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JEE Main Test Series Trigonometry: Previous Year Questions & Solutions and Quick Formulas for Revision
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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 of 7%. Mostly questions asked on Trigonometry can be solved by simply applying trigonometry formulae, however, others will require the application of trigonometry tricks.

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. Spherical trigonometry focuses on curved triangles drawn on the surface of a sphere.

Check Tips for JEE Main Mathematics Preparation

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 Syllabus

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.

  1. 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

  1. 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˚ 

  1. 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

  1. 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

  1. 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

  1. 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

  1. 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

  1. 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

  1. 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

Check JEE Main Sample 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

  1. Trigonometric identities

    1. Sin2 + cos2 = 1

    2. Sec2 = 1 + tan2

    3. cosec2 = 1 + cot2

  1. 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

  1. 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

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