Introduction to Trigonometry — Animated Diagram

📐 Trigonometric Ratios

The Six Ratios — In a right △ABC with ∠B = 90°, for angle A (= θ)

sin θ = O/H cos θ = A/H tan θ = O/A

cosec θ = H/O sec θ = H/A cot θ = A/O

O = Opposite side (BC), A = Adjacent side (AB), H = Hypotenuse (AC). Mnemonic: SOH-CAH-TOA

🔺 Interactive Right Triangle — Drag to explore

Angle θ: 30° Hypotenuse: 5

Key Fact

Ratios depend only on the angle

If two right triangles have the same acute angle θ, they are similar. All trig ratios will be equal regardless of the size of the triangle. This is why trigonometry works!

Reciprocal Pairs

sin ↔ cosec, cos ↔ sec, tan ↔ cot

sin θ × cosec θ = 1
cos θ × sec θ = 1
tan θ × cot θ = 1
Also: tan θ = sin θ / cos θ and cot θ = cos θ / sin θ

Caution

Which side is "opposite"?

The opposite and adjacent sides change depending on which angle you're looking at. The hypotenuse always stays the same — it's always the side opposite the 90° angle.

💡 Indian textbook mnemonic: "Some People Have Curly Brown Hair Through Proper Brushing" → sin = P/H, cos = B/H, tan = P/B (P = perpendicular, B = base, H = hypotenuse)

📊 Standard Angle Values

The sin trick — memorise this one row, derive everything!

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

cos is the same values in reverse order. tan = sin ÷ cos for each angle.

θ	0°	30°	45°	60°	90°
sin θ	0	1/2	1/√2	√3/2	1
cos θ	1	√3/2	1/√2	1/2	0
tan θ	0	1/√3	1	√3	N.D.
cosec θ	N.D.	2	√2	2/√3	1
sec θ	1	2/√3	√2	2	N.D.
cot θ	N.D.	√3	1	1/√3	0

Where do these values come from?

30° and 60° — Take an equilateral triangle of side 2. Drop a perpendicular from a vertex to the opposite side. You get a 30-60-90 triangle with sides 1, √3, 2.
45° — Take an isosceles right triangle with equal legs of 1. By Pythagoras, hypotenuse = √2. This gives a 45-45-90 triangle.
0° and 90° — Limiting cases. As θ → 0°, opposite → 0 and adjacent → hypotenuse. As θ → 90°, opposite → hypotenuse and adjacent → 0.
N.D. = Not Defined (division by zero). tan 90° = sin 90°/cos 90° = 1/0 → undefined.

30-60-90

The "Half Equilateral" Triangle

Side opposite 30° = 1
Side opposite 60° = √3
Hypotenuse = 2

sin 30° = 1/2
cos 30° = √3/2
tan 30° = 1/√3

45-45-90

The Isosceles Right Triangle

Both legs = 1
Hypotenuse = √2

sin 45° = 1/√2 = √2/2
cos 45° = 1/√2 = √2/2
tan 45° = 1/1 = 1

💡 Quick check: As angle increases from 0° to 90°, sin increases (0→1), cos decreases (1→0), and tan increases (0→∞). If your answer violates this pattern, recheck!

🔗 Trigonometric Identities

Three Fundamental Identities (all from Pythagoras!)

① sin²θ + cos²θ = 1

② 1 + tan²θ = sec²θ

③ 1 + cot²θ = cosec²θ

Identity ① is the master identity. Divide by cos²θ → ②. Divide by sin²θ → ③.

Proof of Identity ① : sin²θ + cos²θ = 1

Start with Pythagoras: In right △ABC with ∠B = 90°: BC² + AB² = AC²
Divide both sides by AC² (= H²): BC²/AC² + AB²/AC² = 1
Recognise the ratios: (BC/AC)² + (AB/AC)² = 1
Substitute: sin²A + cos²A = 1

∴ sin²θ + cos²θ = 1 for any acute angle θ ■

Deriving Identity ② from ①

Start: sin²θ + cos²θ = 1
Divide every term by cos²θ: sin²θ/cos²θ + cos²θ/cos²θ = 1/cos²θ
Simplify: tan²θ + 1 = sec²θ

∴ 1 + tan²θ = sec²θ ■ (Similarly, dividing ① by sin²θ gives ③)

Useful Forms

Rearrangements to memorise

sin²θ = 1 − cos²θ
cos²θ = 1 − sin²θ
sec²θ − tan²θ = 1
cosec²θ − cot²θ = 1
(secθ−tanθ)(secθ+tanθ) = 1
(cosecθ−cotθ)(cosecθ+cotθ) = 1

Strategy

How to prove identities

Start from the more complex side
Convert everything to sin and cos
Use sin²θ + cos²θ = 1 liberally
Factor, combine fractions, rationalise
Never cross the = sign (don't assume the result)

📝 Prove: (1 + tan²θ) / (1 + cot²θ) = tan²θ

LHS = (1 + tan²θ) / (1 + cot²θ)
    = sec²θ / cosec²θ     [using identities ② and ③]
    = (1/cos²θ) / (1/sin²θ)
    = sin²θ / cos²θ
    = tan²θ = RHS

∴ LHS = RHS. Hence proved. ■

📝 Prove: (sin θ − cos θ + 1) / (sin θ + cos θ − 1) = 1 / (sec θ − tan θ)

Divide numerator & denominator by cos θ:
LHS = (tan θ − 1 + sec θ) / (tan θ + 1 − sec θ)
= (tan θ + sec θ − 1) / (tan θ − sec θ + 1)
Use sec²θ − tan²θ = 1, so 1 = (sec θ − tan θ)(sec θ + tan θ):
Numerator = (tan θ+sec θ) − (sec²θ−tan²θ)/(sec θ+tan θ)...
Multiply num & denom by (sec θ − tan θ):
= (secθ−tanθ)(tanθ+secθ−1) / ((secθ−tanθ)(tanθ−secθ+1))
= (sec²θ−tan²θ − (secθ−tanθ)) / ((secθ−tanθ)(tanθ−secθ+1))
= (1−secθ+tanθ) / ((secθ−tanθ)(tanθ−secθ+1))
= 1/(secθ−tanθ) = RHS

∴ LHS = RHS. Hence proved. ■

This is HOTS level — key trick is multiplying by the conjugate (sec θ − tan θ).

💡 When stuck on an identity proof, try multiplying numerator and denominator by a conjugate like (1 + sin θ), (sec θ + tan θ), etc.

🔄 Complementary Angles

If A + B = 90° then A and B are complementary

sin(90°−θ) = cos θ cos(90°−θ) = sin θ

tan(90°−θ) = cot θ sec(90°−θ) = cosec θ

Each trig function of an angle = its "co-function" of the complement. The "co" in co-sine literally means "complement's sine"!

Why does this work?

In right △ABC with ∠B = 90°, angles A and C are complementary (A + C = 90°).

sin A = BC/AC and cos C = BC/AC → sin A = cos C = cos(90°−A)
cos A = AB/AC and sin C = AB/AC → cos A = sin C = sin(90°−A)
The "opposite" side for one angle is the "adjacent" side for the other — that's why sin swaps with cos!

Expression	Equals	Reason
sin 72°	cos 18°	72 + 18 = 90
tan 55°	cot 35°	55 + 35 = 90
sec 80°	cosec 10°	80 + 10 = 90
cos 0°	sin 90° = 1	0 + 90 = 90
sin²20° + sin²70°	1	sin²20° + cos²20° = 1

📝 Evaluate: sin 65° / cos 25°

sin 65° = sin(90°−25°) = cos 25°
∴ sin 65° / cos 25° = cos 25° / cos 25° = 1

Answer: 1

📝 If tan A = cot B, prove that A + B = 90°

Given: tan A = cot B
We know: cot B = tan(90°−B)
∴ tan A = tan(90°−B)
∴ A = 90° − B
∴ A + B = 90°

Hence proved. ■

📝 Evaluate: tan 1° · tan 2° · tan 3° · … · tan 89°

Pair them: tan 1°·tan 89°, tan 2°·tan 88°, …
tan k° · tan(90°−k°) = tan k° · cot k° = 1
44 such pairs, each = 1. Middle term = tan 45° = 1
Product = 1⁴⁴ × 1 = 1

Answer: 1

Classic CBSE question — the pairing trick is the key insight.

💡 Whenever you see two angles adding to 90° in a problem, immediately think complementary angle formulas. This simplifies most "evaluate" questions instantly.

✏️ Problem Solving Techniques

Types of Questions in CBSE Exams

Type 1: Given one trig ratio, find all others (use Pythagoras)
Type 2: Evaluate expressions using standard angle values
Type 3: Prove trigonometric identities (LHS = RHS)
Type 4: Simplify using complementary angle relations
Type 5: Find angle θ given a trig equation

📝 Type 1 — Given tan θ = 3/4, find all trig ratios

tan θ = O/A = 3/4
So take: O = 3k, A = 4k (for some positive k)
By Pythagoras: H = √(9k² + 16k²) = √(25k²) = 5k

sin θ = O/H = 3k/5k = 3/5
cos θ = A/H = 4k/5k = 4/5
tan θ = 3/4 (given)
cosec θ = 5/3, sec θ = 5/4, cot θ = 4/3

sin θ = 3/5, cos θ = 4/5, cosec θ = 5/3, sec θ = 5/4, cot θ = 4/3

The 3-4-5 is a Pythagorean triplet. Other common ones: 5-12-13, 8-15-17, 7-24-25.

📝 Type 2 — Evaluate: 2 tan²45° + cos²30° − sin²60°

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

Answer: 2

📝 Type 5 — If sin(A+B) = 1 and cos(A−B) = √3/2, find A and B

sin(A+B) = 1 → A+B = 90° … (i)
cos(A−B) = √3/2 → A−B = 30° … (ii)

Adding (i) and (ii): 2A = 120° → A = 60°
Subtracting: 2B = 60° → B = 30°

A = 60°, B = 30°

Solve simultaneous equations — convert trig equation to angle equation first.

Exam Tip

Common Pythagorean Triplets

3, 4, 5
5, 12, 13
8, 15, 17
7, 24, 25
Multiples also work: 6,8,10 etc.

Common Mistakes

Avoid these errors!

sin²θ ≠ sin(θ²) — it means (sin θ)²
sin(A+B) ≠ sin A + sin B
√(sin²θ + cos²θ) ≠ sin θ + cos θ
tan 90° is undefined, not infinity
1/sin θ = cosec θ, NOT sin⁻¹θ (that's inverse sine)

Quick Check

Range of trig ratios

0 ≤ sin θ ≤ 1
0 ≤ cos θ ≤ 1
0 ≤ tan θ (can be > 1)
cosec θ ≥ 1
sec θ ≥ 1
cot θ ≥ 0

If your answer is sin θ = 2, it's wrong!

💡 Exam strategy: In "prove that" questions, always start from the more complex side. In "evaluate" questions, substitute standard values first, then simplify. In "find angle" questions, match with standard values.