Chapter 15 of 19

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Andhra Pradesh (BIEAP)Class 10 Mathematics★ AP Board (BSEAP) SSC Mathematics · ~8 marks · Board: AP Board

Trigonometry

TRIGONOMETRIC RATIOS (sin, cos, tan). Standard angle values. Fundamental identities. Complementary angle relationships. AP Board Class 10 Mathematics.

⏱ 12 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Define sin, cos, tan, cosec, sec, cot for a right triangle and establish their reciprocal relationships
2Recall the standard angle table for 0°, 30°, 45°, 60°, 90° from memory
3Apply fundamental identities: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, 1 + cot²θ = cosec²θ
4Use complementary angle relations: sin(90°−θ) = cosθ, etc.
5Prove trigonometric identities by simplifying one side using the fundamental identity

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Why this chapter matters

Trigonometry is one of the HIGHEST-SCORING and most-tested chapters in AP SSC Mathematics — typically 6–10 marks per paper. The standard angle table (sin/cos/tan for 0°, 30°, 45°, 60°, 90°) must be memorised — most trigonometry questions in AP SSC are solved by substituting from this table. Proving trigonometric identities is a standard 4-mark question type. Complementary angle relationships allow rapid simplification of complex expressions. This chapter forms the foundation for Class 11 trigonometry.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Trigonometry

"Trigonometry relates ANGLES to SIDES. It lets you measure the HEIGHT of a mountain without climbing it."

Ratios in a Right Triangle

sin θ = Opposite/Hypotenuse. cos θ = Adjacent/Hypotenuse. tan θ = Opposite/Adjacent = sin θ/cos θ. cosec θ = 1/sin θ. sec θ = 1/cos θ. cot θ = 1/tan θ.

Standard Angle Values — MUST Memorise

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	ND

Fundamental Identity: sin²θ + cos²θ = 1

Derived: 1 + tan²θ = sec²θ. 1 + cot²θ = cosec²θ.

Complementary Angles: sin(90°−θ) = cos θ. cos(90°−θ) = sin θ. tan(90°−θ) = cot θ.

Trigonometric Ratios — Detailed

In a right triangle ABC with right angle at B and ∠C = θ:

sin θ = Opposite/Hypotenuse = AB/AC
cos θ = Adjacent/Hypotenuse = BC/AC
tan θ = Opposite/Adjacent = AB/BC
cosec θ = 1/sin θ = AC/AB
sec θ = 1/cos θ = AC/BC
cot θ = 1/tan θ = BC/AB

'Remember the mnemonic: Some People Have Curly Black Hairs Through Proper Brushing. SOH-CAH-TOA is another name for it — Sin=Opp/Hyp, Cos=Adj/Hyp, Tan=Opp/Adj.'

Standard Angle Values Table

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	Not Defined
cosec θ	Not Defined	2	√2	2/√3	1
sec θ	1	2/√3	√2	2	Not Defined
cot θ	Not Defined	√3	1	1/√3	0

'Pattern to memorise sin: sin 0°=√0/2, sin 30°=√1/2, sin 45°=√2/2, sin 60°=√3/2, sin 90°=√4/2. Cos is sin in REVERSE order.'

Trigonometric Identities

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

Derivation: From sin²θ + cos²θ = 1 (dividing by cos²θ): tan²θ + 1 = sec²θ. (Dividing by sin²θ): 1 + cot²θ = cosec²θ.

Worked Examples — Identity Proofs

Example 1: Prove that (sin A + cos A)² + (sin A − cos A)² = 2. LHS = sin²A+cos²A+2sinAcosA + sin²A+cos²A−2sinAcosA = 1+1 = 2 = RHS. ✓

Example 2: Prove that (1 + tan²A) / (1 + cot²A) = tan²A. LHS = sec²A / cosec²A = (1/cos²A) / (1/sin²A) = sin²A/cos²A = tan²A = RHS. ✓

Example 3: Prove that sin⁴θ − cos⁴θ = sin²θ − cos²θ. LHS = (sin²θ+cos²θ)(sin²θ−cos²θ) = 1×(sin²θ−cos²θ) = sin²θ−cos²θ = RHS. ✓

Example 4: Prove that (cosec A − sin A)(sec A − cos A) = 1/(tan A + cot A). LHS = (1/sinA − sinA)(1/cosA − cosA) = ((1−sin²A)/sinA)((1−cos²A)/cosA) = (cos²A/sinA)(sin²A/cosA) = cosA·sinA. RHS = 1/(tanA + cotA) = 1/(sinA/cosA + cosA/sinA) = 1/((sin²A+cos²A)/(sinAcosA)) = 1/(1/(sinAcosA)) = sinAcosA = LHS. ✓

Example 5: Prove that √[(1−sinθ)/(1+sinθ)] = secθ − tanθ. LHS = √[(1−sinθ)(1−sinθ)/(1+sinθ)(1−sinθ)] = √[(1−sinθ)²/(1−sin²θ)] = √[(1−sinθ)²/cos²θ] = (1−sinθ)/cosθ = 1/cosθ − sinθ/cosθ = secθ − tanθ = RHS. ✓

'In identity proofs, convert everything to sin and cos. This is the UNIVERSAL strategy. Then simplify using sin²θ+cos²θ=1.'

Complementary Angle Relationships

For complementary angles (θ and 90°−θ):

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

Example 6: Evaluate: sin 30° cos 60° + cos 30° sin 60°. sin 30°=1/2, cos 60°=1/2, cos 30°=√3/2, sin 60°=√3/2. = (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1.

Alternatively: sin30°cos60°+cos30°sin60° = sin(30°+60°) = sin 90° = 1. ✓

Example 7: Evaluate: tan 45° / (cosec 30° + sec 60° − cot 45°). tan45°=1, cosec30°=2, sec60°=2, cot45°=1. = 1/(2+2−1) = 1/3.

Example 8: Prove: cos A / sin(90°−A) + sin A / cos(90°−A) = 2. sin(90°−A) = cos A, cos(90°−A) = sin A. LHS = cos A/cos A + sin A/sin A = 1 + 1 = 2. ✓

Common Mistakes — Extended

Mixing up sin and cos: In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse. The ratio depends on which angle is θ.
Forgetting the ± when taking square roots: sin²θ=¼ → sinθ=±½, not just +½. However, for acute angles (0°<θ<90°), all ratios are positive.
Assuming all six ratios are independent: They are linked by identities. Knowing sinθ is enough to find all others for an acute angle.
Confusing values at 0° and 90°: tan 0°=0 but tan 90° is undefined (division by zero). sin 0°=0, sin 90°=1.
Complementary angle confusion: sin(90°−θ) = cos θ, NOT sin(θ−90°).
Half-angle mistakes: sin(θ/2) ≠ (sin θ)/2. Trigonometric functions are NOT linear.

AP Exam Focus

Topic	Marks	Frequency
Trigonometric ratios (finding values)	3	Very Common
Identity proofs	4-5	Very Common
Complementary angle evaluation	3-4	Common
Standard angle values	2-3	Very Common
Mixed expression evaluation	4	Common

Self-Test Questions

If sin θ = 3/5, find cos θ and tan θ. (Answer: cos θ = 4/5, tan θ = 3/4)
Evaluate: sin 60° cos 30° + cos 60° sin 30°. (Answer: (√3/2)(√3/2) + (1/2)(1/2) = 3/4+1/4=1)
Prove: (1 + sin θ)/cos θ + cos θ/(1 + sin θ) = 2 sec θ. (Answer: Take LCM: ((1+sinθ)²+cos²θ)/(cosθ(1+sinθ)) = (1+2sinθ+sin²θ+cos²θ)/(cosθ(1+sinθ)) = (2+2sinθ)/(cosθ(1+sinθ)) = 2(1+sinθ)/(cosθ(1+sinθ)) = 2/cosθ = 2 sec θ)
Evaluate: sec 45° + tan 45° − sin 90° − cos 0°. (Answer: √2 + 1 − 1 − 1 = √2 − 1)
If tan A = cot B, prove that A + B = 90°. (Answer: tan A = cot B → tan A = tan(90°−B) → A = 90°−B → A+B=90°)
Find the value of: (sin² 30° + cos² 60° − tan² 45°) / (sec² 60° − cosec² 30°). (Answer: (1/4+1/4−1)/(4−4) = (−1/2)/0 = undefined)
Prove: (tan A + sec A − 1)/(tan A − sec A + 1) = (1 + sin A)/cos A. (Hint: Use identity sec²A−tan²A=1)

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Trigonometric Ratios and Identities

RATIOS: sin θ = Opp/Hyp. cos θ = Adj/Hyp. tan θ = Opp/Adj. cosec θ = 1/sinθ. sec θ = 1/cosθ. cot θ = 1/tanθ. STANDARD TABLE: sin 0°=0, sin 30°=1/2, sin 45°=1/√2, sin 60°=√3/2, sin 90°=1. cos is sin read in REVERSE (cos 0°=1, cos 90°=0). tan 30°=1/√3, tan 45°=1, tan 60°=√3, tan 90°=undefined. IDENTITIES: sin²θ+cos²θ=1 → cos²θ=1−sin²θ, sin²θ=1−cos²θ. 1+tan²θ=sec²θ. 1+cot²θ=cosec²θ. COMPLEMENTARY: sin(90−θ)=cosθ, cos(90−θ)=sinθ, tan(90−θ)=cotθ, sec(90−θ)=cosecθ.

MEMORY FOR STANDARD TABLE: 'sin goes 0, 1, 2, 3, 4 — under the root — divided by 2': sin 0°=√0/2=0, sin 30°=√1/2=1/2, sin 45°=√2/2=1/√2, sin 60°=√3/2, sin 90°=√4/2=1. cos is the REVERSE of sin. PROVING IDENTITIES: Start with the MORE COMPLEX side. Convert everything to sin and cos. Use sin²+cos²=1 and derived forms. DO NOT move terms across the = sign.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Writing tan 90° = 0 instead of undefined

✓ tan 90° = sin 90°/cos 90° = 1/0 = UNDEFINED (not 0). The value goes to infinity as θ approaches 90°. Similarly: cot 0° = cos 0°/sin 0° = 1/0 = UNDEFINED. cosec 0° = 1/sin 0° = 1/0 = UNDEFINED. sec 90° = 1/cos 90° = UNDEFINED. Remember: any ratio with 0 in denominator is UNDEFINED, not zero.

WATCH OUT

✗ Writing sin²θ + cos²θ = 1 as sin θ + cos θ = 1

✓ The fundamental identity is sin²θ + cos²θ = 1 (SQUARED). NOT sin θ + cos θ = 1 (unsquared — this is generally false). Example: sin 30° + cos 30° = 1/2 + √3/2 ≠ 1. But sin²30° + cos²30° = 1/4 + 3/4 = 1 ✓.

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Trigonometry?

2 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

2 questions~2 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•SIX TRIG RATIOS: sin θ = O/H. cos θ = A/H. tan θ = O/A. cosec θ = H/O = 1/sin θ. sec θ = H/A = 1/cos θ. cot θ = A/O = 1/tan θ. Also: tan θ = sin θ/cos θ. cot θ = cos θ/sin θ. H = hypotenuse, O = opposite side, A = adjacent side.
•STANDARD ANGLE TABLE: sin 0°=0, sin 30°=½, sin 45°=1/√2, sin 60°=√3/2, sin 90°=1. cos is sin IN REVERSE: cos 0°=1, cos 30°=√3/2, cos 45°=1/√2, cos 60°=½, cos 90°=0. tan: 0, 1/√3, 1, √3, undefined. MEMORY: sin increases from 0 to 1; cos decreases from 1 to 0; tan increases from 0 to ∞.
•THREE FUNDAMENTAL IDENTITIES: (1) sin²θ + cos²θ = 1 (derives: sin²θ = 1−cos²θ and cos²θ = 1−sin²θ). (2) 1 + tan²θ = sec²θ (derives: tan²θ = sec²θ−1 and sec²θ−tan²θ=1). (3) 1 + cot²θ = cosec²θ (derives: cot²θ = cosec²θ−1 and cosec²θ−cot²θ=1). These three come from Pythagoras theorem divided by different sides.
•COMPLEMENTARY ANGLE RELATIONS: 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. KEY USE: sin 35°/cos 55° = sin 35°/sin 35° = 1 (since cos 55° = cos(90°−35°) = sin 35°).
•PROVING IDENTITIES — METHOD: (1) Start with the MORE COMPLEX side. (2) Express everything in terms of sin and cos. (3) Apply the relevant identity (sin²+cos²=1 and its derived forms). (4) Factorise or simplify. (5) Show it equals the other side. NEVER transfer terms across the equality sign during a proof.
•FINDING ALL RATIOS FROM ONE: If sin θ = 3/5 (given), draw a right triangle: O=3, H=5. By Pythagoras: A = √(H²−O²) = √(25−9) = 4. Then: cos θ=4/5, tan θ=3/4, cosec θ=5/3, sec θ=5/4, cot θ=4/3. This drawing method is reliable — always draw the triangle.
•EVALUATION QUESTIONS: Substitute standard angle values. sin²60° = (√3/2)² = 3/4. 2tan 45° = 2×1 = 2. cos²30° = (√3/2)² = 3/4. Compute step by step — substituting all values before simplifying reduces errors.
•UNDEFINED VALUES: tan 90° = undefined (sin 90° / cos 90° = 1/0). cot 0° = undefined (cos 0° / sin 0° = 1/0). sec 90° = undefined. cosec 0° = undefined. ANY ratio with zero in denominator → UNDEFINED, not zero.
•SPECIAL PROPERTY: sin θ × cosec θ = 1 (always). cos θ × sec θ = 1 (always). tan θ × cot θ = 1 (always). These come from the reciprocal definitions and are useful shortcuts in simplification.
•RANGE OF TRIG FUNCTIONS: −1 ≤ sin θ ≤ 1 and −1 ≤ cos θ ≤ 1 for all θ. So sin²θ ≤ 1 and cos²θ ≤ 1. If a question gives sin θ = 5/3 (> 1), it is IMPOSSIBLE for a real angle θ. This range check can verify problem setup.

Andhra Pradesh (BIEAP) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Engineering and architecture — angles and forces

Every structural engineer uses trig ratios to resolve forces into horizontal and vertical components. A cable at angle θ to the horizontal carrying tension T has horizontal component T cos θ and vertical component T sin θ. Roof pitch, bridge cable angles, and crane arm positions are all designed using sin and cos. The standard trig table values (30°, 45°, 60°) correspond to common engineering angles: 45° for symmetric loads, 60° for steep inclines, 30° for gentle slopes.

Sound, light, and waves

All wave phenomena (sound, light, radio waves, water waves) are described mathematically by sinusoidal functions: y = A sin(ωt + φ). The amplitude A, frequency ω, and phase φ are trig parameters. Recording studios, noise-cancelling headphones, and 5G antenna design all require deep understanding of sin and cos. The Fourier Transform — the mathematical tool that converts any signal into its frequency components — is entirely built on trig identities. Class 10 trigonometry is the starting point for all of signal processing.

Satellite navigation and angle of elevation

When tracking a satellite, ground stations measure the ELEVATION ANGLE (angle of elevation to the satellite above the horizon) and AZIMUTH (compass direction). The satellite's altitude can be calculated using trigonometry: altitude = distance × tan(elevation angle). The same principle is used for radio telescopes, radar systems, and anti-aircraft artillery. The angle of elevation concept from Class 10 applications of trigonometry is the exact geometry used in satellite communication.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Standard angle evaluation (2 marks): substitute values from the table FIRST, then simplify. Write each substitution explicitly: 'sin 30° = 1/2, cos 60° = 1/2, ...' before combining. Substitution errors happen when values are recalled during calculation rather than written first.
Identity proof (4 marks): write 'LHS =' on one line, work downward showing each step. Write '= RHS' as the final line. Do NOT write LHS = RHS at the start — that is the conclusion, not the method. AP Board examiners look for step-by-step working; each algebraic step earns marks.
Find all ratios from given ratio (2 marks): draw a right triangle. Label the sides using the given ratio (O, A, H). Use Pythagoras to find the missing side. List all 6 ratios from the triangle. This drawing method is much more reliable than algebraic manipulation.
Complementary angle simplification (2 marks): look for pairs θ and (90°−θ) in the expression. Replace using complementary relations: sin(90°−θ)=cosθ, etc. Then simplify. Common AP SSC patterns: sin θ/cos(90°−θ) = sin θ/sin θ = 1.
Identity selection: when starting a proof, write the THREE fundamental identities at the top of the answer space. Having them visible prevents mid-proof memory lapses and saves time.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Research the unit circle definition of trigonometry — extending sin and cos beyond 0°–90° to all angles. On the unit circle (radius 1 centred at origin), the coordinates of a point at angle θ from the positive x-axis are (cos θ, sin θ). This definition extends sin and cos to angles >90°, <0°, and beyond 360°. Research how this leads to the four quadrants with different signs for sin/cos, the periodicity of trig functions, and eventually to complex numbers and Euler's formula e^(iθ) = cos θ + i sin θ.
Investigate Fourier Analysis — any periodic wave (no matter how complex) can be decomposed into a sum of simple sin and cos waves at different frequencies. Joseph Fourier (1822) proved this, and it revolutionised mathematics, physics, and engineering. Every mp3 audio file, JPEG image, and mobile phone signal uses the Fast Fourier Transform (FFT) to compress and transmit data efficiently. The FFT computes millions of sin/cos additions per second — the Class 10 trig table is the conceptual foundation.
Explore the origin of the word 'sine' — it comes from the Sanskrit word 'jya' meaning 'bowstring' (the chord of a circle). Indian mathematicians (Aryabhata, 499 CE; Brahmagupta, 628 CE) were among the first to systematically tabulate trigonometric values and use them in astronomy. The word was later mistranslated through Arabic ('jiba') into Latin ('sinus' = bay or bosom). Research how medieval Indian mathematics contributed to the global development of trigonometry centuries before it reached Europe.
Research the triple angle identities: sin 3θ = 3 sin θ − 4 sin³ θ and cos 3θ = 4 cos³ θ − 3 cos θ. These were used by ancient mathematicians to trisect angles approximately and to compute cube roots. The Bhaskara I approximation sin(x°) ≈ 4x(180−x)/(40500−x(180−x)) (valid within 1% for 0° to 180°) was a remarkable 7th-century Indian formula — more accurate than contemporary European methods.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

AP Board SSC (Class 10)Very High — Trigonometry (identity proofs and evaluations) is one of the most heavily tested chapters in AP SSC Mathematics, typically 6–10 marks

JEE Main and AdvancedVery High — Trigonometry (identity proofs, equations, inverse trig) is among the top 3 most tested topics in JEE Mathematics

NTSE (Mathematics)High — Trigonometric ratios and identities are tested in NTSE Stage I and II

AP EAPCETVery High — Trigonometry (functions, identities, equations, inverse) forms a major chapter in EAPCET Mathematics

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FIVE-STEP METHOD: (1) Choose the MORE COMPLEX side (usually the one with fractions, products, or more terms). (2) Convert all terms to sin and cos using the ratio definitions. (3) Simplify using algebra — common denominators, factoring, expanding brackets. (4) Substitute sin²θ+cos²θ=1 (or derived forms: 1−sin²θ=cos²θ, sec²θ−tan²θ=1, etc.) at the right moment. (5) Show you have reached the simpler side. CRITICAL RULE: NEVER move terms from LHS to RHS or vice versa — this is NOT proving, this is rearranging. You must simplify ONE side to equal the other independently. If stuck: try expanding the simpler side and working upward to the complex form.

Because cos θ = sin(90°−θ) — cosine of any angle equals sine of the complementary angle. As θ increases from 0° to 90°, (90°−θ) decreases from 90° to 0°. So cos reads the same values as sin but in REVERSE ORDER: sin 0°=0 and cos 90°=0 (same value, opposite ends). sin 30°=1/2 and cos 60°=1/2 (same value, complementary angles). This is not a coincidence — it is the definition of complementary trig ratios. Memory trick: in a right triangle, sin of one acute angle = cos of the OTHER acute angle (since the two acute angles add to 90°).

Match the FORM of what you want to prove with the right identity: (1) If the target has sec² or tan², think 1+tan²=sec². (2) If it has cosec² or cot², think 1+cot²=cosec². (3) If it has a mix of sin and cos, think sin²+cos²=1. (4) If you see (1−sin θ)(1+sin θ), recognise it equals cos²θ (difference of squares + identity). (5) If you see (1−cos θ)(1+cos θ), it equals sin²θ. (6) If you see (sec θ−tan θ)(sec θ+tan θ), it equals 1 (from sec²−tan²=1). Recognising these patterns comes from practice — the first 10 proofs feel hard; after 20, you see the patterns automatically.

For ACUTE ANGLES in a right triangle (and for real angles in general), NO. sin θ = Opposite/Hypotenuse. The hypotenuse is ALWAYS the longest side in a right triangle, so Opposite ≤ Hypotenuse → Opposite/Hypotenuse ≤ 1 → sin θ ≤ 1. Similarly cos θ = Adjacent/Hypotenuse ≤ 1. Since all sides are positive, sin θ ≥ 0 and cos θ ≥ 0 for acute angles. Therefore 0 ≤ sin θ ≤ 1 and 0 ≤ cos θ ≤ 1 for 0° ≤ θ ≤ 90°. However, tan θ and sec θ can be greater than 1 (tan 60° = √3 > 1, sec 60° = 2 > 1). If a problem states sin θ = 7/5, this is impossible and the problem has an error.

First find the missing side: A = √(13²−5²) = √(169−25) = √144 = 12. The smallest angle is opposite the smallest side (5). So O=5, A=12, H=13. sin θ = 5/13. cos θ = 12/13. tan θ = 5/12. cosec θ = 13/5. sec θ = 13/12. cot θ = 12/5. Verify: sin²θ+cos²θ = 25/169 + 144/169 = 169/169 = 1 ✓. Also: 1+tan²θ = 1+25/144 = 169/144 = sec²θ = (13/12)² ✓.

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