Trigonometry Formulas for SSC CGL
The trigonometry formulas SSC CGL tests most are: the standard values table (0°, 30°, 45°, 60°, 90°), the three Pythagorean identities (sin²θ + cos²θ = 1, sec²θ − tan²θ = 1, cosec²θ − cot²θ = 1), complementary-angle rules (sin(90° − θ) = cos θ), maximum and minimum values (a sin θ + b cos θ lies between −√(a² + b²) and +√(a² + b²)) and heights and distances. Learn these and you can solve almost every Tier 1 trigonometry question.
🔥 Key Exam Takeaway: sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2. If sec θ + tan θ = k, then sec θ − tan θ = 1/k. Max of a sin θ + b cos θ = √(a² + b²). Min of a² sec²θ + b² cosec²θ = (a + b)². tan 1° × tan 2° × … × tan 89° = 1.
SSC CGL Tier 1 usually has 2 to 4 questions on trigonometry and heights and distances, and Tier 2 has more. Every value and answer below was checked numerically in Python. Maths here is written in plain text (sin²θ means (sin θ)²).
Trigonometric Ratios Formula (Basic Definitions)
In a right-angled triangle with angle θ, P = perpendicular (opposite side), B = base (adjacent side), H = hypotenuse.
| Ratio | Formula | Reciprocal |
|---|---|---|
| sin θ | P ÷ H | cosec θ = H ÷ P |
| cos θ | B ÷ H | sec θ = H ÷ B |
| tan θ | P ÷ B = sin θ ÷ cos θ | cot θ = B ÷ P = cos θ ÷ sin θ |
Memory trick: "Pandit Badri Prasad / Har Har Bole": sin = P/H, cos = B/H, tan = P/B.
Pythagorean triplets to recognise instantly: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (20, 21, 29). If tan θ = 3/4, then sin θ = 3/5 and cos θ = 4/5 (for acute θ).
Trigonometry Values Table (0° to 90°)
| Angle | 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 | Not defined |
| cot | Not defined | √3 | 1 | 1/√3 | 0 |
| sec | 1 | 2/√3 | √2 | 2 | Not defined |
| cosec | Not defined | 2 | √2 | 2/√3 | 1 |
Trick for sin values: write √0/2, √1/2, √2/2, √3/2, √4/2 for 0°, 30°, 45°, 60°, 90°. That gives 0, 1/2, 1/√2, √3/2, 1. Cos is the same list in reverse.
Useful extra values (Tier 2): sin 15° = (√6 − √2)/4, cos 15° = (√6 + √2)/4, tan 15° = 2 − √3, sin 75° = (√6 + √2)/4, tan 75° = 2 + √3.
Trigonometric Identities for SSC CGL
| Identity | Rearranged forms |
|---|---|
| sin²θ + cos²θ = 1 | sin²θ = 1 − cos²θ; cos²θ = 1 − sin²θ |
| sec²θ − tan²θ = 1 | (sec θ + tan θ)(sec θ − tan θ) = 1 |
| cosec²θ − cot²θ = 1 | (cosec θ + cot θ)(cosec θ − cot θ) = 1 |
| tan θ × cot θ = 1 | sin θ × cosec θ = 1; cos θ × sec θ = 1 |
Most-used shortcut: if sec θ + tan θ = k, then sec θ − tan θ = 1/k. Adding: sec θ = (k + 1/k)/2; subtracting: tan θ = (k − 1/k)/2.
Complementary Angle Formulas (90° − θ)
| Formula | Example |
|---|---|
| sin(90° − θ) = cos θ | sin 70° = cos 20° |
| cos(90° − θ) = sin θ | cos 50° = sin 40° |
| tan(90° − θ) = cot θ | tan 65° = cot 25° |
| sec(90° − θ) = cosec θ | sec 10° = cosec 80° |
Results SSC loves:
- If A + B = 90°, then sin A = cos B, tan A × tan B = 1, sin²A + sin²B = 1.
- tan 1° × tan 2° × tan 3° × … × tan 89° = 1 (pairs like tan 1° × tan 89° = 1, and tan 45° = 1).
- sin²1° + sin²2° + … + sin²89° = 44.5 (44 pairs give 1 each, plus sin²45° = 1/2).
Sum, Difference and Double Angle Formulas (Tier 2)
| Formula | Expression |
|---|---|
| sin(A + B) | sin A cos B + cos A sin B |
| sin(A − B) | sin A cos B − cos A sin B |
| cos(A + B) | cos A cos B − sin A sin B |
| cos(A − B) | cos A cos B + sin A sin B |
| tan(A + B) | (tan A + tan B) ÷ (1 − tan A tan B) |
| sin 2A | 2 sin A cos A = 2 tan A ÷ (1 + tan²A) |
| cos 2A | cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A = (1 − tan²A) ÷ (1 + tan²A) |
| tan 2A | 2 tan A ÷ (1 − tan²A) |
| sin 3A | 3 sin A − 4 sin³A |
| cos 3A | 4 cos³A − 3 cos A |
Maximum and Minimum Value of Trigonometric Functions
| Expression | Minimum | Maximum |
|---|---|---|
| sin θ, cos θ | −1 | 1 |
| sin²θ, cos²θ | 0 | 1 |
| a sin θ + b cos θ | −√(a² + b²) | √(a² + b²) |
| sin θ × cos θ | −1/2 | 1/2 |
| tan²θ + cot²θ | 2 | No maximum |
| a² sec²θ + b² cosec²θ | (a + b)² | No maximum |
| a² tan²θ + b² cot²θ | 2ab | No maximum |
Example: maximum of 3 sin θ + 4 cos θ = √(9 + 16) = 5; minimum = −5. Example: minimum of 9 sec²θ + 16 cosec²θ = (3 + 4)² = 49.
Heights and Distances Formulas
| Situation | Formula |
|---|---|
| Tower of height h, observer at distance d, angle of elevation θ | tan θ = h ÷ d |
| Angle of depression from top equals angle of elevation from bottom | Alternate angles are equal |
| Two points on the same side at angles α and β (α > β), distance between them x | x = h(cot β − cot α) |
| Ladder of length L at angle θ with the ground | Height reached = L sin θ; foot distance = L cos θ |
| Shadow length s of a pole of height h | tan(sun's elevation) = h ÷ s |
Standard ratios: at 30°, height : distance = 1 : √3; at 45°, 1 : 1; at 60°, √3 : 1.
Trigonometry Questions for SSC CGL with Solutions
Q1. If sin θ + cos θ = √2 (0° < θ < 90°), find tan θ + cot θ. Solution: This holds only for θ = 45°. tan 45° + cot 45° = 1 + 1 = 2.
Q2. If tan θ = 3/4 (θ acute), find sin θ and cos θ. Solution: P = 3, B = 4, H = 5. sin θ = 3/5, cos θ = 4/5.
Q3. Find sin 30° + cos 60°. Solution: 1/2 + 1/2 = 1.
Q4. Find the maximum value of 3 sin θ + 4 cos θ. Solution: √(3² + 4²) = 5.
Q5. Find the minimum value of 9 sec²θ + 16 cosec²θ. Solution: (√9 + √16)² = (3 + 4)² = 49.
Q6. Find the minimum value of tan²θ + cot²θ. Solution: 2 (at θ = 45°).
Q7. The angle of elevation of the top of a tower from a point 30 m from its foot is 60°. Find its height. Solution: h = 30 × tan 60° = 30√3 m ≈ 51.96 m.
Q8. From two points on the same side of that tower (height 30√3 m), the angles of elevation are 30° and 60°. How far apart are the points? Solution: x = h(cot 30° − cot 60°) = 30√3 × (√3 − 1/√3) = 30√3 × 2/√3 = 60 m.
Q9. If sec θ + tan θ = 5, find sin θ. Solution: sec θ − tan θ = 1/5. sec θ = (5 + 1/5)/2 = 13/5, tan θ = (5 − 1/5)/2 = 12/5. sin θ = tan θ ÷ sec θ = 12/13.
Q10. Find tan 1° × tan 2° × … × tan 89°. Solution: 1.
Q11. Find sin²1° + sin²2° + … + sin²89°. Solution: 44 pairs (sin²θ + cos²θ = 1) + sin²45° = 44 + 1/2 = 44.5.
Q12. If θ = 30°, find (1 − tan²θ) ÷ (1 + tan²θ). Solution: This equals cos 2θ = cos 60° = 1/2.
Q13. Find sin 60° cos 30° + cos 60° sin 30°. Solution: sin(60° + 30°) = sin 90° = 1.
Q14. If cosec θ − cot θ = 1/3, find cos θ. Solution: cosec θ + cot θ = 3. cosec θ = (3 + 1/3)/2 = 5/3, cot θ = (3 − 1/3)/2 = 4/3. cos θ = cot θ ÷ cosec θ = 4/5.
Q15. Prove that (sec A + tan A)(1 − sin A) = cos A. Solution: (1/cos A + sin A/cos A)(1 − sin A) = (1 + sin A)(1 − sin A)/cos A = (1 − sin²A)/cos A = cos²A/cos A = cos A. ✔
More Trigonometry Questions (SSC CGL Tier 1 Pattern)
Q16. If sin A = 1/2 and cos B = 1/2 (A, B acute), find A + B. Solution: A = 30°, B = 60°, so A + B = 90°.
Q17. Find sin²30° + cos²45° + tan²60°. Solution: 1/4 + 1/2 + 3 = 15/4.
Q18. If tan θ + cot θ = 2, find tan¹⁰θ + cot¹⁰θ. Solution: tan θ + 1/tan θ = 2 only when tan θ = 1. So the value is 1 + 1 = 2.
Q19. If x = a cos θ and y = b sin θ, find x²/a² + y²/b². Solution: cos²θ + sin²θ = 1.
Q20. Find cos 1° × cos 2° × … × cos 90°. Solution: One factor is cos 90° = 0, so the product is 0.
Trigonometric Values Beyond 90° (ASTC Rule)
For angles in the second, third and fourth quadrants, use "All Silver Tea Cups": All ratios are positive in quadrant I, Sine (and cosec) in II, Tan (and cot) in III, Cos (and sec) in IV.
| Angle | Rule | Example |
|---|---|---|
| 180° − θ | sin positive, cos and tan negative | sin 150° = sin 30° = 1/2; cos 150° = −√3/2 |
| 180° + θ | tan positive, sin and cos negative | tan 225° = tan 45° = 1; sin 210° = −1/2 |
| 360° − θ | cos positive, sin and tan negative | cos 300° = cos 60° = 1/2; sin 330° = −1/2 |
SSC CGL Tier 1 rarely goes beyond 90°, but Tier 2 and CPO sometimes do, so keep this one table in your revision sheet.
Common Mistakes in Trigonometry Questions
| Mistake | Fix |
|---|---|
| Writing tan 90° = 1 or ∞ as a number | tan 90° is not defined |
| Treating sin²θ as sin(θ²) | sin²θ = (sin θ)² |
| Forgetting the sign range | sin and cos lie between −1 and 1, so sin θ = 1.2 is impossible |
| Using the wrong side in heights and distances | Angle of elevation is at the observer; tan = height ÷ horizontal distance |
| Applying a max-min shortcut without checking it is reachable | Check the equality case (for example tan²θ = cot²θ at 45°) |
Trigonometry Practice Questions (with Answers)
- sin²60° + cos²60° = ? → 1
- cos 60° + sin 30° + tan 45° = ? → 2
- Maximum value of sin θ cos θ? → 1/2
- If sin θ = cos θ (θ acute), find 2tan²θ + sin²θ − 1. → 3/2
- A 20 m ladder makes 60° with the ground. How high up the wall does it reach? → 10√3 m
- A tower 50√3 m high casts a 50 m shadow. Find the sun's elevation. → 60°
- If 4 sin²θ = 3 (θ acute), find θ. → 60°
- tan 15° = ? → 2 − √3
Trigonometry FAQs
Which trigonometry formulas are most important for SSC CGL?
The values table, the three Pythagorean identities, complementary-angle rules, the sec + tan and cosec + cot reciprocal tricks, max-min formulas and heights and distances.
How many trigonometry questions come in SSC CGL Tier 1?
Usually 2 to 4, including heights and distances. Tier 2 has more, along with geometry and mensuration.
What is the easiest way to remember the trigonometry table?
Write √0/2, √1/2, √2/2, √3/2, √4/2 for sin 0° to 90°. Reverse it for cos. tan = sin ÷ cos.
What is the maximum value of a sin θ + b cos θ?
√(a² + b²). The minimum is −√(a² + b²).
Is there a trigonometry formulas PDF for SSC CGL?
This page works as a formula sheet; use your browser's "Save as PDF" to keep an offline copy.
Related topics: revise algebra formulas for SSC CGL, mensuration formulas for SSC CGL and height and distance for SSC CGL next.
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