In this guide
Trigonometry is often the section where well-prepared SSC CGL candidates gain the most time. Questions look complicated — long expressions of sines, cosines and secants — but they almost always collapse under one of three identities, or under a simple substitution such as θ = 45°. If you know the value table and the identities by heart, trigonometry becomes a sprint.
Ratios in a right triangle
For an angle θ in a right triangle:
| Ratio | Definition | Reciprocal |
|---|---|---|
| sin θ | perpendicular/hypotenuse | cosec θ |
| cos θ | base/hypotenuse | sec θ |
| tan θ | perpendicular/base = sin θ/cos θ | cot θ |
Worked example: If tan θ = 5/12, find sin θ.
Perpendicular = 5, base = 12, hypotenuse = 13. sin θ = 5/13.
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 | not defined |
The three identities
- sin²θ + cos²θ = 1
- 1 + tan²θ = sec²θ
- 1 + cot²θ = cosec²θ
From identity 2: sec²θ − tan²θ = 1, so (sec θ + tan θ)(sec θ − tan θ) = 1.
Worked example: If sec θ + tan θ = 3, find sec θ − tan θ.
It is the reciprocal: 1/3.
Adding: 2 sec θ = 3 + 1/3 = 10/3, so sec θ = 5/3.
Similarly, (cosec θ + cot θ)(cosec θ − cot θ) = 1.
Complementary angles
- sin(90° − θ) = cos θ; cos(90° − θ) = sin θ
- tan(90° − θ) = cot θ; sec(90° − θ) = cosec θ
Worked example: Find tan 10° × tan 20° × tan 70° × tan 80°.
tan 10° × tan 80° = tan 10° × cot 10° = 1; likewise tan 20° × tan 70° = 1. Product = 1.
Worked example: If sin 3θ = cos(θ − 2°), find θ (with 3θ acute).
3θ + θ − 2 = 90 → 4θ = 92 → θ = 23°.
The substitution method
When a question gives an expression valid for all θ (an identity), put θ = 45° or another convenient value and test the options.
Worked example: Simplify (1 − sin²θ)(1 + tan²θ).
Using the identities: cos²θ × sec²θ = 1. Check at θ = 45°: (1 − ½)(1 + 1) = 1. ✓
Maximum and minimum values
- sin θ and cos θ lie between −1 and 1.
- a sin θ + b cos θ lies between −√(a² + b²) and +√(a² + b²).
- For a sin²θ + b cos²θ, the values lie between a and b.
Worked example: Find the maximum value of 3 sin θ + 4 cos θ.
√(9 + 16) = 5.
Worked example: Find the minimum value of 9 tan²θ + 4 cot²θ.
By AM ≥ GM: minimum = 2√(9 × 4) = 12.
Useful results
- sin θ + cos θ = √2 → θ = 45°.
- If sin θ + sin²θ = 1, then cos²θ = sin θ, so cos²θ + cos⁴θ = 1.
- sin²θ + cos²θ appears everywhere — look for it first.
Common traps
| Trap | Correct approach |
|---|---|
| Confusing sin and cos values | Learn the table in both directions |
| Forgetting tan 90° is undefined | Watch for division by zero |
| Expanding long expressions | Try an identity or substitution first |
Practice
- If cos θ = 3/5, find tan θ.
- Find the value of sin²30° + cos²60° + tan²45°.
- If cosec θ − cot θ = 1/4, find cosec θ + cot θ.
- Find sin 25° cos 65° + cos 25° sin 65°.
- Find the maximum value of 5 sin θ + 12 cos θ.
- Simplify (sec²θ − 1)(cosec²θ − 1).
Answers: 1. 4/3. 2. 3/2. 3. 4. 4. 1. 5. 13. 6. 1.
A note on dates and numbers. Exam patterns, vacancies and schedules change from year to year. Always confirm the current details in the latest notification on the Staff Selection Commission website .
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