JEE Main Test Series Trigonometry: Previous Year Questions & Solution

Trigonometry is one of the most important topics in JEE Main exam. Every year about 1-3 questions are asked from this topic. The overall weightage of Trigonometry in JEE Main Question Paper is 7%. Some of the topics in Trignometry include Identities of Trignometry and Trigonometric Equations, Functions of Trignometry, Properties of Inverse trigonometric functions, and Problems of Heights and Distances.  Check JEE Main Mathematics Syllabus 

• Trigonometry is the branch of mathematics that deals with the study of lengths of triangles and relationships between sides and angles. It involves the study of statistics, calculus, and linear algebra. There are basically two parts of trigonometry namely, Plane trigonometry and Spherical trigonometry. 
• Plane trigonometry focuses on the connection between sides of angles and triangles in JEE Main mathematics. Spherical trigonometry focuses on curved triangles drawn on the surface of a sphere. Most questions asked on Trigonometry can be solved by simply applying trigonometry formulae, however, others will require the application of trigonometry tricks.
• Trigonometric Ratios and concepts are used in questions from Algebra and Calculus (as a direct function or as a way to simplify functions). Questions can be asked from trigonometric identities, equations, heights and distances, and inverse trigonometric functions. Check JEE Main Mathematics Preparation

Also Read

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Trigonometry Test Series for JEE with Solutions

JEE Questions in Trigonometry often test advanced concepts related to trigonometric identities, equations, and applications. Some common types of problems include:

• Trigonometric Equations: Questions that require solving equations like \sin x = a, \cos x = b, or more complex trigonometric expressions, including multiple angle and periodicity-based solutions.
• Trigonometric Identities: Problems that involve simplifying or proving standard identities (e.g., sum-to-product, double angle, and reduction formulas) or deriving results using complex identities.
• Solving Triangles: Applications of the Law of Sines and Law of Cosines, along with problems involving the use of trigonometric ratios to solve non-right angled triangles and applying properties of triangles in coordinate geometry.
• Height and Distance Problems: These problems often involve real-world scenarios such as calculating the height of a building, distance between two points, or angles of elevation and depression, frequently requiring the use of advanced trigonometric concepts.
• Multiple Angle and Inverse Trigonometric Functions: Problems that involve inverse functions, transformations of angles (e.g., solving \tan(2x) = 1), or using inverse trigonometric functions to find exact values or angles in specified ranges.

These questions aim to test not only basic knowledge but also the ability to apply advanced techniques and solve complex problems efficiently.

The video by Unacademy provides multiple problems and solutions for JEE Main trigonometry. 

Also Read

• How to Prepare for JEE Main Maths?
• JEE Main 2026 Preparation Guide

JEE Main Test Series Trigonometry: Previous Year Questions and Solutions

JEE Main aspirants can refer to the previous years’ questions listed below, to get an idea of the type of questions asked about Trigonometry.

1. In ∆PQR, if 3sin P + 4cos Q = 6 and 4sin Q + 3cos P = 1, then the angle R is equal to?

   1. 56

   2. 6

   3. 4

   4. 34

Answer: Option b: 6

Solution: 

Let, 3sin P + 4cos Q = 6 … (1)

4sin Q + 3 cos P = 1 … (2)

• (1)2 + (2)2 = (9) (1) + 16 (1) + 24

• Sin P Cos Q + Cos P Sin Q = 36 + 1

• 25 + 24 sin (P + Q) = 37

• 24 sin (P + Q) = 12

• Sin (P + Q) = ½

• P + Q = 6, R = 56

• If P + Q = π, then equation (1) and (2) is not satisfied.

2. The sides of a triangle are 3x + 4y, 4x + 3y, and 5x + 5y, where x, y > 0, then the triangle is?

   1. Right angled

   2. Obtuse angled

   3. Equilateral

   4. None of these

Answer: Option b: Obtuse angled

Solution:

Let a = 3x + 4y, b = 4x + 3y, and c = 5x + 5y

As x, y > 0, c = 5x + 5y is the largest side

• Hence, C is the largest angle

• Now, cos C = (3x +4y)2 + (4x + 3y)2 + (5x + 5y)2 2 3x + 4y (4x + 3y)

• = - 2xy 2 3x + 4y (4x + 3y) < 0

• So, C is an obtuse angle

3. The number of solutions for tan x + sec x = 2cos x in [0, 2π) is?

   1. 2

   2. 3

   3. 0

   4. 1

Answer: Option b: 3

Solution:

The given equation is tan x + sec x = 2 cos x

• Sin x + 1 = 2cos2 x

• 2sin2 x + sin x – 1 = 0

• (2 sin x – 1) (sin x + 1) = 0

• Sin x = ½, -1

• x = 30˚, 150˚, 270˚ 

4. In a triangle ABC, let ∠C = 2. If r is in radius and R is the circumradius of the triangle ABC, then 2(r + R) is equal to?

   1. b + c

   2. a + b 

   3. a + b + c

   4. c + a

Answer: Option b: a + b

Solution:

2r + 2R = c + 2ab(a + b + c)

• c2 + c a + b + 2 aba + b + c

• (a + b)2 + c a + ba + b + c

• (a + b + c) a + ba + b + c

• a+ b

5. If A = sin2 x + cos4 x = then for all real x:

   1. 1 ≤ A ≤ 2

   2. 34 ≤ A ≤ 1316

   3. 34 ≤ A ≤ 1

   4. 1316 ≤ A ≤ 1

Answer: Option c: 34 ≤ A ≤ 1

Solution:

A = 1 – cos2 x + cos4 x

• 1 - cos2 x (1 - cos2 x)

• 1 - cos2 x sin2 x 

• 1 – ¼ (sin2 2x)

• 1 – ¼ (1) ≤ A ≤ 1 – ¼ (0)

• ¾ ≤ A ≤ 1

6. The equation esin x- e-sin x-4=0 has?

   1. Infinite number of real roots

   2. No real roots

   3. Exactly one real root

   4. Exactly four real roots

Answer: Option b: No real roots

Solution: esin x- e-sin x-4=0

• Let esin x=t

• t – 1t = 4

• t2 – 4t – 1 = 0

• t = 4 ± √202 = 2 ± √5

• esin x= 2 ± √5

7. Let A and B denote statements 

A: cos α + cos β + cos γ = 0 and

B: sin α + sin β + sin γ = 0 

If cos (β – γ) + cos (γ – α) + cos (α - β) = - 32

1. A is true and B is false

2. A is false and B is true

3. Both A and B are true

4. Both A and B are false

Answer: Option c: Both A and B are true

Solution: 

cos (β – γ) + cos (γ – α) + cos (α - β) = - 32

• 2 [cos (β – γ) + cos (γ – α) + cos (α - β)] + 3 = 0

• 2 [cos (β – γ) + cos (γ – α) + cos (α - β)] + sin2 α + cos2 α + sin2 β + cos2 β + sin2 γ + cos2 γ = 0

• (sin α + sin β + sin γ)2 + (cos α + cos β + cos γ)2 = 0

8. If 0 ≤ x < 2π then the number of real values of x, which satisfy the equation cos x + cos 2x + cos 3x + cos 4x = 0 is:

   1. 9

   2. 5

   3. 7

   4. 3

Answer: Option c: 7

Solution:

2 cos 2x cos x + 2 cos 3x cos x = 0

• 2 cos x (cos 2x + cos 3x) = 0

• 2 cos x 2 cos 5x/2 cos x/2 = 0

• x = 2, 3π2,, 5 , 3π5, 7π5, 9π5

• 7 solutions

9. If in a triangle ABC, the altitudes from the vertices A, B, C on the opposite sides are in H. P, then sin A, sin B, sin C are in:

   1. G.P

   2. A.P

   3. Arithmetic – Geometric Progression

   4. H.P

Answer: Option b: A.P

Solution: ∆ ½ p1a = ½ p2b = ½ p3c

• p1, p2, p3 are in H.P

• 2∆a, 2∆b, 2∆c are in H.P

• 1a, 1b, 1c are in H.P

• a, b, c is in A.P

• Sin A, sin B, sin C are in A.P

10. If sum of all solutions of the equation 8 cos x. (cos (6+x) . cos 6+x – ½ ) = 1, in [0, ] is k, then k equals to?

    1. 139

    2. 89

    3. 209

    4. 23

Answer: Option a: 139

Solution:

8 cos x (cos2 6 - sin2 x – ½ = 1

• 8 cos x (¼ - (1 - cos2 x)) = 1

• 8 cos x (cos2 x – ¾) = 1

• 2 cos 3x = 1

• cos 3x = ½

• 3x + 2n ± 3, n I

• x = 2nπ3 ± 9 , in x [0, ]

• x = 9 , 2π3 + 9 , 2π3 - 9

• k = 139

Question 11: The general solution of sin x − 3 sin2x + sin3x = cos x − 3 cos2x + cos3x is _________.

Solution:

sinx − 3 sin2x + sin3x = cosx − 3 cos2x + cos3x

⇒ 2 sin2x cosx − 3 sin2x − 2 cos2x cosx + 3 cos2x = 0

⇒ sin2x (2cosx − 3) − cos2x (2 cosx − 3) = 0

⇒ (sin2x − cos2x) (2 cosx − 3) = 0

⇒ sin2x = cos2x

⇒ tan 2x = 1

2x = nπ + (π / 4 )

x = nπ / 2 + π / 8

Question 12: If sec 4θ − sec 2θ = 2, then the general value of θ is __________.

Solution:

sec 4θ − sec 2θ = 2 ⇒ cos 2θ − cos 4θ = 2 cos 4θ cos 2θ

⇒ −cos 4θ = cos 6θ

⇒ 2 cos 5θ cos θ = 0

When cos 5θ = 0, 5θ = (2n + 1)π/2

So θ  = nπ/5 + π/10

= (2n + 1)π/10

When cos θ = 0, θ = (2n+1)π/2.

Question 13: If tan (cot x) = cot (tan x), then sin 2x = ___________.

Solution:

tan (cot x) = cot (tan x) ⇒ tan (cot x) = tan (π / 2 − tan x)

cot x = nπ + π / 2 − tanx

⇒ cot x + tan x = nπ + π / 2

1/sin x cos x = nπ + π / 2

1/sin 2x = nπ/2  + π / 4

⇒ sin2x = 2 / [nπ + {π / 2}]

= 4 / {(2n + 1) π}

Question 14: If the solution for θ of cospθ + cosqθ = 0, p > 0, q > 0 are in A.P., then numerically the smallest common difference of A.P. is ___________.

Solution:

Given cospθ = −cosqθ = cos (π + qθ)

pθ = 2nπ ± (π + qθ), n ∈ I

θ = [(2n + 1)π] / [p − q] or [(2n − 1)π] / [p + q], n ∈ I

Both the solutions form an A.P. θ = [(2n + 1)π] / [p − q] gives us an A.P. with common difference 2π / [p − q] and θ = [(2n − 1)π] / [p + q] gives us an A.P. with common difference = 2π / [p + q].

Certainly, {2π / [p + q]} < {∣2π / [p − q]∣}.

Question 15: If α , β are different values of x satisfying a cosx + b sinx = c, then tan ([α + β] / 2) = ______________.

Solution:

a cosx + b sinx = c ⇒ a {[(1 − tan² (x / 2)] / [1 + tan² (x / 2)]} + 2b {[tan (x / 2) / 1 + tan² (x / 2)} = c

⇒ (a + c) * tan² [x / 2] − 2b tan [x / 2] + (c − a) = 0

This equation has roots tan [α / 2] and tan [β / 2].

Therefore, tan [α / 2] + tan [β / 2] = 2b / [a + c] and tan [α / 2] * tan [β / 2]

= [c − a] / [a + c]

Now

tan ((α + β)/2) = {tan [α / 2] + tan [β / 2]} / {1 − tan [α / 2] * tan [β / 2]}

= {[2b] / [a + c]} / {1− ([c − a] / [a + c])}

= b/a

Question 16: In a triangle, the length of the two larger sides are 10 cm and 9 cm, respectively. If the angles of the triangle are in arithmetic progression, then the length of the third side in cm can be _________.

Solution:

We know that in a triangle larger the side, larger the angle.

Since angles ∠A, ∠B and ∠C are in AP.

Hence, ∠B = 60^o cosB = [a² + c² −b²] / [2ac]

⇒ 1 / 2 = [100 + a² − 81] / [20a]

⇒ a² + 19 = 10a

⇒ a² − 10a + 19 = 0

a = 10 ± (√[100 − 76] / [2])

⇒ a = 5 ± √6

Question 17: In triangle ABC, if ∠A = 45∘, ∠B = 75∘, then a + c√2 = __________.

Solution:

∠C = 180^o − 45^o − 75^o = 60^o

a/sin A = b/sin b = c/sin C

a/sin 45 = b/sin 75 = c/sin 60

=> √2a = 2√2b/(√3+1) = 2c/√3

=> a = 2b/(√3+1)

c = √6b/(√3+1)

a+√2c = [2b/(√3+1)] + [√12b/(√3+1)]

Solving, we get

= 2b

Question 18: If cos⁻¹ p + cos⁻¹ q + cos⁻¹ r = π then p² + q² + r² + 2pqr = ________.

Solution:

Given cos⁻¹ p + cos⁻¹ q + cos⁻¹ r = π

cos⁻¹ p + cos⁻¹ q = π – cos⁻¹ r

cos⁻¹ (pq – √(1 – p²) √(1 – q²) = cos^-1 (-r)

(pq – √(1 – p²) √(1 – q²) = -r

(pq + r) = √(1 – p²) √(1 – q²)

squaring

(pq + r)² = (1 – p²) (1 – q²)

p²q² + 2pqr + r² = 1 – p² – q² + p²q²

p² + q² + r² + 2pqr = 1

Question 19: tan [(π / 4) + (1 / 2) * cos⁻¹ (a / b)] + tan [(π / 4) − (1 / 2) cos^{−1 .}(a / b)] = _________.

Solution:

tan [(π / 4) + (1 / 2) * cos⁻¹ (a / b)] + tan [(π / 4) − (1 / 2) cos^{−1 .}(a / b)]

Let (1 / 2) * cos⁻¹ (a / b) = θ

⇒ cos 2θ = a / b

Thus, tan [{π / 4} + θ] + tan [{π / 4} − θ] = [(1 + tanθ) / (1 − tanθ)]+ ([1 − tanθ] / [1 + tanθ])

= [(1 + tanθ)² + (1 − tanθ)²] / [(1 − tan²θ)]

= [1 + tan²θ + 2tanθ + 1 + tan²θ − 2tanθ] / [(1 – tan²θ)]

= 2 (1 + tan²θ) / [(1 – tan²θ)]

= 2 sec²θ cos²θ/(cos²θ – sin²θ)

= 2 /cos2θ

= 2 / [a / b]

= 2b / a

Question 20: The number of real solutions of tan⁻¹ √[x (x + 1)] + sin⁻¹ [√(x² + x + 1)] = π / 2 is ___________.

Solution:

tan⁻¹ √[x (x + 1)] + sin⁻¹ [√(x² + x + 1)] = π / 2

tan⁻¹ √[x (x + 1)] is defined when x (x + 1) ≥ 0 ..(i)

sin⁻¹ [√x² + x + 1] is defined when 0 ≤ x (x + 1) + 1 ≤ 1 or x² + x + 1  ≥ 1 ..(ii)

From (i) and (ii), x (x + 1) = 0 or x = 0 and -1.

Hence, the number of solution is 2.

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Check JEE Main Practice Papers 

JEE Main Trigonometry: Quick Formulas

Most of the questions asked on Trigonometry topic in JEE Mains exam can be solved by applying the formulas directly. Hence, students need to remember all the major formulae from this chapter. Here’s a list of the important formulae for JEE Main Trigonometry:

1. Trigonometric ratios of acute angles:

   1. Sin = ph cos = bh tan = pb

   2. cosec = hp sec = hb cot = bp

2. Trigonometric identities

   1. Sin2 + cos2 = 1

   2. Sec2 = 1 + tan2

   3. cosec2 = 1 + cot2

3. Trigonometric Ratios of compound angles

   1. Sin (A ± B) = sin A Cos B ± cos A Sin B

   2. cos (A ± B) = cos A Cos B ∓ sin A Sin B

   3. tan (A ± B) = tan A B 1+Atan B

   4. cot (A ± B) = cot Acot B 1 B A

   5. sin (A + B) sin (A - B) = sin2 A – Sin2 B = cos2 B – cos2 A

   6. cos (A + B) cos (A - B) = cos2 A - Sin2 B = cos2 B - sin2 A

   7. sin (A + B + C) = sin A Cos B Cos C + sin B CosA Cos C + sin C Cos A Cos B – sin A sin B sin C

   8. cos (A + B + C) = cos A Cos B Cos C – cos A Sin B Sin C – cos B Sin A Sin C – cos C sin A sin B

   9. tan (A + B + C) = tan A +B +C -tan Atan Btan C 1 -Atan B -tan Btan C-tan Ctan A

4. Transformation formulae in Trigonometry:

   1. 2sin A Cos B = sin (A + B) + Sin (A – B)

   2. 2sin B Cos A = sin (A + B) - Sin (A – B)

   3. 2 cos A Cos B = cos (A + B) + cos (A – B)

   4. 2 sin A Sin B = cos (A – B) – cos (A + B)

   5. Sin A + Sin B = 2 Sin (A + B2) Cos (A- B2)

   6. Sin A - Sin B = 2 Cos (A + B2) Sin (A- B2)

   7. Cos A + Cos B = 2 Cos (A + B2) Cos (A- B2)

   8. Cos A - Cos B = 2 Sin (A + B2) Sin (A- B2)

Check JEE Main Study Notes on Trigonometry