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

  • 1Define the six trigonometric ratios of an acute angle in a right triangle
  • 2Recall and use the exact values of sin, cos and tan at 0°, 30°, 45°, 60° and 90°
  • 3Apply the identities sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, 1 + cot²θ = cosec²θ to prove and evaluate
  • 4Use complementary-angle relations such as sin(90°−θ) = cosθ to simplify expressions
  • 5Find all trigonometric ratios of an angle when one of them is given
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Why this chapter matters
Trigonometry is a high-weight unit in the RBSE Class 10 paper and the foundation for the next chapter (heights & distances) and all of Class 11–12 trigonometry. The standard-angle table and the identity sin²θ + cos²θ = 1 are used in almost every later problem — getting them automatic pays off for years.

Introduction to Trigonometry — RBSE Class 10 (Mathematics)

Trigonometry literally means "triangle measurement". Stand at the foot of the Qutub Minar, measure the angle up to its top and your distance from it, and trigonometry hands you the height — without ever climbing. This chapter is the dictionary that translates angles into ratios of sides, and back.


1. The six trigonometric ratios

Take a right-angled triangle and focus on one acute angle, call it θ (theta). The three sides get names relative to θ:

  • Hypotenuse — the side opposite the right angle (the longest side).
  • Opposite (perpendicular) — the side facing θ.
  • Adjacent (base) — the remaining side, next to θ.

The six ratios:

and their reciprocals:

Two relationships fall straight out:

Mnemonic: "Pandit Badri Prasad, Har Har Bhole" → sin = P/H, cos = B/H, tan = P/B (Perpendicular, Base, Hypotenuse). Many students prefer SOH-CAH-TOA.

The ratios depend only on the angle, not on the size of the triangle — all triangles with the same angle θ give the same ratios. That is why a fixed table of values works.


2. The standard-angle table (memorise this)

θ30°45°60°90°
sin θ01/21/√2√3/21
cos θ1√3/21/√21/20
tan θ01/√31√3not defined

Memory trick for sin: write 0, 1, 2, 3, 4 under the angles, divide each by 4, then take the square root → = 0, 1/2, 1/√2, √3/2, 1. For cos, read the same row backwards. For tan, divide sin by cos.

The reciprocal ratios (cosec, sec, cot) are just 1 ÷ these. Note tan 90° and cot 0° are not defined (division by zero), and so are sec 90° and cosec 0°.


3. The fundamental identities

For any acute angle θ:

The first is just Pythagoras' theorem in disguise (divide by ). The other two come from dividing the first by and respectively. These three identities are the workhorses for proving trig identities and for finding one ratio when another is given.

Note: means , not .


4. Trigonometric ratios of complementary angles

Two angles are complementary if they add up to 90°. In a right triangle, the two acute angles are complementary, and this swaps the ratios:

So , , and (since ). These let you simplify expressions instantly — a favourite 1–2 mark question.


5. Finding all ratios from one

If you're given one ratio, you can find the rest. Example: given , find and .

  • , so opp = 4k, adj = 3k.
  • Hypotenuse .
  • , .

(Alternatively, use to get , then .)


6. Closing thought

This chapter is mostly about fluency, not difficulty. Lock in three things and almost every question opens up:

  1. the definitions (SOH-CAH-TOA + the three reciprocals),
  2. the standard-angle table, recallable in seconds, and
  3. the three identities, with as the keystone.

Everything builds on this — the next chapter (Some Applications of Trigonometry) uses these exact ratios to compute heights and distances, and Class 11 extends θ beyond 90° into the full circle. For the RBSE board, expect a "find the value", a "prove the identity" and a "complementary angle" question — all directly from the boxes above.

Key formulas & results

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

Primary ratios
sinθ = opp/hyp, cosθ = adj/hyp, tanθ = opp/adj
SOH-CAH-TOA.
Reciprocal ratios
cosecθ = 1/sinθ, secθ = 1/cosθ, cotθ = 1/tanθ
The three reciprocals.
Quotient relations
tanθ = sinθ/cosθ, cotθ = cosθ/sinθ
Connect tan/cot to sin and cos.
Pythagorean identity
sin²θ + cos²θ = 1
The keystone identity.
Secant identity
1 + tan²θ = sec²θ
Divide the keystone by cos²θ.
Cosecant identity
1 + cot²θ = cosec²θ
Divide the keystone by sin²θ.
Complementary angles
sin(90°−θ) = cosθ, tan(90°−θ) = cotθ, sec(90°−θ) = cosecθ
Swaps co-ratios.
⚠️

Common mistakes & fixes

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

WATCH OUT
Reading sin²θ as sin(θ²)
sin²θ means (sinθ)², i.e. square the ratio. sin(θ²) would square the angle — completely different and wrong.
WATCH OUT
Mixing up opposite and adjacent for a given angle
'Opposite' faces the angle θ; 'adjacent' is next to it. If you change which acute angle you look at, opposite and adjacent swap.
WATCH OUT
Misremembering the standard table
Use the 0,1,2,3,4 → divide by 4 → square root trick for sin; reverse it for cos. tan = sin/cos. tan 90° is NOT defined.
WATCH OUT
Writing tan 90° = 0 or = 1
tan 90° is NOT DEFINED (cos 90° = 0, so we'd divide by zero). The same goes for sec 90°, cosec 0° and cot 0°.
WATCH OUT
Forgetting complementary relations swap the co-ratio
sin(90°−θ) = cosθ, not sinθ. The 'co' partner appears: sin↔cos, tan↔cot, sec↔cosec.
WATCH OUT
Assuming ratios change with triangle size
For a fixed angle θ, all right triangles give the SAME ratios (similar triangles). Only the angle matters.

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Standard values
Write the value of sin 30° + cos 60°.
Show solution
Step 1 — sin 30° = 1/2, cos 60° = 1/2. Step 2 — Sum = 1/2 + 1/2 = 1. ✦ Answer: 1.
Q2EASY· Complementary
Evaluate sin 35° / cos 55°.
Show solution
Step 1 — cos 55° = cos(90° − 35°) = sin 35°. Step 2 — So sin 35° / cos 55° = sin 35° / sin 35° = 1. ✦ Answer: 1.
Q3EASY· Reciprocal
If sin θ = 3/5, find cosec θ.
Show solution
Step 1 — cosec θ = 1/sin θ. Step 2 — = 1 ÷ (3/5) = 5/3. ✦ Answer: cosec θ = 5/3.
Q4MEDIUM· From one ratio
If tan θ = 4/3, find sin θ and cos θ.
Show solution
Step 1 — tan θ = opp/adj = 4/3 → opp = 4k, adj = 3k. Step 2 — hyp = √((4k)² + (3k)²) = √(25k²) = 5k. Step 3 — sin θ = 4k/5k = 4/5; cos θ = 3k/5k = 3/5. ✦ Answer: sin θ = 4/5, cos θ = 3/5.
Q5MEDIUM· Evaluate
Evaluate 2 tan²45° + cos²30° − sin²60°.
Show solution
Step 1 — tan 45° = 1, cos 30° = √3/2, sin 60° = √3/2. Step 2 — 2(1)² + (√3/2)² − (√3/2)² = 2 + 3/4 − 3/4. Step 3 — = 2. ✦ Answer: 2.
Q6MEDIUM· Identity use
If sec θ = 13/12, find tan θ using an identity.
Show solution
Step 1 — Use 1 + tan²θ = sec²θ. Step 2 — tan²θ = sec²θ − 1 = (13/12)² − 1 = 169/144 − 144/144 = 25/144. Step 3 — tan θ = 5/12 (taking the positive value for an acute angle). ✦ Answer: tan θ = 5/12.
Q7HARD· Prove identity
Prove that (1 − sin θ)/(1 + sin θ) = (sec θ − tan θ)².
Show solution
Step 1 — Start from the RHS: (sec θ − tan θ)² = (1/cos θ − sin θ/cos θ)² = ((1 − sin θ)/cos θ)². Step 2 — = (1 − sin θ)² / cos²θ. Replace cos²θ = 1 − sin²θ = (1 − sin θ)(1 + sin θ). Step 3 — = (1 − sin θ)² / [(1 − sin θ)(1 + sin θ)] = (1 − sin θ)/(1 + sin θ) = LHS. ✦ Answer: RHS simplifies to the LHS, hence proved. ∎
Q8HARD· Prove identity
Prove that (sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan²θ + cot²θ.
Show solution
Step 1 — Expand: sin²θ + 2 sinθ cosecθ + cosec²θ + cos²θ + 2 cosθ secθ + sec²θ. Step 2 — sinθ·cosecθ = 1 and cosθ·secθ = 1, so the cross terms give 2 + 2 = 4. Step 3 — (sin²θ + cos²θ) = 1; cosec²θ = 1 + cot²θ; sec²θ = 1 + tan²θ. Step 4 — Total = 1 + 4 + (1 + cot²θ) + (1 + tan²θ) = 7 + tan²θ + cot²θ = RHS. ✦ Answer: LHS = 7 + tan²θ + cot²θ, hence proved. ∎
Q9HARD· Complementary
Without using tables, evaluate (cos 38° cos 52° − sin 38° sin 52°).
Show solution
Step 1 — cos 52° = cos(90° − 38°) = sin 38°; sin 52° = sin(90° − 38°) = cos 38°. Step 2 — Substitute: cos 38°·sin 38° − sin 38°·cos 38°. Step 3 — = 0. ✦ Answer: 0.
Q10HARD· Mixed
If 3 cot θ = 4, find the value of (5 sin θ − 3 cos θ)/(5 sin θ + 3 cos θ).
Show solution
Step 1 — cot θ = 4/3 ⇒ tan θ = 3/4. So opp = 3k, adj = 4k, hyp = 5k. Step 2 — sin θ = 3/5, cos θ = 4/5. Step 3 — Numerator: 5(3/5) − 3(4/5) = 3 − 12/5 = 3/5. Step 4 — Denominator: 5(3/5) + 3(4/5) = 3 + 12/5 = 27/5. Step 5 — Ratio = (3/5)/(27/5) = 3/27 = 1/9. ✦ Answer: 1/9.

5-minute revision

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

  • Six ratios: sin = opp/hyp, cos = adj/hyp, tan = opp/adj; cosec, sec, cot are their reciprocals.
  • tan θ = sinθ/cosθ; cot θ = cosθ/sinθ.
  • Standard table: sin 0,30,45,60,90 = 0, 1/2, 1/√2, √3/2, 1; cos is the reverse; tan = sin/cos.
  • tan 90°, sec 90°, cosec 0°, cot 0° are NOT defined.
  • Identities: sin²θ + cos²θ = 1; 1 + tan²θ = sec²θ; 1 + cot²θ = cosec²θ.
  • Complementary: sin(90°−θ)=cosθ, tan(90°−θ)=cotθ, sec(90°−θ)=cosecθ.
  • Ratios depend only on the angle, not the triangle's size (similar triangles).
  • Given one ratio, build a right triangle with sides as multiples of k to find the others.

Rajasthan (RBSE) marks blueprint

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

Typical chapter weightage: 6–8 marks

Question typeMarks eachTypical countWhat it tests
MCQ / very short11–2Standard values, complementary simplification, reciprocal
Short answer21–2Find ratios from one given; evaluate an expression
Short answer31Prove an identity; complementary-angle evaluation
Long answer40–1Combined ratio-and-identity problem
Prep strategy
  • Memorise the standard-angle table cold using the √(n/4) trick
  • Practise 10+ identity proofs — start from the messier side and use sin²+cos²=1
  • Drill complementary-angle simplifications (sin/cos pairs adding to 90°)
  • Master 'given one ratio, find the rest' using a right triangle with sides k

Where this shows up in the real world

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

Heights and distances

Measuring the height of a tower or the width of a river from a ground angle is the very next chapter — pure trigonometric ratios.

Navigation and GPS

Bearings, triangulation and satellite positioning all rest on trigonometric ratios of angles.

Construction and surveying

Surveyors use theodolites to measure angles and compute distances and slopes using sin, cos and tan.

Architecture

Roof pitches, ramp gradients and staircase angles are designed using trigonometric ratios.

Physics — waves and motion

Projectile components, oscillations and wave equations are written in terms of sine and cosine.

Astronomy

The distances to nearby stars and the sizes of celestial bodies are estimated using angle measurements and trigonometry.

Exam strategy

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

  1. Write the standard-angle table in the margin at the start of the exam — instant recall, no slips.
  2. For identity proofs, start from the more complicated side and convert everything to sin and cos.
  3. Use sin²θ + cos²θ = 1 to swap between sin and cos whenever a proof stalls.
  4. For complementary problems, convert one ratio so both angles match, then cancel.
  5. When given one ratio, draw the right triangle and use Pythagoras to find the third side.
  6. Take positive roots only (acute angles) and keep surds exact unless asked to approximate.

Going beyond the textbook

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

  • Extending the ratios to angles beyond 90° using the unit circle (Class 11).
  • Compound-angle formulas sin(A±B), cos(A±B) and their many consequences.
  • Proving the standard values geometrically from 30-60-90 and 45-45-90 triangles.
  • Trigonometric equations and the periodic nature of sine and cosine.

Where else this chapter is tested

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

RBSE Class 10 Board (BSER Ajmer)High — values, identity proof and complementary question every year
NTSE / state scholarshipMedium — ratio and standard-value MCQs
JEE FoundationVery high — the entire Class 11–12 trigonometry rests on this
Maths Olympiad (NMTC)Medium — identity manipulation and clever evaluations

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Yes. RBSE prescribes the NCERT Mathematics textbook for Class 10, so the chapter, the standard-angle table and the exercises are identical. RBSE (BSER Ajmer) sets the exam pattern and marking.

tan θ = sin θ / cos θ. At 90°, cos 90° = 0, so tan 90° would require dividing by zero, which is undefined. The same reasoning makes sec 90°, cosec 0° and cot 0° undefined.

Write 0, 1, 2, 3, 4 under 0°,30°,45°,60°,90°. Divide each by 4 and take the square root to get sin values: 0, 1/2, 1/√2, √3/2, 1. Reverse the row for cos. Then tan = sin ÷ cos.

Two angles that add to 90°. For them, a ratio of one equals the co-ratio of the other: sin(90°−θ) = cos θ, tan(90°−θ) = cot θ, and so on. This lets you simplify expressions like sin 35°/cos 55° to 1.

In Class 10 we work with acute angles (0° to 90°), where all six ratios are positive. So always take the positive value, e.g. tan θ = +5/12, not −5/12.
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Last reviewed on 15 June 2026. Written and reviewed by subject-matter experts — read about our process.
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