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JEE Advanced Previous Year Questions (2021 - 2024): Trigonometric Ratios, Functions & Equations | Mathematics (Maths) for JEE Main & Advanced PDF Download

Q: 4. What are some common mistakes to avoid when solving trigonometric problems in JEE Advanced?

Ans. Common mistakes include:1. Forgetting to consider the range of angles when solving equations.2. Misapplying trigonometric identities, leading to incorrect simplifications.3. Ignoring the periodic nature of trigonometric functions, which might result in missing solutions.4. Failing to check for extraneous solutions after solving trigonometric equations.

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 Page 1


 
Q1: Let 
p
2
 < x < p be such that cot x = 
-5
v11
. Then ( sin 
11x
2
 ) (sin 6x - cos 6x) + ( cos 
11x
2
 ) (sin 6x + cos 6x) is
equal to:
(a) 
v11 - 1
2 v3
(b) 
v11 + 1
2 v3
(c) 
v11 + 1
3 v2
(d) 
v11 - 1
3 v2
 [JEE Advanced 2024 Paper 1]
Ans: (b)
Given the information, let's start by analyzing the trigonometric relationships involving:
cot x = 
-5
v11
 where 
p
2
 < x < p.
We know that cot x is the reciprocal of tan x. Hence,
cot x = 
cos x
sin x
 = 
-5
v11
From the above, it follows that:
cos x = -5k and sin x = v11k
For some constant k. Using the Pythagorean identity:
cos² x + sin² x = 1
Substituting the values of cos x and sin x into this identity:
(-5k)² + (v11k)² = 1
25k² + 11k² = 1
36k² = 1
k² = 
1
36
k = 
1
6
 or k = 
-1
6
Page 2


 
Q1: Let 
p
2
 < x < p be such that cot x = 
-5
v11
. Then ( sin 
11x
2
 ) (sin 6x - cos 6x) + ( cos 
11x
2
 ) (sin 6x + cos 6x) is
equal to:
(a) 
v11 - 1
2 v3
(b) 
v11 + 1
2 v3
(c) 
v11 + 1
3 v2
(d) 
v11 - 1
3 v2
 [JEE Advanced 2024 Paper 1]
Ans: (b)
Given the information, let's start by analyzing the trigonometric relationships involving:
cot x = 
-5
v11
 where 
p
2
 < x < p.
We know that cot x is the reciprocal of tan x. Hence,
cot x = 
cos x
sin x
 = 
-5
v11
From the above, it follows that:
cos x = -5k and sin x = v11k
For some constant k. Using the Pythagorean identity:
cos² x + sin² x = 1
Substituting the values of cos x and sin x into this identity:
(-5k)² + (v11k)² = 1
25k² + 11k² = 1
36k² = 1
k² = 
1
36
k = 
1
6
 or k = 
-1
6
Therefore, we have two sets of values:
cos x = -5/6 and sin x = v11/6
Given that x is in the interval ( 
p
2
 , p), where sine is positive and cosine is negative, we take:
cos x = -5/6, sin x = v11/6
Now consider the expression:
(sin 
11x
2
 ) (sin 6x - cos 6x) + (cos 
11x
2
 ) (sin 6x + cos 6x)
Using trigonometric identities, we simplify the expression step by step.
Thus, the correct option is:
Option B
v11 + 1
2v3
Q1: Consider an obtuse-angled triangle ABC in which the difference between the largest and the smallest
angle is p/2 and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on
a circle of radius 1.
Then the inradius of the triangle ABC is : _______. [JEE Advanced 2023 Paper 2]
Ans: 0.25
In radius r = 
?
S
 = 
a
2R(sin A + sin B + sin C)
r = 
a
sin (p/2 - 2C) sin (p/2 + C) + sin C
= 
a
cos 2C + cos C + sin C
= 
a
cos 2C + v1 + sin 2C
= 
3v7 / 16
v(7/4) + v(7/2)
 = 
1
4
? r = 
1
4
 = 0.25
? r = 0.25
Page 3


 
Q1: Let 
p
2
 < x < p be such that cot x = 
-5
v11
. Then ( sin 
11x
2
 ) (sin 6x - cos 6x) + ( cos 
11x
2
 ) (sin 6x + cos 6x) is
equal to:
(a) 
v11 - 1
2 v3
(b) 
v11 + 1
2 v3
(c) 
v11 + 1
3 v2
(d) 
v11 - 1
3 v2
 [JEE Advanced 2024 Paper 1]
Ans: (b)
Given the information, let's start by analyzing the trigonometric relationships involving:
cot x = 
-5
v11
 where 
p
2
 < x < p.
We know that cot x is the reciprocal of tan x. Hence,
cot x = 
cos x
sin x
 = 
-5
v11
From the above, it follows that:
cos x = -5k and sin x = v11k
For some constant k. Using the Pythagorean identity:
cos² x + sin² x = 1
Substituting the values of cos x and sin x into this identity:
(-5k)² + (v11k)² = 1
25k² + 11k² = 1
36k² = 1
k² = 
1
36
k = 
1
6
 or k = 
-1
6
Therefore, we have two sets of values:
cos x = -5/6 and sin x = v11/6
Given that x is in the interval ( 
p
2
 , p), where sine is positive and cosine is negative, we take:
cos x = -5/6, sin x = v11/6
Now consider the expression:
(sin 
11x
2
 ) (sin 6x - cos 6x) + (cos 
11x
2
 ) (sin 6x + cos 6x)
Using trigonometric identities, we simplify the expression step by step.
Thus, the correct option is:
Option B
v11 + 1
2v3
Q1: Consider an obtuse-angled triangle ABC in which the difference between the largest and the smallest
angle is p/2 and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on
a circle of radius 1.
Then the inradius of the triangle ABC is : _______. [JEE Advanced 2023 Paper 2]
Ans: 0.25
In radius r = 
?
S
 = 
a
2R(sin A + sin B + sin C)
r = 
a
sin (p/2 - 2C) sin (p/2 + C) + sin C
= 
a
cos 2C + cos C + sin C
= 
a
cos 2C + v1 + sin 2C
= 
3v7 / 16
v(7/4) + v(7/2)
 = 
1
4
? r = 
1
4
 = 0.25
? r = 0.25
  
 
Q1. Let ?? and ?? be real numbers such that -
?? ?? < ?? < ?? < ?? <
?? ?? . 
If ?? ???? ? ( ?? + ?? ) =
?? ?? and ?? ???? ? ( ?? - ?? ) =
?? ?? , then the greatest integer less than or equal 
to (
?? ?? ?? ? ?? ???? ?? ? ?? +
???? ?? ? ?? ?? ?? ?? ? ?? +
???? ?? ? ?? ?? ?? ?? ? ?? +
?? ?? ?? ? ?? ???? ?? ? ?? )
?? is   [JEE Advanced 2022 Paper 2 Online] 
Ans: 1 
Given, sin ? ( ?? + ?? ) =
1
3
 
and c o s ? ( ?? - ?? ) =
2
3
 
Let, ?? =
s i n ? ?? c o s ? ?? +
c o s ? ?? s i n ? ?? +
c o s ? ?? s i n ? ?? +
s i n ? ?? c o s ? ?? 
=
sin ? ?? c o s ? ?? +
c o s ? ?? sin ? ?? +
c o s ? ?? sin ? ?? +
sin ? ?? c o s ? ?? 
? =
sin ? ?? sin ? ?? + c o s ? ?? c o s ? ?? sin ? ?? c o s ? ?? +
c o s ? ?? c o s ? ?? + sin ? ?? sin ? ?? sin ? ?? c o s ? ?? ? =
c o s ? ( ?? - ?? )
sin ? ?? c o s ? ?? +
c o s ? ( ?? - ?? )
sin ? ?? c o s ? ?? ? = c o s ? ( ?? - ?? ) [
2
2 s i n ? ?? c o s ? ?? +
2
2 s i n ? ?? c o s ? ?? ]
? =
2
3
[
2
sin ? 2 ?? +
2
sin ? 2 ?? ]
? =
4
3
[
1
sin ? 2 ?? +
1
sin ? 2 ?? ]
? =
4
3
[
sin ? 2 ?? + sin ? 2 ?? sin ? 2 ?? sin ? 2 ?? ]
? =
4 × 2
3
[
2 s i n ? (
2 ?? + 2 ?? 2
) c o s ? (
2 ?? - 2 ?? 2
)
2 s i n ? 2 ?? sin ? 2 ?? ]
 
=
16
3
[
sin ? ( ?? + ?? ) c o s ? ( ?? - ?? )
c o s ? ( 2 ?? - 2 ?? ) - c o s ? ( 2 ?? + 2 ?? )
] 
Page 4


 
Q1: Let 
p
2
 < x < p be such that cot x = 
-5
v11
. Then ( sin 
11x
2
 ) (sin 6x - cos 6x) + ( cos 
11x
2
 ) (sin 6x + cos 6x) is
equal to:
(a) 
v11 - 1
2 v3
(b) 
v11 + 1
2 v3
(c) 
v11 + 1
3 v2
(d) 
v11 - 1
3 v2
 [JEE Advanced 2024 Paper 1]
Ans: (b)
Given the information, let's start by analyzing the trigonometric relationships involving:
cot x = 
-5
v11
 where 
p
2
 < x < p.
We know that cot x is the reciprocal of tan x. Hence,
cot x = 
cos x
sin x
 = 
-5
v11
From the above, it follows that:
cos x = -5k and sin x = v11k
For some constant k. Using the Pythagorean identity:
cos² x + sin² x = 1
Substituting the values of cos x and sin x into this identity:
(-5k)² + (v11k)² = 1
25k² + 11k² = 1
36k² = 1
k² = 
1
36
k = 
1
6
 or k = 
-1
6
Therefore, we have two sets of values:
cos x = -5/6 and sin x = v11/6
Given that x is in the interval ( 
p
2
 , p), where sine is positive and cosine is negative, we take:
cos x = -5/6, sin x = v11/6
Now consider the expression:
(sin 
11x
2
 ) (sin 6x - cos 6x) + (cos 
11x
2
 ) (sin 6x + cos 6x)
Using trigonometric identities, we simplify the expression step by step.
Thus, the correct option is:
Option B
v11 + 1
2v3
Q1: Consider an obtuse-angled triangle ABC in which the difference between the largest and the smallest
angle is p/2 and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on
a circle of radius 1.
Then the inradius of the triangle ABC is : _______. [JEE Advanced 2023 Paper 2]
Ans: 0.25
In radius r = 
?
S
 = 
a
2R(sin A + sin B + sin C)
r = 
a
sin (p/2 - 2C) sin (p/2 + C) + sin C
= 
a
cos 2C + cos C + sin C
= 
a
cos 2C + v1 + sin 2C
= 
3v7 / 16
v(7/4) + v(7/2)
 = 
1
4
? r = 
1
4
 = 0.25
? r = 0.25
  
 
Q1. Let ?? and ?? be real numbers such that -
?? ?? < ?? < ?? < ?? <
?? ?? . 
If ?? ???? ? ( ?? + ?? ) =
?? ?? and ?? ???? ? ( ?? - ?? ) =
?? ?? , then the greatest integer less than or equal 
to (
?? ?? ?? ? ?? ???? ?? ? ?? +
???? ?? ? ?? ?? ?? ?? ? ?? +
???? ?? ? ?? ?? ?? ?? ? ?? +
?? ?? ?? ? ?? ???? ?? ? ?? )
?? is   [JEE Advanced 2022 Paper 2 Online] 
Ans: 1 
Given, sin ? ( ?? + ?? ) =
1
3
 
and c o s ? ( ?? - ?? ) =
2
3
 
Let, ?? =
s i n ? ?? c o s ? ?? +
c o s ? ?? s i n ? ?? +
c o s ? ?? s i n ? ?? +
s i n ? ?? c o s ? ?? 
=
sin ? ?? c o s ? ?? +
c o s ? ?? sin ? ?? +
c o s ? ?? sin ? ?? +
sin ? ?? c o s ? ?? 
? =
sin ? ?? sin ? ?? + c o s ? ?? c o s ? ?? sin ? ?? c o s ? ?? +
c o s ? ?? c o s ? ?? + sin ? ?? sin ? ?? sin ? ?? c o s ? ?? ? =
c o s ? ( ?? - ?? )
sin ? ?? c o s ? ?? +
c o s ? ( ?? - ?? )
sin ? ?? c o s ? ?? ? = c o s ? ( ?? - ?? ) [
2
2 s i n ? ?? c o s ? ?? +
2
2 s i n ? ?? c o s ? ?? ]
? =
2
3
[
2
sin ? 2 ?? +
2
sin ? 2 ?? ]
? =
4
3
[
1
sin ? 2 ?? +
1
sin ? 2 ?? ]
? =
4
3
[
sin ? 2 ?? + sin ? 2 ?? sin ? 2 ?? sin ? 2 ?? ]
? =
4 × 2
3
[
2 s i n ? (
2 ?? + 2 ?? 2
) c o s ? (
2 ?? - 2 ?? 2
)
2 s i n ? 2 ?? sin ? 2 ?? ]
 
=
16
3
[
sin ? ( ?? + ?? ) c o s ? ( ?? - ?? )
c o s ? ( 2 ?? - 2 ?? ) - c o s ? ( 2 ?? + 2 ?? )
] 
? =
16
3
[
1
3
×
2
3
( 2 c o s
2
? ( ?? - ?? ) - 1 ) - ( 1 - 2 sin
2
? ( ?? + ?? ) )
]
? =
32
27
[
1
2 ×
4
9
- 2 + 2 ×
1
9
]
? =
32
27
[
9
8 - 18 + 2
]
? =
32
27
[
9
- 8
]
? = -
4
3
? ? ?? 2
=
16
9
= 1 . 77
? ? [ ?? 2
] = 1
 
 
Q2. Consider the following lists : 
List-I List-II 
(I) { ?? ? [ -
?? ?? ?? ,
?? ?? ?? ] : ?? ???? ? ?? + ?? ???? ? ?? = ?? } (P) has two elements 
(II) { ?? ? [ -
?? ?? ????
,
?? ?? ????
] : v ?? ?? ?? ?? ? ?? ?? = ?? } (Q) has three elements 
(III) { ?? ? [ -
?? ?? ?? ,
?? ?? ?? ] : ?? ?? ???? ? ( ?? ?? ) = v ?? } (R) has four elements 
(IV) { ?? ? [ -
?? ?? ?? ,
?? ?? ?? ] : ?? ???? ? ?? - ?? ???? ? ?? = ?? } (S) has five elements 
 (T) has six elements 
 
The correct option is: 
A (I) ? ( ?? ); (II) ? ( ?? ) ; ( ?????? ) ? ( ?? ); (IV) ? ( ?? ) 
(B) ( ?? ) ? ( ?? ) ; ( ???? ) ? ( ?? ) ; ( ?????? ) ? ( ?? ) ; ( ???? ) ? ( ?? ) 
(C) ( ?? ) ? ( ?? ) ; ( ???? ) ? ( ?? ); (III) ? ( ?? ); (IV) ? ( ?? ) 
(D) (I) ? ( ?? ) ; ( ???? ) ? ( ?? ) ; ( ?????? ) ? ( ?? ) ; ( ???? ) ? ( ?? )    
[JEE Advanced 2022 Paper 1 Online] 
Ans: (b) 
Solving all question one by one we get, 
 ( i ) { ?? ? [
- 2 ?? 3
,
2 ?? 3
] , c o s ? ?? + sin ? ?? = 1 }
c o s ? ?? + sin ? ?? = 1
 
Page 5


 
Q1: Let 
p
2
 < x < p be such that cot x = 
-5
v11
. Then ( sin 
11x
2
 ) (sin 6x - cos 6x) + ( cos 
11x
2
 ) (sin 6x + cos 6x) is
equal to:
(a) 
v11 - 1
2 v3
(b) 
v11 + 1
2 v3
(c) 
v11 + 1
3 v2
(d) 
v11 - 1
3 v2
 [JEE Advanced 2024 Paper 1]
Ans: (b)
Given the information, let's start by analyzing the trigonometric relationships involving:
cot x = 
-5
v11
 where 
p
2
 < x < p.
We know that cot x is the reciprocal of tan x. Hence,
cot x = 
cos x
sin x
 = 
-5
v11
From the above, it follows that:
cos x = -5k and sin x = v11k
For some constant k. Using the Pythagorean identity:
cos² x + sin² x = 1
Substituting the values of cos x and sin x into this identity:
(-5k)² + (v11k)² = 1
25k² + 11k² = 1
36k² = 1
k² = 
1
36
k = 
1
6
 or k = 
-1
6
Therefore, we have two sets of values:
cos x = -5/6 and sin x = v11/6
Given that x is in the interval ( 
p
2
 , p), where sine is positive and cosine is negative, we take:
cos x = -5/6, sin x = v11/6
Now consider the expression:
(sin 
11x
2
 ) (sin 6x - cos 6x) + (cos 
11x
2
 ) (sin 6x + cos 6x)
Using trigonometric identities, we simplify the expression step by step.
Thus, the correct option is:
Option B
v11 + 1
2v3
Q1: Consider an obtuse-angled triangle ABC in which the difference between the largest and the smallest
angle is p/2 and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on
a circle of radius 1.
Then the inradius of the triangle ABC is : _______. [JEE Advanced 2023 Paper 2]
Ans: 0.25
In radius r = 
?
S
 = 
a
2R(sin A + sin B + sin C)
r = 
a
sin (p/2 - 2C) sin (p/2 + C) + sin C
= 
a
cos 2C + cos C + sin C
= 
a
cos 2C + v1 + sin 2C
= 
3v7 / 16
v(7/4) + v(7/2)
 = 
1
4
? r = 
1
4
 = 0.25
? r = 0.25
  
 
Q1. Let ?? and ?? be real numbers such that -
?? ?? < ?? < ?? < ?? <
?? ?? . 
If ?? ???? ? ( ?? + ?? ) =
?? ?? and ?? ???? ? ( ?? - ?? ) =
?? ?? , then the greatest integer less than or equal 
to (
?? ?? ?? ? ?? ???? ?? ? ?? +
???? ?? ? ?? ?? ?? ?? ? ?? +
???? ?? ? ?? ?? ?? ?? ? ?? +
?? ?? ?? ? ?? ???? ?? ? ?? )
?? is   [JEE Advanced 2022 Paper 2 Online] 
Ans: 1 
Given, sin ? ( ?? + ?? ) =
1
3
 
and c o s ? ( ?? - ?? ) =
2
3
 
Let, ?? =
s i n ? ?? c o s ? ?? +
c o s ? ?? s i n ? ?? +
c o s ? ?? s i n ? ?? +
s i n ? ?? c o s ? ?? 
=
sin ? ?? c o s ? ?? +
c o s ? ?? sin ? ?? +
c o s ? ?? sin ? ?? +
sin ? ?? c o s ? ?? 
? =
sin ? ?? sin ? ?? + c o s ? ?? c o s ? ?? sin ? ?? c o s ? ?? +
c o s ? ?? c o s ? ?? + sin ? ?? sin ? ?? sin ? ?? c o s ? ?? ? =
c o s ? ( ?? - ?? )
sin ? ?? c o s ? ?? +
c o s ? ( ?? - ?? )
sin ? ?? c o s ? ?? ? = c o s ? ( ?? - ?? ) [
2
2 s i n ? ?? c o s ? ?? +
2
2 s i n ? ?? c o s ? ?? ]
? =
2
3
[
2
sin ? 2 ?? +
2
sin ? 2 ?? ]
? =
4
3
[
1
sin ? 2 ?? +
1
sin ? 2 ?? ]
? =
4
3
[
sin ? 2 ?? + sin ? 2 ?? sin ? 2 ?? sin ? 2 ?? ]
? =
4 × 2
3
[
2 s i n ? (
2 ?? + 2 ?? 2
) c o s ? (
2 ?? - 2 ?? 2
)
2 s i n ? 2 ?? sin ? 2 ?? ]
 
=
16
3
[
sin ? ( ?? + ?? ) c o s ? ( ?? - ?? )
c o s ? ( 2 ?? - 2 ?? ) - c o s ? ( 2 ?? + 2 ?? )
] 
? =
16
3
[
1
3
×
2
3
( 2 c o s
2
? ( ?? - ?? ) - 1 ) - ( 1 - 2 sin
2
? ( ?? + ?? ) )
]
? =
32
27
[
1
2 ×
4
9
- 2 + 2 ×
1
9
]
? =
32
27
[
9
8 - 18 + 2
]
? =
32
27
[
9
- 8
]
? = -
4
3
? ? ?? 2
=
16
9
= 1 . 77
? ? [ ?? 2
] = 1
 
 
Q2. Consider the following lists : 
List-I List-II 
(I) { ?? ? [ -
?? ?? ?? ,
?? ?? ?? ] : ?? ???? ? ?? + ?? ???? ? ?? = ?? } (P) has two elements 
(II) { ?? ? [ -
?? ?? ????
,
?? ?? ????
] : v ?? ?? ?? ?? ? ?? ?? = ?? } (Q) has three elements 
(III) { ?? ? [ -
?? ?? ?? ,
?? ?? ?? ] : ?? ?? ???? ? ( ?? ?? ) = v ?? } (R) has four elements 
(IV) { ?? ? [ -
?? ?? ?? ,
?? ?? ?? ] : ?? ???? ? ?? - ?? ???? ? ?? = ?? } (S) has five elements 
 (T) has six elements 
 
The correct option is: 
A (I) ? ( ?? ); (II) ? ( ?? ) ; ( ?????? ) ? ( ?? ); (IV) ? ( ?? ) 
(B) ( ?? ) ? ( ?? ) ; ( ???? ) ? ( ?? ) ; ( ?????? ) ? ( ?? ) ; ( ???? ) ? ( ?? ) 
(C) ( ?? ) ? ( ?? ) ; ( ???? ) ? ( ?? ); (III) ? ( ?? ); (IV) ? ( ?? ) 
(D) (I) ? ( ?? ) ; ( ???? ) ? ( ?? ) ; ( ?????? ) ? ( ?? ) ; ( ???? ) ? ( ?? )    
[JEE Advanced 2022 Paper 1 Online] 
Ans: (b) 
Solving all question one by one we get, 
 ( i ) { ?? ? [
- 2 ?? 3
,
2 ?? 3
] , c o s ? ?? + sin ? ?? = 1 }
c o s ? ?? + sin ? ?? = 1
 
? ? sin ? (
?? 4
+ ?? ) =
1
v 2
? ?
?? 4
+ ?? = ???? + ( - 1 )
?? ?? 4
? ? ?? = ???? + ( - 1 )
?? ?? 4
-
?? 4
 
So, ?? ? { 0 ,
?? 2
} 
? ?? has 2 elements ? ?? 
(ii) { ?? ? (
- 5 ?? 18
,
5 ?? 18
) , v 3 t a n ? 3 ?? = 1 } 
? t a n ? 3 ?? =
1
v 3
 
? ?? = ???? +
?? 6
 
? ?? =
????
3
+
?? 18
 
So ?? ? {
?? 18
,
- 5 ?? 18
} 
? ?? has 2 elements ? ?? 
(iii) { ?? ? [
- 6 ?? 5
,
6 ?? 5
] , 2 c o s ? 2 ?? = v 3 } 
? ? 2 c o s ? 2 ?? = v 3
? ? c o s ? 2 ?? =
v 3
2
? ? 2 ?? = 2 ???? ±
?? 6
? ? ?? = ???? ±
?? 12
 So ?? ? { ±
?? 12
, ?? ±
?? 12
, - ?? ±
?? 12
}
 
So ?? ? { ±
?? 12
, ?? ±
?? 12
, - ?? ±
?? 12
} 
? ?? has 6 elements ? T 
 ( i v) { ?? ? [
- 7 ?? 4
,
7 ?? 4
] , sin ? ?? - c o s ? ?? = 1 }
? ? sin ? ?? - c o s ? ?? = 1
? ? sin ? ( ?? -
?? 4
) =
1
v 2
? ? ?? -
?? 4
= ???? + ( - 1 )
?? ?? 4
? ? ?? = ???? + ( - 1 )
?? ?? 4
+
?? 4
 
So, 
?? ? {
?? 2
,
- 3 ?? 2
, ?? , - ?? } 
? ?? has 4 elements ? R

	Mathematics (Maths) for JEE Main & Advanced 172 videos\|472 docs\|154 tests

Mathematics (Maths) for JEE Main & Advanced

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FAQs on JEE Advanced Previous Year Questions (2021 - 2024): Trigonometric Ratios, Functions & Equations - Mathematics (Maths) for JEE Main & Advanced

1. What are the key trigonometric ratios that every student should know for JEE Advanced?

Ans. The key trigonometric ratios to know for JEE Advanced are sine (sin), cosine (cos), tangent (tan), cosecant (csc), secant (sec), and cotangent (cot). These ratios are defined for an angle θ in a right-angled triangle as follows: - sin(θ) = opposite side/hypotenuse - cos(θ) = adjacent side/hypotenuse - tan(θ) = opposite side/adjacent side - csc(θ) = 1/sin(θ) - sec(θ) = 1/cos(θ) - cot(θ) = 1/tan(θ)

2. How do trigonometric functions behave in different quadrants?

Ans. Trigonometric functions have specific signs based on the quadrant in which the angle resides: - In Quadrant I: sin, cos, and tan are positive. - In Quadrant II: sin is positive, while cos and tan are negative. - In Quadrant III: tan is positive, while sin and cos are negative. - In Quadrant IV: cos is positive, while sin and tan are negative. Understanding these signs is crucial for solving problems in JEE Advanced.

3. What techniques can be used to solve trigonometric equations in JEE Advanced?

Ans. To solve trigonometric equations in JEE Advanced, several techniques can be employed: 1. Using fundamental identities (e.g., Pythagorean, angle sum, and difference identities). 2. Converting all trigonometric functions to sine and cosine. 3. Factoring and simplifying the equation. 4. Using substitution methods for complex equations. 5. Graphical methods to find intersections of functions if needed.

4. What are some common mistakes to avoid when solving trigonometric problems in JEE Advanced?

Ans. Common mistakes include: 1. Forgetting to consider the range of angles when solving equations. 2. Misapplying trigonometric identities, leading to incorrect simplifications. 3. Ignoring the periodic nature of trigonometric functions, which might result in missing solutions. 4. Failing to check for extraneous solutions after solving trigonometric equations.

5. How can understanding the unit circle help in solving trigonometric problems for JEE Advanced?

Ans. Understanding the unit circle is crucial as it provides the values of sine and cosine for commonly used angles (0°, 30°, 45°, 60°, 90°, etc.). It helps visualize the relationships between angles and their corresponding trigonometric values, making it easier to solve problems involving angle addition, subtraction, and transformations. Additionally, it aids in identifying the signs of trigonometric functions in different quadrants.

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