📑 Topics on this page (Jump to Topic):
📌 1. Standard Values
| θ | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan | 0 | 1/√3 | 1 | √3 | undefined |
- sinθ = perpendicular/hypotenuse, cosθ = base/hypotenuse, tanθ = perpendicular/base; cosec, sec, cot are reciprocals.
📌 2. Identities
📘 Formulas:
- sin²θ + cos²θ = 1
- sec²θ − tan²θ = 1 ⇒ if secθ + tanθ = k then secθ − tanθ = 1/k
- cosec²θ − cot²θ = 1
- sin(90° − θ) = cosθ, tan(90° − θ) = cotθ, sec(90° − θ) = cosecθ
- Max of a sinθ + b cosθ = √(a² + b²), min = −√(a² + b²)
- If A + B = 90° then tanA·tanB = 1, sin²A + sin²B = 1
💡 Shortcut Trick: Value-putting: substitute θ = 45° or 30° and match options — very fast in SSC.
⚠️ Exam Traps:
- sin²θ means (sinθ)², not sin(θ²).
📌 3. Degree and Radian
📘 Formulas:
- π radian = 180°; 1 radian = 180°/π ≈ 57.3° (≈ 57°16′ taking π = 22/7)
- Degree → radian: × π/180; radian → degree: × 180/π
- 1° = 60′ (minutes), 1′ = 60″ (seconds)
- Arc length l = rθ and sector area = ½r²θ (θ in radians)
- Clock: minute hand 6° per minute, hour hand 0.5° per minute
| Degree | 30° | 45° | 60° | 90° | 120° | 180° | 270° | 360° |
|---|---|---|---|---|---|---|---|---|
| Radian | π/6 | π/4 | π/3 | π/2 | 2π/3 | π | 3π/2 | 2π |
⚠️ Exam Traps:
- In l = rθ always keep θ in radians; convert first if degrees are given.
- For angles with minutes (e.g. 40°20′) first convert 20′ = 1/3°.
📌 4. Practice MCQs
📝 Practice MCQs (prepared by SF Prep — not PYQs)
Q1. The value of 4 sin²30° + cos²0° is: Practice
Correct answer: (B)
Solution: Substituting standard values, 4 sin²30° + cos²0° = 2.
Solution: Substituting standard values, 4 sin²30° + cos²0° = 2.
Q2. The value of sec 40° / cosec 50° is: Practice
Correct answer: (C)
Solution: cosec(90° − θ) = secθ → cosec 50° = sec 40° → ratio = 1.
Solution: cosec(90° − θ) = secθ → cosec 50° = sec 40° → ratio = 1.
Q3. The maximum value of 6 sinθ + 8 cosθ is: Practice
Correct answer: (D)
Solution: Maximum = √(6² + 8²) = 10.
Solution: Maximum = √(6² + 8²) = 10.
Q4. If secθ + tanθ = 5, then secθ − tanθ = ? Practice
Correct answer: (A)
Solution: sec²θ − tan²θ = 1 → (secθ − tanθ) = 1/5 = 1/5.
Solution: sec²θ − tan²θ = 1 → (secθ − tanθ) = 1/5 = 1/5.
Q5. If sinθ = 3/5 (θ acute), then cosecθ = ? Practice
Correct answer: (B)
Solution: Perpendicular 3, hypotenuse 5 → base = 4; cosecθ = 5/3.
Solution: Perpendicular 3, hypotenuse 5 → base = 4; cosecθ = 5/3.
Q6. sin⁴θ − cos⁴θ + cos²θ = ? Practice
Correct answer: (C)
Solution: Simplifying with identities (sin² + cos² = 1, sec² − tan² = 1) gives sin²θ.
Solution: Simplifying with identities (sin² + cos² = 1, sec² − tan² = 1) gives sin²θ.
Q7. tan 35° × tan 55° + sin²35° + sin²55° = ? Practice
Correct answer: (D)
Solution: A + B = 90° → tanA·tanB = 1 and sin²A + sin²B = 1 → 1 + 1 = 2.
Solution: A + B = 90° → tanA·tanB = 1 and sin²A + sin²B = 1 → 1 + 1 = 2.
Q8. The value of tan 45° + cot 45° is: Practice
Correct answer: (A)
Solution: Substituting standard values, tan 45° + cot 45° = 2.
Solution: Substituting standard values, tan 45° + cot 45° = 2.
Q9. The value of sec 15° / cosec 75° is: Practice
Correct answer: (B)
Solution: cosec(90° − θ) = secθ → cosec 75° = sec 15° → ratio = 1.
Solution: cosec(90° − θ) = secθ → cosec 75° = sec 15° → ratio = 1.
Q10. The maximum value of 7 sinθ + 24 cosθ is: Practice
Correct answer: (C)
Solution: Maximum = √(7² + 24²) = 25.
Solution: Maximum = √(7² + 24²) = 25.
Q11. If secθ + tanθ = 4, then secθ − tanθ = ? Practice
Correct answer: (D)
Solution: sec²θ − tan²θ = 1 → (secθ − tanθ) = 1/4 = 1/4.
Solution: sec²θ − tan²θ = 1 → (secθ − tanθ) = 1/4 = 1/4.
Q12. If sinθ = 20/29 (θ acute), then cosecθ = ? Practice
Correct answer: (A)
Solution: Perpendicular 20, hypotenuse 29 → base = 21; cosecθ = 29/20.
Solution: Perpendicular 20, hypotenuse 29 → base = 21; cosecθ = 29/20.
Q13. (1 + tan²θ) cos²θ = ? Practice
Correct answer: (B)
Solution: Simplifying with identities (sin² + cos² = 1, sec² − tan² = 1) gives 1.
Solution: Simplifying with identities (sin² + cos² = 1, sec² − tan² = 1) gives 1.
Q14. The value of cos²30° − sin²30° is: Practice
Correct answer: (C)
Solution: Substituting standard values, cos²30° − sin²30° = 1/2.
Solution: Substituting standard values, cos²30° − sin²30° = 1/2.
Q15. The value of tan 15° / cot 75° is: Practice
Correct answer: (D)
Solution: cot(90° − θ) = tanθ → cot 75° = tan 15° → ratio = 1.
Solution: cot(90° − θ) = tanθ → cot 75° = tan 15° → ratio = 1.
Q16. The maximum value of 8 sinθ + 15 cosθ is: Practice
Correct answer: (A)
Solution: Maximum = √(8² + 15²) = 17.
Solution: Maximum = √(8² + 15²) = 17.
Q17. If secθ + tanθ = 3/2, then secθ − tanθ = ? Practice
Correct answer: (B)
Solution: sec²θ − tan²θ = 1 → (secθ − tanθ) = 1/3/2 = 2/3.
Solution: sec²θ − tan²θ = 1 → (secθ − tanθ) = 1/3/2 = 2/3.
Q18. tanθ × cotθ + sin²θ = ? Practice
Correct answer: (C)
Solution: Simplifying with identities (sin² + cos² = 1, sec² − tan² = 1) gives 1 + sin²θ.
Solution: Simplifying with identities (sin² + cos² = 1, sec² − tan² = 1) gives 1 + sin²θ.
Q19. tan 40° × tan 50° + sin²40° + sin²50° = ? Practice
Correct answer: (D)
Solution: A + B = 90° → tanA·tanB = 1 and sin²A + sin²B = 1 → 1 + 1 = 2.
Solution: A + B = 90° → tanA·tanB = 1 and sin²A + sin²B = 1 → 1 + 1 = 2.
Q20. The value of cosec 30° + sec 60° is: Practice
Correct answer: (A)
Solution: Substituting standard values, cosec 30° + sec 60° = 4.
Solution: Substituting standard values, cosec 30° + sec 60° = 4.
Q21. The angles of a triangle are in the ratio 1 : 2 : 3. The largest angle in radians is: Practice
Correct answer: (C)
Solution: Angles 30°, 60°, 90° → 90° = π/2.
Solution: Angles 30°, 60°, 90° → 90° = π/2.
Q22. Convert 7π/12 radian into degrees. Practice
Correct answer: (A)
Solution: 7π/12 × 180/π = 7 × 15 = 105°.
Solution: 7π/12 × 180/π = 7 × 15 = 105°.
Q23. In a circle of radius 21 cm, the length of an arc subtending 60° at the centre is (π = 22/7): Practice
Correct answer: (D)
Solution: l = rθ = 21 × π/3 = 21 × 22/21 = 22 cm.
Solution: l = rθ = 21 × π/3 = 21 × 22/21 = 22 cm.
Q24. Convert 40°20′ into radians. Practice
Correct answer: (B)
Solution: 40°20′ = 40⅓° = 121/3° → 121/3 × π/180 = 121π/540.
Solution: 40°20′ = 40⅓° = 121/3° → 121/3 × π/180 = 121π/540.
Q25. An arc of length 15 cm in a circle of radius 10 cm subtends what angle at the centre? Practice
Correct answer: (D)
Solution: θ = l/r = 15/10 = 1.5 radian.
Solution: θ = l/r = 15/10 = 1.5 radian.