JEE Main Trigonometry Questions with Answer Keys PDF

JEE Main Trigonometry questions must be practised consistently and regularly. The formulae in Trigonometry must be memorised very well. Trigonometry is a high-weightage unit in the JEE Main 2026 Mathematics syllabus, and candidates can score well in it by practising the main questions regularly.

The NTA has conducted the JEE Main Session 1 examination from Jan 21 to Jan 29, 2026. Session 2 is scheduled from Apr 2 to Apr 9, 2026. The admit cards for the other exams of Session 2 will be released shortly.

JEE Main Trigonometry Questions with Answer Keys - Download PDF

The JEE Main Trigonometry questions with solutions asked in JEE Main 2026 are detailed and added to the PDF below. Students can download the PDF directly by clicking the link.

JEE Main Trigonometry Questions with Answer Keys

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Top JEE Main Trigonometry Questions with Answers 

Top JEE Main Trigonometry Questions with answers are provided below step-wise for students; these questions are frequently asked in JEE Main 2026. The weightage of trigonometry in the exam is almost 4 to 5 questions asked in each shift.

Question 1: The general solution of sin x − 3 sin²x + sin³x = cos x − 3 cos²x + cos³x is?

Answer: 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 sec 4θ − sec 2θ = 2, then the general value of θ is?

Answer: 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.

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Question 3: If tan (cot x) = cot (tan x), then sin 2x ?

Answer: 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 4: If the solutions for θ of cospθ + cosqθ = 0, p > 0, q > 0 are in A.P., then numerically the smallest common difference of A.P. is?

Answer: 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]∣}.

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Question 5: If α, β are different values of x satisfying a cosx + b sinx = c, then tan ([α + β] / 2) =?

Answer: a cost + b six = 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 6: In a triangle, the lengths of the two larger sides are 10 cm and 9 cm, respectively. If the angles of the triangle are in arithmetic progression, then what can the length of the third side be in cm?

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

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

Hence, ∠B = 60^o cost = [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

Question7: In triangle ABC, if ∠A = 45∘, ∠B = 75∘, then a + c√2 =?

Answer: ∠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 8: If cos−1 p + cos−1 q + cos−1 r = π then p² + q² + r² + 2pqr = ?

Answer: 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

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Question 9: tan [(π / 4) + (1 / 2) * cos−1 (a / b)] + tan [(π / 4) − (1 / 2) cos−1 .(a / b)] = ?

Answer: 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: The number of real solutions of tan−1 √[x (x + 1)] + sin−1 [√(x2 + x + 1)] = π / 2 is ?

Answer: 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 solutions is 2.