JEE Main Trigonometry Important Questions

JEE Main Trigonometry Important Questions - Getting ready to give the JEE Mains 2026? Want to learn how to successfully clear the Trigonometry section given in the exam? We have got you covered. You can ace the JEE Main 2026 by practicing and solving the trigonometry section of the JEE Mains Previous Year Question Papers. While attempting the trigonometry part of the question paper, one needs to have a solid understanding of the principles of trigonometry and efficient problem-solving skills. Trigonometry involves the study of calculus, linear algebra, and statistics.
Also Read: JEE Main 2026 Admit Card

JEE Main Trigonometry Important Questions

Solving the JEE Main Trigonometry Important Questions is not an easy task. Given below are some important questions along with their step-by-step solution for students to practice and ace their JEE Mains. These questions will also help you in managing your time when taking the JEE Mains 2026 Exam.

Question 1: 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 2 : 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 3 : If tan3θ−1tan3θ+1=3√, then the general value of θ is

Solution :

tan3θ−1 / tan3θ+1=3√ Þ tan3θ−tan(π/4) / 1+tan3θ.tan(π/4)

= √3Þ tan(3θ−π4)=tanπ3 Þ 3θ−(π/4)

= nπ+(π/3) Þ 3 θ=π+7π12Þ θ

= nπ3+7π36.

Question 4: The equation 3cosx+4sinx=6

Solution: 3cosx+4sinx=6 Þ

35 cos x+45 sin x=65 Þ cos(x−θ)=65,

[where θ=cos−1(3/5)]

So, that equation has no solution.

Question 5: 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 6: 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 is _________.

Solution:

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

Since angles ∠A, ∠B, and ∠C are in AP.

Hence, ∠B = 60o 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 7: In triangle ABC, if ∠A = 45∘, ∠B = 75∘, then a + c√2 = __________.

Solution:

∠C = 180o − 45o − 75o = 60o

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 8 : If sin2θ=cosθ,0<θ<π, then the possible values of θ are

A) 90,60,30

B) 90,150,60

C) 90,45,150

D) 90,30,150

Solution : sin2θ=cosθ⇒cosθ=cos(π/2−2θ) ⇒ θ

= 2nπ±(π/2−2θ)⇒θ±2θ=2nπ±π/2 i.e.

3 θ=2nπ+π/2⇒θ = 13(2nπ+π2) and −θ=2nπ−π/2⇒θ=−(2nπ−π/2)

Hence values of θ between 0 and π are π/6, π/2, 5π/6 i.e., 30,90,150°

Question 9. 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 10 : sec2 (tan−1 2) + cosec2 (cot−1 3) = _________.

Solution:

Let (tan−1 2) = α

⇒ tan α = 2 and cot−1 3 = β

⇒ cot β = 3

sec2 (tan−1 2) + cosec2 (cot−1 3)

= sec2 α + cosec2β

= 1 + tan2α + 1 + cot2β

= 2 + (2)2 + (3)2

= 15

Trigonometric Ratios

In trigonometry, there are six basic ratios that help in building a relationship between the ratios of sides of a right triangle with the angle. Given below are the six ratios that one needs to know :

• sin θ = Perpendicular / Hypotenuse
• cos θ = Base / Hypotenuse
• tan θ = Perpendicular / Base
• cot θ = 1/tan θ = Base / Perpendicular
• sec θ = 1/cos θ = Hypotenuse / Base
• cosec θ = 1/sinθ = Hypotenuse / Perpendicular

Also, check

Important Trigonometric Formulas

1. Trigonometry Ratio Formulas

• Sinθ = Opposite Side / Hypotenuse
• Cosθ = Adjacent Side / Hypotenuse
• Tanθ = Opposite Side / Adjacent Side
• Cotθ = 1/tanθ = Adjacent Side / Opposite Side
• Secθ = 1/cosθ = Hypotenuse / Adjacent Side
• Cosecθ = 1/sinθ = Hypotenuse / Opposite Side

2. Trigonometry Formulas ( Pythagorean Identities )

• sin²θ + cos²θ = 1
• tan2θ + 1 = sec2θ
• cot2θ + 1 = cosec2θ

3. Sine & Cosine Law

• a/sinA = b/sinB = c/sinC
• c2 = a2 + b2 – 2ab cos C
• a2 = b2 + c2 – 2bc cos A
• b2 = a2 + c2 – 2ac cos B

About Trigonometry

One of the most important branches in mathematics, Trigonometry, is the study of the relation between the ratios of the sides of a right-angled triangle and their angles. The trigonometric ratios used to examine this connection are sine, cosine, tangent, cotangent, secant, and cosecant. The term trigonometry is a 16th-century Latin derivation of the Greek mathematician Hipparchus' notion. The word trigonometry is formed by combining the words 'Trigonon', which means triangle and 'Metron', which means measure.
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JEE Main Mathematics Question Paper

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