? ?? + sin
2
? ?? = 1 
? sin ? ?? = c os
2
? ?? ? tan ? ?? = c os ? ?? 
? Given expression 
= 2 c os
12
? ?? + 6 [ c os
10
? ?? + c os
8
? ?? ] + 2 c os
6
? ?? 
= 2 [ sin
6
? ?? + 3 sin
5
? ?? + 3 sin
4
? ?? + sin
3
? ?? ] 
Page 3


JEE Main Previous Year Questions (2025): 
Trigonometric Ratios, Functions & 
Equations 
Q1: The value of ( ?? ???? ? ????
°
) ( ?? ?? ?? ? ????
°
?? ?? ?? ? ????
°
- ?? ) is 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 0 
B. 2 / 3 
C. 1 
D. 3 / 2 
Ans: C 
Solution: 
sin ? 70
°
( c o t ? 10
°
c ot ? 70
°
- 1 ) 
?
c os ? ( 80
°
)
sin ? 10
= 1 
 
Q2: Let the range of the function 
?? ( ?? ) = ?? + ???? ?? ?? ?? ? ?? · ?? ?? ?? ? (
?? ?? - ?? ) · ?? ?? ?? ? (
?? ?? + ?? ) · ?? ???? ? ?? ?? · ?? ?? ?? ? ?? ?? , ?? ? ?? be [ ?? , ?? ]. Then the 
distance of the point ( ?? , ?? ) from the line ?? ?? + ?? ?? + ???? = ?? is : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. 11 
B. 10 
C. 8 
D. 9 
Ans: A 
Solution: 
?? ( ?? ) ? = 6 + 16 (
1
4
c os ? 3 ?? ) s i n ? 3 ?? · c os ? 6 ?? ? = 6 + 4c os ? 3 ?? sin ? 3 ?? c os ? 6 ?? ? = 6 + sin ? 12 ?? 
Range of ?? ( ?? ) is [5,7] 
( ?? , ?? ) = ( 5 , 7 ) 
distance = |
15 + 28 + 12
5
| = 11 
Q3: If ?
?? = ?? ????
? {
?? ?? ?? ?? ? (
?? ?? + ( ?? - ?? )
?? ?? ) ?? ?? ?? ? (
?? ?? +
????
?? )
} = ?? v ?? + ?? , ?? , ?? ? ?? , then ?? ?? + ?? ?? is equal to : 
JEE Main 2025 (Online) 28th January Evening Shift 
Options: 
A. 10 
B. 4 
C. 8 
D. 2 
 
Ans: C 
Solution: 
1
sin ?
?? 6
?
?? = 1
13
?
sin ? [ (
?? 4
+
r ?? 6
) - (
?? 4
) - ( r - 1 )
?? 6
]
sin ? (
?? 4
+ ( r - 1 )
?? 6
) sin ? (
?? 4
+
r ?? 6
)
 
1
sin ?
?? 6
?
r = 1
13
? ( c o t ? (
?? 4
+ ( r - 1 )
?? 6
) - c ot ? (
?? 4
+
r ?? 6
) ) 
= 2 v 3 - 2 = ?? v 3 + b 
So a
2
+ b
2
= 8 
 
Q4: If ?? ???? ? ?? + ?? ???? ?? ? ?? = ?? , ?? ? ( ?? ,
?? ?? ), then 
( ?? ?? ?? ????
? ?? + ?????? ????
? ?? ) + ?? ( ?? ?? ?? ????
? ?? + ?????? ????
? ?? + ?? ?? ?? ?? ? ?? + ?????? ?? ? ?? ) + ( ?? ?? ?? ?? ? ?? + ?????? ?? ? ?? ) is equal to: 
JEE Main 2025 (Online) 29th January Evening Shift 
Options: 
A. 3 
B. 4 
C. 2 
D. 1 
 
Ans: C 
Solution: 
sin ? ?? + sin
2
? ?? = 1 
? sin ? ?? = c os
2
? ?? ? tan ? ?? = c os ? ?? 
? Given expression 
= 2 c os
12
? ?? + 6 [ c os
10
? ?? + c os
8
? ?? ] + 2 c os
6
? ?? 
= 2 [ sin
6
? ?? + 3 sin
5
? ?? + 3 sin
4
? ?? + sin
3
? ?? ] 
= 2 sin
3
? ?? [ ( sin ? ?? + 1 )
3
] 
= 2 [ sin
2
? ?? + sin ? ?? ]
3
 
= 2 
 
Q5: If ???? ?? ???? ?? ? ?? + ???? ?? ?? ?? ?? ? ?? = ?? , then the value of 
???? ?? ?? ?? ?? ?? ?? ?? + ?? ??????
?? ? ?? ???? ??????
?? ? ?? is 
JEE Main 2025 (Online) 4th April Morning Shift 
Options: 
A. 
2
5
 
B. 
3
5
 
C. 
1
5
 
D. 
3
4
 
Ans: A 
Solution: 
? 10 sin
4
? ?? + 15 c os
4
? ?? = 6
? ? 10 sin
4
? ?? + 10 c o s
4
? ?? + 5 c os
4
? ?? = 6
? ? 10 [ ( sin
2
? ?? + c os
2
? ?? )
2
- 2 sin
2
? ?? c os
2
? ?? ] + 5 c os
4
? ?? = 6
? ? 10 - 20 ( 1 - c os
2
? ?? ) c os
2
? ?? + 5 c os
4
? ?? = 6
 L et c os
2
? ?? = ?? 10 - 20 ( ?? - ?? 2
) + 5 ?? 2
= 6
? ? 25 ?? 2
- 20 ?? + 4 = 0
? ( 5 ?? - 2 )
2
= 0 ? ?? =
2
5
? ? c os
2
? ?? =
2
5
? sin
2
? ?? =
3
5
? s ec
2
? ?? =
5
2
, c os ec
2
?? =
5
3
27 c os ec
6
?? + 8 s ec
6
? ?? 16 s ec
8
? ?? =
27 (
5
3
)
3
+ 8 (
5
2
)
3
16 (
5
2
)
4
? =
5
3
+ 5
3
5
4
=
2 . 5
3
5
4
=
2
5
 
 
 
 
Q6: If for ?? ? [ -
?? ?? , ?? ], the points ( ?? , ?? ) = ( ???????? ? ( ?? +
?? ?? ) , ???????? ? ( ?? +
?? ?? ) ) lie on ???? + ???? + ???? +
?? = ?? , then ?? ?? + ?? ?? + ?? ?? is equal to : 
Page 4


JEE Main Previous Year Questions (2025): 
Trigonometric Ratios, Functions & 
Equations 
Q1: The value of ( ?? ???? ? ????
°
) ( ?? ?? ?? ? ????
°
?? ?? ?? ? ????
°
- ?? ) is 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 0 
B. 2 / 3 
C. 1 
D. 3 / 2 
Ans: C 
Solution: 
sin ? 70
°
( c o t ? 10
°
c ot ? 70
°
- 1 ) 
?
c os ? ( 80
°
)
sin ? 10
= 1 
 
Q2: Let the range of the function 
?? ( ?? ) = ?? + ???? ?? ?? ?? ? ?? · ?? ?? ?? ? (
?? ?? - ?? ) · ?? ?? ?? ? (
?? ?? + ?? ) · ?? ???? ? ?? ?? · ?? ?? ?? ? ?? ?? , ?? ? ?? be [ ?? , ?? ]. Then the 
distance of the point ( ?? , ?? ) from the line ?? ?? + ?? ?? + ???? = ?? is : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. 11 
B. 10 
C. 8 
D. 9 
Ans: A 
Solution: 
?? ( ?? ) ? = 6 + 16 (
1
4
c os ? 3 ?? ) s i n ? 3 ?? · c os ? 6 ?? ? = 6 + 4c os ? 3 ?? sin ? 3 ?? c os ? 6 ?? ? = 6 + sin ? 12 ?? 
Range of ?? ( ?? ) is [5,7] 
( ?? , ?? ) = ( 5 , 7 ) 
distance = |
15 + 28 + 12
5
| = 11 
Q3: If ?
?? = ?? ????
? {
?? ?? ?? ?? ? (
?? ?? + ( ?? - ?? )
?? ?? ) ?? ?? ?? ? (
?? ?? +
????
?? )
} = ?? v ?? + ?? , ?? , ?? ? ?? , then ?? ?? + ?? ?? is equal to : 
JEE Main 2025 (Online) 28th January Evening Shift 
Options: 
A. 10 
B. 4 
C. 8 
D. 2 
 
Ans: C 
Solution: 
1
sin ?
?? 6
?
?? = 1
13
?
sin ? [ (
?? 4
+
r ?? 6
) - (
?? 4
) - ( r - 1 )
?? 6
]
sin ? (
?? 4
+ ( r - 1 )
?? 6
) sin ? (
?? 4
+
r ?? 6
)
 
1
sin ?
?? 6
?
r = 1
13
? ( c o t ? (
?? 4
+ ( r - 1 )
?? 6
) - c ot ? (
?? 4
+
r ?? 6
) ) 
= 2 v 3 - 2 = ?? v 3 + b 
So a
2
+ b
2
= 8 
 
Q4: If ?? ???? ? ?? + ?? ???? ?? ? ?? = ?? , ?? ? ( ?? ,
?? ?? ), then 
( ?? ?? ?? ????
? ?? + ?????? ????
? ?? ) + ?? ( ?? ?? ?? ????
? ?? + ?????? ????
? ?? + ?? ?? ?? ?? ? ?? + ?????? ?? ? ?? ) + ( ?? ?? ?? ?? ? ?? + ?????? ?? ? ?? ) is equal to: 
JEE Main 2025 (Online) 29th January Evening Shift 
Options: 
A. 3 
B. 4 
C. 2 
D. 1 
 
Ans: C 
Solution: 
sin ? ?? + sin
2
? ?? = 1 
? sin ? ?? = c os
2
? ?? ? tan ? ?? = c os ? ?? 
? Given expression 
= 2 c os
12
? ?? + 6 [ c os
10
? ?? + c os
8
? ?? ] + 2 c os
6
? ?? 
= 2 [ sin
6
? ?? + 3 sin
5
? ?? + 3 sin
4
? ?? + sin
3
? ?? ] 
= 2 sin
3
? ?? [ ( sin ? ?? + 1 )
3
] 
= 2 [ sin
2
? ?? + sin ? ?? ]
3
 
= 2 
 
Q5: If ???? ?? ???? ?? ? ?? + ???? ?? ?? ?? ?? ? ?? = ?? , then the value of 
???? ?? ?? ?? ?? ?? ?? ?? + ?? ??????
?? ? ?? ???? ??????
?? ? ?? is 
JEE Main 2025 (Online) 4th April Morning Shift 
Options: 
A. 
2
5
 
B. 
3
5
 
C. 
1
5
 
D. 
3
4
 
Ans: A 
Solution: 
? 10 sin
4
? ?? + 15 c os
4
? ?? = 6
? ? 10 sin
4
? ?? + 10 c o s
4
? ?? + 5 c os
4
? ?? = 6
? ? 10 [ ( sin
2
? ?? + c os
2
? ?? )
2
- 2 sin
2
? ?? c os
2
? ?? ] + 5 c os
4
? ?? = 6
? ? 10 - 20 ( 1 - c os
2
? ?? ) c os
2
? ?? + 5 c os
4
? ?? = 6
 L et c os
2
? ?? = ?? 10 - 20 ( ?? - ?? 2
) + 5 ?? 2
= 6
? ? 25 ?? 2
- 20 ?? + 4 = 0
? ( 5 ?? - 2 )
2
= 0 ? ?? =
2
5
? ? c os
2
? ?? =
2
5
? sin
2
? ?? =
3
5
? s ec
2
? ?? =
5
2
, c os ec
2
?? =
5
3
27 c os ec
6
?? + 8 s ec
6
? ?? 16 s ec
8
? ?? =
27 (
5
3
)
3
+ 8 (
5
2
)
3
16 (
5
2
)
4
? =
5
3
+ 5
3
5
4
=
2 . 5
3
5
4
=
2
5
 
 
 
 
Q6: If for ?? ? [ -
?? ?? , ?? ], the points ( ?? , ?? ) = ( ???????? ? ( ?? +
?? ?? ) , ???????? ? ( ?? +
?? ?? ) ) lie on ???? + ???? + ???? +
?? = ?? , then ?? ?? + ?? ?? + ?? ?? is equal to : 
JEE Main 2025 (Online) 7th April Morning Shift 
Options: 
A. 75 
B. 96 
C. 80 
D. 72 
Ans: A 
Solution: 
 L et ?? = ?? +
?? 3
? ?? = ?? -
?? 6
?? = 3tan ? ( ?? +
?? 3
) = 3tan ? ( ?? +
?? 6
)
?? = 2tan ? ?? tan ? ( ?? +
?? 6
) =
tan ? ?? +
1
v 3
1 - tan ? ?? ·
1
v 3
?? 3
=
?? 2
+
1
v 3
1 -
?? 2
·
1
v 3
? ? ? ?? =
3 ( ?? v 3 + 2 )
2 v 3 - ?? ???? + ???? + ???? + ?? = 0
3 (
?? v 3 + 2
2 v 3 - ?? ) + ?? ( 3
( ?? v 3 + 2 )
( 2 v 3 - ?? )
) + ???? + ?? = 0
? = ( 3 v 3 - ?? ) ?? 2
+ ( 6 + 3 v 3 ?? + 2 v 3 ?? - ?? ) ?? ? + ( 6 ?? + 2 v 3 ?? ) = 0
 
For this identity to hold for all ?? , coefficients must be 0 
? ? ? ?? = 3 v 3
?? = - ?? v 3
6 + 3 v 3 ?? + ( 2 v 3 ) ( 3 v 3 ) + ?? v 3 = 0
? ? ?? = - 2 v 3
? ? ?? = 6
?? 2
+ ?? 2
+ ?? 2
= 75
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FAQs on JEE Main Previous Year Questions (2025): Trigonometric Ratios, Functions & Equations

1. What are the basic trigonometric ratios and how are they defined?
Ans. The basic trigonometric ratios are sine (sin), cosine (cos), tangent (tan), cosecant (csc), secant (sec), and cotangent (cot). They are defined based on the sides of a right triangle. For an angle θ in a right triangle: - sin(θ) = Opposite side/Hypotenuse - cos(θ) = Adjacent side/Hypotenuse - tan(θ) = Opposite side/Adjacent side - csc(θ) = 1/sin(θ) = Hypotenuse/Opposite side - sec(θ) = 1/cos(θ) = Hypotenuse/Adjacent side - cot(θ) = 1/tan(θ) = Adjacent side/Opposite side.
2. How do the trigonometric functions behave for special angles like 0°, 30°, 45°, 60°, and 90°?
Ans. The values of trigonometric functions at these special angles are: - sin(0°) = 0, cos(0°) = 1, tan(0°) = 0 - sin(30°) = 1/2, cos(30°) = √3/2, tan(30°) = 1/√3 - sin(45°) = √2/2, cos(45°) = √2/2, tan(45°) = 1 - sin(60°) = √3/2, cos(60°) = 1/2, tan(60°) = √3 - sin(90°) = 1, cos(90°) = 0, tan(90°) is undefined.
3. What is the significance of the unit circle in trigonometry?
Ans. The unit circle, which is a circle with a radius of 1 centered at the origin of the coordinate plane, is significant in trigonometry because it provides a geometric interpretation of the trigonometric functions. The x-coordinate of a point on the unit circle represents the cosine of the angle, while the y-coordinate represents the sine of the angle. This helps in understanding the periodic nature of these functions, their values for different angles, and their relationships with each other.
4. How can trigonometric identities be used to simplify expressions?
Ans. Trigonometric identities are equations that involve trigonometric functions and are true for all values of the variable. They can be used to simplify expressions by substituting one side of the identity for another. Common identities include the Pythagorean identities, reciprocal identities, and angle sum/difference identities. For example, using the identity sin²(θ) + cos²(θ) = 1 can simplify expressions involving sine and cosine.
5. What are the solutions to basic trigonometric equations, and how can they be found?
Ans. Basic trigonometric equations such as sin(x) = a, cos(x) = b, and tan(x) = c can be solved by finding the angles that satisfy these equations within a given interval. Solutions can be found using inverse trigonometric functions and considering the periodic nature of the trigonometric functions. For example, if sin(x) = 1/2, the solutions in the interval [0, 2π) would be x = π/6 and x = 5π/6, considering the behavior of the sine function.
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