Take timed practice tests on Trigonometry for JEE Main and JEE Advanced with session-wise drills, score review, and explanation-led revision.
Six 20-question timed sessions plus one 60-question module test. Each question is original and calibrated from the uploaded material pattern without copying PDF wording.
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1. The value of sin²30° + cos²30° is:
Explanation: By the fundamental Pythagorean identity, sin²θ + cos²θ = 1 for every angle θ.
2. The value of sin 60° is:
Explanation: sin 60° = √3/2. Standard value — must be memorised along with sin 30° = 1/2, sin 45° = 1/√2.
3. The value of tan 45° is:
Explanation: tan 45° = sin 45°/cos 45° = (1/√2)/(1/√2) = 1.
4. sin(90° − θ) is equal to:
Explanation: sin(90° − θ) = cos θ. This co-function identity swaps sine and cosine when the angle is complementary.
5. If sec θ = 2, then cos θ =
Explanation: sec θ = 1/cos θ, so cos θ = 1/sec θ = 1/2.
6. 1 + tan²θ equals:
Explanation: Dividing sin²θ + cos²θ = 1 by cos²θ gives tan²θ + 1 = sec²θ.
7. In the second quadrant (90° < θ < 180°), which of the following is positive?
Explanation: In Q2: sin is positive; cos, tan, sec, cot are negative. Use CAST (or All-Sin-Tan-Cos) mnemonic.
8. cos(180° + θ) equals:
Explanation: 180° + θ is in Q3 (for 0
9. sin(A + B) equals:
Explanation: This is the standard sum formula: sin(A + B) = sin A cos B + cos A sin B.
10. sin 2θ in terms of sin θ and cos θ is:
Explanation: sin 2θ = sin(θ + θ) = sin θ cos θ + cos θ sin θ = 2 sin θ cos θ.
11. cos 2θ can be written as:
Explanation: cos 2θ = cos(θ+θ) = cos²θ − sin²θ. It can also be written as 2cos²θ − 1 or 1 − 2sin²θ.
12. 2 sin A cos B equals:
Explanation: Using product-to-sum: 2 sin A cos B = sin(A+B) + sin(A−B).
13. sin C + sin D equals:
Explanation: Sum-to-product: sin C + sin D = 2 sin((C+D)/2) cos((C−D)/2).
14. sin 3θ in terms of sin θ is:
Explanation: sin 3θ = 3 sin θ − 4 sin³θ. Derived by expanding sin(2θ + θ).
15. tan(θ/2) in terms of sin θ and cos θ is:
Explanation: Both forms are valid: tan(θ/2) = sin θ/(1 + cos θ) = (1 − cos θ)/sin θ. Verify by cross-multiplying.
16. The value of cos 20° cos 40° cos 80° is:
Explanation: Using the identity: cos θ cos(60°−θ) cos(60°+θ) = cos(3θ)/4. With θ = 20°: cos 20° cos 40° cos 80° = cos 60°/4 = (1/2)/4 = 1/8.
17. The maximum value of 3 sin θ + 4 cos θ is:
Explanation: a sin θ + b cos θ has maximum value √(a²+b²). Here: √(9+16) = √25 = 5.
18. The minimum value of 9 sec²θ + 16 cosec²θ is:
Explanation: Let f = 9 sec²θ + 16 cosec²θ = 9(1+tan²θ) + 16(1+cot²θ) = 25 + 9t + 16/t where t = tan²θ. By AM-GM: 9t + 16/t ≥ 2√(9×16) = 24. Minimum f = 25 + 24 = 49.
19. (sin A + cos A)² + (sin A − cos A)² equals:
Explanation: Expanding: (sin²A + 2sinAcosA + cos²A) + (sin²A − 2sinAcosA + cos²A) = 1 + 1 = 2.
20. If tan A = 1/2 and tan B = 1/3, then tan(A + B) equals:
Explanation: tan(A+B) = (tan A + tan B)/(1 − tan A tan B) = (1/2 + 1/3)/(1 − 1/6) = (5/6)/(5/6) = 1. So A + B = 45°.
21. The general solution of sin θ = 0 is:
Explanation: sin θ = 0 at all integer multiples of π: θ = nπ where n ∈ Z.
22. The general solution of cos θ = 0 is:
Explanation: cos θ = 0 at 90°, 270°, ... i.e. θ = (2n+1)π/2.
23. The general solution of sin θ = 1/2 is:
Explanation: General solution of sin θ = sin α is θ = nπ + (−1)ⁿ α. Here α = π/6, so θ = nπ + (−1)ⁿ π/6.
24. The general solution of cos θ = 1/2 is:
Explanation: General solution of cos θ = cos α is θ = 2nπ ± α. Here α = π/3, so θ = 2nπ ± π/3.
25. The number of solutions of 2 sin x = 1 in [0, 2π] is:
Explanation: sin x = 1/2 → x = π/6 and x = 5π/6 in [0, 2π]. These are 2 solutions.
26. Solutions of tan θ = √3 in [0°, 360°] are:
Explanation: tan θ = √3 = tan 60°. tan is positive in Q1 and Q3. Solutions: 60° and 60° + 180° = 240°.
27. The number of solutions of 2 sin²x + sin x − 1 = 0 in [0, 2π] is:
Explanation: Factor: (2 sin x − 1)(sin x + 1) = 0. So sin x = 1/2 (gives x = π/6, 5π/6) or sin x = −1 (gives x = 3π/2). Total: 3 solutions.
28. The general solution of sin x + cos x = 1 is:
Explanation: Write as √2 sin(x + π/4) = 1 → sin(x + π/4) = 1/√2. So x + π/4 = 2nπ + π/4 (giving x = 2nπ) or x + π/4 = 2nπ + 3π/4 (giving x = 2nπ + π/2).
29. The number of solutions of tan x = x in (0, π/2) is:
Explanation: tan x grows from 0 to +∞ on (0, π/2) while x grows from 0 to π/2. They intersect infinitely many... Actually, tan x and x intersect only once in (0, π/2) counting distinct intersections. But y = tan x starts above y = x for small x? At x→0: tan x ≈ x + x³/3 > x. At x→π/2: tan x → ∞ > x. So tan x > x on entire (0, π/2) → 0 solutions. Correct answer: 0.
30. The number of values of x in [0, 2π] satisfying sin x > cos x is:
Explanation: sin x > cos x → sin x − cos x > 0 → √2 sin(x − π/4) > 0 → sin(x − π/4) > 0. This holds when x − π/4 ∈ (0, π) i.e. x ∈ (π/4, 5π/4).
31. If sin θ = sin φ, then one relation between θ and φ is:
Explanation: General solution when sin θ = sin φ is θ = nπ + (−1)ⁿ φ. This covers both θ = φ and θ = π − φ.
32. The principal value of arcsin(1/2) is:
Explanation: arcsin(1/2) gives the angle in [−π/2, π/2] whose sine is 1/2. That angle is π/6 (30°).
33. In triangle ABC, the sine rule states:
Explanation: The sine rule relates each side to the sine of its opposite angle: a/sin A = b/sin B = c/sin C = 2R (circumradius).
34. In triangle ABC, a² = b² + c² − 2bc cos A. This is known as:
Explanation: The cosine rule expresses each side in terms of the other two and the included angle.
35. The area of triangle ABC using the sine formula is:
Explanation: Area = (1/2) × base × height = (1/2) bc sin A. This requires two sides and the included angle.
36. The circumradius R of triangle ABC satisfies:
Explanation: From sine rule: 2R = a/sin A so R = a/(2 sin A). Also R = abc/(4Δ) where Δ is the area. Both are correct.
37. The inradius r of a triangle is given by:
Explanation: Inradius r = Area/semi-perimeter = Δ/s, where s = (a+b+c)/2.
38. In a triangle, if a, b and angle A are given with a < b, the ambiguous case gives:
Explanation: When a b sin A, two triangles are possible (the 'ambiguous case' of SSA). The altitude from the given side can hit the base in two ways.
39. Which of the following is/are CORRECT? (I) sin(x+y) can exceed 1 (II) |sin x + cos x| ≤ √2 (III) sin x · cos x ≤ 1/2
Explanation: (I) False: sin(x+y) ≤ 1 always. (II) |sin x + cos x| = √2 |sin(x+π/4)| ≤ √2. True. (III) sin x cos x = (sin 2x)/2 ≤ 1/2. True. So II and III.
40. If tan A − tan B = x and cot B − cot A = y, then cot(A − B) equals:
Explanation: cot(A−B) = (cot A cot B + 1)/(cot A − cot B). Given: cot B − cot A = y → cot A − cot B = −y. Also tan A − tan B = x. Since cot A − cot B = −(tan B − tan A)/(tan A tan B), working through: cot(A−B) = 1/x + 1/y.
41. The value of sin 10° sin 50° sin 70° is:
Explanation: Use identity: sin θ sin(60°−θ) sin(60°+θ) = sin(3θ)/4. With θ = 10°: sin 10° sin 50° sin 70° = sin 30°/4 = (1/2)/4 = 1/8.
42. If sin A = 3/5 and A is in the second quadrant, cos 2A is:
Explanation: cos A = −4/5 (negative in Q2). cos 2A = 1 − 2sin²A = 1 − 2(9/25) = 1 − 18/25 = 7/25.
43. The value of cos 2π/7 + cos 4π/7 + cos 6π/7 is:
Explanation: The sum of cosines of angles in arithmetic progression: sum = (1/2)(sin(nα/2)/sin(α/2)) cos(first + last)/2. For 7th roots of unity: cos(2π/7) + cos(4π/7) + ... + cos(12π/7) = −1. Taking the real parts symmetrically: cos(2π/7) + cos(4π/7) + cos(6π/7) = −1/2.
44. cos 36° − cos 72° equals:
Explanation: cos 36° = (√5+1)/4 × 2 = (1+√5)/4. cos 72° = (√5−1)/4 × 2. cos36° − cos72° = [(1+√5)−(√5−1)]/4 = 2/4 = 1/2.
45. The number of solutions of sin²x + cos⁴x = 1 in [0, 2π] is:
Explanation: sin²x + cos⁴x = 1 → sin²x = 1 − cos⁴x = (1−cos²x)(1+cos²x) = sin²x(1+cos²x). Either sin²x = 0 or 1+cos²x = 1 → cos²x = 0. So sin x = 0 (x = 0, π, 2π) or cos x = 0 (x = π/2, 3π/2). In [0, 2π]: 0, π/2, π, 3π/2, 2π — but 0 and 2π share endpoints depending on interval interpretation. Distinct values: 4.
46. If A + B + C = π (angles of a triangle), then tan A + tan B + tan C equals:
Explanation: A+B+C = π → A+B = π−C. tan(A+B) = tan(π−C) = −tan C. (tanA+tanB)/(1−tanA tanB) = −tan C. Cross-multiplying: tan A + tan B + tan C = tan A tan B tan C.
47. In a triangle, cot(A/2) + cot(B/2) + cot(C/2) equals:
Explanation: This is a standard triangle result: cot(A/2) + cot(B/2) + cot(C/2) = s/r, where s = semi-perimeter and r = inradius.
48. The range of f(x) = 2 sin x + 3 cos x is:
Explanation: a sin x + b cos x has range [−√(a²+b²), √(a²+b²)]. Here: √(4+9) = √13. Range = [−√13, √13].
49. The period of |sin x| is:
Explanation: sin x has period 2π. Taking absolute value halves the period: |sin x| has period π.
50. sin A sin(60°−A) sin(60°+A) simplifies to:
Explanation: This is the identity: sin θ sin(60°−θ) sin(60°+θ) = sin 3θ/4. Used frequently in JEE to simplify triple products.
51. The value of (sin 90° + cos 0°) is:
Explanation: sin 90° = 1, cos 0° = 1. Sum = 2.
52. cos 2A in terms of tan A is:
Explanation: cos 2A = (cos²A − sin²A)/(cos²A + sin²A) = (1−tan²A)/(1+tan²A) after dividing numerator and denominator by cos²A.
53. The equation sin x = x/100 has approximately how many solutions in (−∞, ∞)?
Explanation: sin x oscillates between −1 and 1. The line y = x/100 intersects the sine curve when −1 ≤ x/100 ≤ 1 → −100 ≤ x ≤ 100. In this range, there are about 100/(π) ≈ 31.8 cycles. Each full cycle gives 2 crossings (going up and down). Approximately 2 × 100/π + 1 ≈ 63 solutions.
54. In a triangle ABC, if a = 2, b = 3 and C = 60°, then c equals:
Explanation: By cosine rule: c² = a² + b² − 2ab cos C = 4 + 9 − 2(2)(3)(1/2) = 13 − 6 = 7. So c = √7.
55. If sin(A−B)/sin(A+B) = (a−b)/(a+b) — which trigonometric rule does this represent?
Explanation: This is Napier's analogy or the tangent rule in a triangle. It relates angle differences to side ratios.
56. The maximum value of sin x(1 + cos x) is:
Explanation: f(x) = sin x + sin x cos x = sin x + (sin 2x)/2. Differentiating: cos x + cos 2x = 0 → cos x + 2cos²x − 1 = 0 → 2cos²x + cos x − 1 = 0 → cos x = 1/2 or −1. At cos x = 1/2 (x = π/3): f = (√3/2)(1 + 1/2) = 3√3/4.
57. sin 75° equals:
Explanation: sin 75° = sin(45°+30°) = sin45°cos30° + cos45°sin30° = (1/√2)(√3/2) + (1/√2)(1/2) = (√3+1)/(2√2) = (√6+√2)/4.