Trigonometry Shortcuts: All Formulas and Tricks for SSC CGL in One Place

Trigonometry is the topic most SSC CGL aspirants either fear or skip — and both responses are mistakes. Skipping it forfeits 4–6 guaranteed marks in Tier 1 and significantly more in Tier 2. Fearing it is unnecessary because SSC CGL trigonometry is far more restricted in scope than school trigonometry — it tests a small, well-defined set of values, identities, and height-distance templates repeatedly.

A student who masters the standard angle values, five core identities, the complementary angle relationship, and the four height-distance configurations can solve every SSC CGL trigonometry question ever set — without advanced mathematics, without calculus, and without the anxiety that comes from treating trigonometry as an infinite subject.

This guide is that complete, restricted, exam-focused trigonometry package — every concept you need, nothing you do not, with worked examples throughout.

Part 1: Standard Angle Values — The Foundation of Everything

Every trigonometry question in SSC CGL either directly uses these values or reduces to them. There is no shortcut to memorization — but there is a shortcut to memorizing them.

The Hand Trick — Memorizing sin Values Instantly

Hold your left hand in front of you with fingers spread. Number your fingers 0 to 4 from thumb to little finger representing angles 0°, 30°, 45°, 60°, 90°.

For sin of any standard angle:

sin θ = √(finger number) ÷ 2

• sin 0° = √0 ÷ 2 = 0
• sin 30° = √1 ÷ 2 = 1/2
• sin 45° = √2 ÷ 2
• sin 60° = √3 ÷ 2
• sin 90° = √4 ÷ 2 = 1

For cos: read the same table in reverse order
cos 0° = 1, cos 30° = √3/2, cos 45° = √2/2, cos 60° = 1/2, cos 90° = 0

Complete Standard Values Table

Reciprocal Relationships — Never Confuse Again

Quotient Relationships

tan θ = sin θ ÷ cos θ

cot θ = cos θ ÷ sin θ

Part 2: The Five Core Identities

These five identities cover every identity-based question in SSC CGL. Master them in this order.

Identity 1: Pythagorean Identity (The Most Important)

sin²θ + cos²θ = 1

Derived forms:

sin²θ = 1 − cos²θ

cos²θ = 1 − sin²θ

(sin θ + cos θ)² = 1 + 2 sinθ cosθ

(sin θ − cos θ)² = 1 − 2 sinθ cosθ

Identity 2: Tangent Pythagorean Identity

1 + tan²θ = sec²θ

Derived forms:

sec²θ − tan²θ = 1

(sec θ + tan θ) × (sec θ − tan θ) = 1

Exam shortcut: If sec θ + tan θ = x, then sec θ − tan θ = 1/x

Identity 3: Cotangent Pythagorean Identity

1 + cot²θ = cosec²θ

Derived forms:

cosec²θ − cot²θ = 1

(cosec θ + cot θ) × (cosec θ − cot θ) = 1

Exam shortcut: If cosec θ + cot θ = x, then cosec θ − cot θ = 1/x

Identity 4: Complementary Angle Relationships

sin(90° − θ) = cos θ

cos(90° − θ) = sin θ

tan(90° − θ) = cot θ

cot(90° − θ) = tan θ

sec(90° − θ) = cosec θ

cosec(90° − θ) = sec θ

Memory rule: Any trig function of (90° − θ) = its co-function of θ.
sin ↔ cos, tan ↔ cot, sec ↔ cosec

Identity 5: Negative Angle Relationships

sin(−θ) = −sin θ → odd function

cos(−θ) = cos θ → even function

tan(−θ) = −tan θ → odd function

Part 3: Solving Identity-Based Questions — The Substitution Shortcut

For questions like "if sin θ + cos θ = √2, find sin θ × cos θ" — always use algebraic manipulation of the Pythagorean identity rather than solving for θ.

Worked Example 1:
If sin θ + cos θ = √2, find sin θ × cos θ.

• Square both sides: sin²θ + 2 sinθ cosθ + cos²θ = 2
• 1 + 2 sinθ cosθ = 2
• sinθ cosθ = 1/2

Worked Example 2:
If sec θ + tan θ = 3, find sec θ − tan θ and then find sec θ and tan θ.

• sec θ − tan θ = 1/3 (reciprocal identity)
• Adding: 2 sec θ = 3 + 1/3 = 10/3 → sec θ = 5/3
• Subtracting: 2 tan θ = 3 − 1/3 = 8/3 → tan θ = 4/3

Worked Example 3:
If sin θ − cos θ = 1/5, find sin θ + cos θ.

• (sin θ + cos θ)² = 1 + 2 sinθ cosθ
• (sin θ − cos θ)² = 1 − 2 sinθ cosθ = 1/25
• Adding both: (sinθ + cosθ)² + (sinθ − cosθ)² = 2
• (sinθ + cosθ)² = 2 − 1/25 = 49/25
• sin θ + cos θ = 7/5

Worked Example 4:
If tan θ + cot θ = 2, find tan²θ + cot²θ.

• (tan θ + cot θ)² = tan²θ + 2 + cot²θ = 4
• tan²θ + cot²θ = 4 − 2 = 2

Part 4: The θ = 45° Substitution Trick

For expressions involving sums of trig functions that look complex, substituting θ = 45° often simplifies to a verifiable answer in seconds — useful for checking options in MCQ format.

Example: Which option equals (1 + tan²θ) ÷ tan²θ?

• At θ = 45°: tan 45° = 1
• Expression = (1+1) ÷ 1 = 2
• Check options: cosec²θ at 45° = (√2)² = 2 ✓
• Answer: cosec²θ

Example: Simplify sin⁴θ + cos⁴θ + 2 sin²θ cos²θ

• At θ = 45°: (1/√2)⁴ + (1/√2)⁴ + 2 × (1/2) × (1/2) = 1/4 + 1/4 + 1/2 = 1
• This equals (sin²θ + cos²θ)² = 1² = 1 always ✓

Part 5: Complementary Angle Sum Shortcuts

SSC CGL frequently tests sums like:
sin 1° × sin 89° + sin 2° × sin 88° + … + sin 44° × sin 46° + sin²45°

The pattern: sin k° = cos(90° − k°)

So sin 1° × sin 89° = sin 1° × cos 1° = (1/2) sin 2°

More useful direct pattern: sin²θ + cos²θ = 1 always

Worked Example:
sin²10° + sin²20° + sin²30° + sin²40° + sin²50° + sin²60° + sin²70° + sin²80° + sin²90°

• Pair: sin²10° + sin²80° = sin²10° + cos²10° = 1
• Pair: sin²20° + sin²70° = 1
• Pair: sin²30° + sin²60° = (1/2)² + (√3/2)² = 1/4 + 3/4 = 1
• Pair: sin²40° + sin²50° = 1
• Remaining: sin²90° = 1
• Total = 4 × 1 + 1 = 5

Part 6: Height and Distance — The 4 Templates

Height and distance problems use trigonometry to find heights of towers, widths of rivers, and distances between points. Every SSC CGL height-distance question fits one of four templates.

Template 1: Angle of Elevation — Single Observer

Setup: Observer at ground level looks up at an object. Angle of elevation = θ.

tan θ = Height ÷ Distance

Worked Example:
From a point 30 m away from the base of a tower, the angle of elevation of the top is 30°. Height?

• tan 30° = h ÷ 30
• 1/√3 = h ÷ 30
• h = 30/√3 = 10√3 ≈ 17.32 m

Template 2: Two Observers — Same Side

Setup: Two observers at distances d₁ and d₂ from tower base see top at angles α and β (α > β since closer observer sees steeper angle).

h = (d₂ − d₁) × tan α × tan β ÷ (tan α − tan β)

Worked Example:
From two points 20 m and 30 m away from a tower, angles of elevation are 60° and 45°. Height?

• h = (30 − 20) × tan 60° × tan 45° ÷ (tan 60° − tan 45°)
• = 10 × √3 × 1 ÷ (√3 − 1)
• = 10√3 ÷ (√3 − 1) × (√3 + 1)/(√3 + 1)
• = 10√3(√3 + 1) ÷ 2 = 5√3(√3 + 1) = 5(3 + √3) = 15 + 5√3 m

Template 3: Angle of Depression

Observer at height looks down at object. Angle of depression = θ.

Key property: Angle of depression from A to B = Angle of elevation from B to A (alternate interior angles with horizontal).

Worked Example:
From top of a 60 m cliff, angle of depression of a boat is 30°. Distance of boat from base?

• tan 30° = 60 ÷ d
• 1/√3 = 60 ÷ d
• d = 60√3 ≈ 103.9 m

Template 4: Two Angles — Finding Height Without Given Distance

Setup: From a point, angle of elevation = α. After moving d meters toward the tower, angle = β.

h = d × tan α × tan β ÷ (tan β − tan α)

Worked Example:
A person walks 20 m toward a tower. Angle of elevation changes from 30° to 60°. Height?

• h = 20 × tan 30° × tan 60° ÷ (tan 60° − tan 30°)
• = 20 × (1/√3) × √3 ÷ (√3 − 1/√3)
• = 20 × 1 ÷ (2/√3)
• = 20√3 ÷ 2 = 10√3 m

Part 7: Allied Angle Values (Beyond 90°)

SSC CGL Tier 2 occasionally tests angles beyond 90°. Use these rules:

Memory aid — ASTC (All Students Take Calculus):

• All positive in Q1
• Sin positive in Q2
• Tan positive in Q3
• Cos positive in Q4

Standard reductions:

• sin(180° − θ) = sin θ
• cos(180° − θ) = −cos θ
• sin(180° + θ) = −sin θ
• cos(360° − θ) = cos θ

Worked Example: sin 150°

• 150° = 180° − 30° → sin(180° − 30°) = sin 30° = 1/2

Worked Example: cos 210°

• 210° = 180° + 30° → cos(180° + 30°) = −cos 30° = −√3/2

One-Page Formula Reference

Pythagorean Identities

• sin²θ + cos²θ = 1
• 1 + tan²θ = sec²θ
• 1 + cot²θ = cosec²θ

Reciprocal Product Shortcuts

• (sec θ + tan θ) × (sec θ − tan θ) = 1
• (cosec θ + cot θ) × (cosec θ − cot θ) = 1
• sin θ × cosec θ = 1
• cos θ × sec θ = 1
• tan θ × cot θ = 1

Sum/Difference Formulas (Tier 2)

• sin(A+B) = sin A cos B + cos 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)

Double Angle Formulas (Tier 2)

• sin 2θ = 2 sin θ cos θ
• cos 2θ = cos²θ − sin²θ = 1 − 2sin²θ = 2cos²θ − 1
• tan 2θ = 2 tan θ ÷ (1 − tan²θ)