Trigonometry — Ratios, Identities, and Aptitude Shortcuts | SSC, Placement

Quick Concept Summary

Trigonometry in aptitude tests focuses on: 1. Values of sin, cos, tan at standard angles (0°, 30°, 45°, 60°, 90°) 2. Pythagorean identities 3. Complementary angle rules 4. Height and distance problems (using tan, sin, cos)

The Six Trigonometric Ratios

In a right-angled triangle with angle θ:

Standard Angle Values — Memorise This Table

Memory trick for sin row (0° to 90°): √0/2, √1/2, √2/2, √3/2, √4/2 = 0, 1/2, 1/√2, √3/2, 1

cos is sin in reverse. tan = sin/cos.

Pythagorean Identities

These are the most tested. Learn all three forms of each:

Identity 1:

• sin²θ + cos²θ = 1
• sin²θ = 1 − cos²θ
• cos²θ = 1 − sin²θ

Identity 2:

• 1 + tan²θ = sec²θ
• sec²θ − tan²θ = 1
• sec²θ − 1 = tan²θ

Identity 3:

• 1 + cot²θ = cosec²θ
• cosec²θ − cot²θ = 1
• cosec²θ − 1 = cot²θ

Complementary Angle Rules

For angles A + B = 90° (complementary):

Worked Examples

Example 1 — Standard Values

Q: Evaluate: 2sin²30° + cos²60° − tan²45°

sin 30° = 1/2 → sin²30° = 1/4

cos 60° = 1/2 → cos²60° = 1/4

tan 45° = 1 → tan²45° = 1

Result = 2(1/4) + 1/4 − 1 = 1/2 + 1/4 − 1 = 3/4 − 1 = −1/4

Example 2 — Identity Simplification

Q: Simplify: (sin²θ + cos²θ) / (1 + tan²θ)

Numerator = sin²θ + cos²θ = 1 (identity 1)

Denominator = 1 + tan²θ = sec²θ (identity 2)

Result = 1 / sec²θ = cos²θ

Example 3 — Complementary Angle

Q: If A + B = 90°, simplify: sin A / cos B + cos A / sin B

Since A + B = 90°, we have B = 90° − A.

cos B = cos(90° − A) = sin A

sin B = sin(90° − A) = cos A

Substituting:

sin A / sin A + cos A / cos A = 1 + 1 = 2

Example 4 — Given one ratio, find others

Q: If sin θ = 3/5, find cos θ and tan θ.

sin²θ + cos²θ = 1

(3/5)² + cos²θ = 1 → 9/25 + cos²θ = 1 → cos²θ = 16/25 → cos θ = 4/5

tan θ = sin θ / cos θ = (3/5) / (4/5) = 3/4

Example 5 — Height and Distance

Q: A tower casts a shadow of 20 m. The angle of elevation of the sun is 30°. Find the height of the tower.

tan(angle of elevation) = Height / Shadow length

tan 30° = h / 20

1/√3 = h / 20

h = 20/√3 = 20√3/3 ≈ 11.55 m

Or rationalised: 20√3 / 3 m

Example 6 — Two-angle Height Problem

Q: From a point on the ground, the angle of elevation of the top of a building is 45°. From a point 20 m closer, it is 60°. Find the height of the building.

Let height = h, closer distance = d.

From far point: tan 45° = h/(d+20) → 1 = h/(d+20) → h = d+20 ... (1)

From close point: tan 60° = h/d → √3 = h/d → h = d√3 ... (2)

From (1) and (2): d + 20 = d√3 → 20 = d(√3 − 1) → d = 20/(√3−1)

Rationalise: d = 20(√3+1)/((√3)²−1²) = 20(√3+1)/2 = 10(√3+1)

h = d√3 = 10√3(√3+1) = 10(3 + √3) = 30 + 10√3 m

Shortcuts and Tricks

Trick 1 — (sec²θ − tan²θ) and (cosec²θ − cot²θ)

Both always equal 1. Frequently tested in SSC as direct simplification.

Trick 2 — sin θ × cosec θ = 1, cos θ × sec θ = 1, tan θ × cot θ = 1

These products always equal 1 — useful for quick simplification.

Trick 3 — sin 45° = cos 45° = 1/√2

When you see 45° in a height-distance problem, the height equals the distance (tan 45° = 1).

Trick 4 — tan 60° = √3 ≈ 1.73, tan 30° = 1/√3 ≈ 0.577

Commit to memory — height-distance problems almost always use 30°, 45°, or 60°.

Exam Pattern

SSC CGL/CHSL: 3–5 questions per paper from trigonometry. Mix of identity simplification and standard angle evaluation. Occasionally height-distance (1 question).

Bank PO: Rarely tested directly, but appears in some IBPS papers.

Placement (TCS, Infosys): May appear in quantitative section. Usually standard angle values or basic identity simplification.

Practice Questions