Class 10 Mathematics Notes

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Introduction to Trigonometry Class 10 Maths Notes

Trigonometry begins here with six ratios taken from a right-angled triangle and one acute angle. This page fixes the opposite, adjacent and hypotenuse convention that every later question depends on, proves the ratios depend on the angle alone, gives the values at 0°, 30°, 45°, 60° and 90°, and shows how to choose the ratio a problem needs.

Class:10Subject:MathematicsUnit:VCovers:CBSE 2024-25

8 Key Formulas

DWritten byDeep Narayan

UpdatedOctober 2026

Key Concept Summary

Which of the six ratios do not exist, and at which angle?

Only two of the six are undefined, and only at the two ends of the range. At 0° the opposite side is zero, so cosec 0° = 1 / sin 0° and cot 0° = cos 0° / sin 0° both divide by zero and do not exist. At 90° the adjacent side is zero, so sec 90° = 1 / cos 90° and tan 90° = sin 90° / cos 90° both divide by zero and do not exist.

Why Trigonometry Starts From a Right Triangle

Take any right-angled triangle and mark one of its two acute angles as θ. Three sides are available and three lengths can be measured, and the useful discovery is that the ratio of any two of them, for a fixed θ, is always the same. That constant is a trigonometric ratio, and the collection of the six such ratios is what the chapter defines.

The angle θ must be acute, that is strictly between 0° and 90°. The right angle itself cannot be used as θ.
A ratio such as 3 / 4 is the same number however it is written, so √3 / 3 and 1 / √3 are both acceptable for tan 30°.
The ratios are pure numbers. They carry no unit, which is why they can multiply and divide a length directly.
Every question in this unit, and every heights-and-distances problem in the last trigonometry chapter, ends at one of these ratios.

No tables will be supplied

In the 3-hour examination no trigonometric tables are provided, so every numerical answer must come from the standard values at 30°, 45° and 60°. That is exactly why these three angles and no others appear in heights-and-distances questions this year.

Opposite, Adjacent and Hypotenuse

Every ratio is built from the same three sides, and the whole chapter fails if you misidentify them. Name the sides relative to the marked angle θ and not relative to the triangle in general, because the side opposite θ in one triangle is the adjacent side in another.

The hypotenuse is always the side opposite the right angle. It is the longest side and it is the denominator in sin θ, cos θ, cosec θ and sec θ.
The side opposite θ is the side touching θ only at the θ vertex. It is the numerator in sin θ and tan θ and the denominator in cosec θ and cot θ.
The adjacent side is the other of the two shorter sides. It is the numerator in cos θ and cot θ and the denominator in tan θ and sec θ.
A memory aid that holds in every triangle: sin, cosec and the hypotenuse go together, as do cos, sec and the adjacent side.
In a right triangle the two sides other than the hypotenuse are the legs, and both of them are legs — there is no second hypotenuse.

The one identification error that spoils everything

Do not call the hypotenuse the adjacent side. If the angle θ is measured from the base, the base is adjacent and the vertical side is opposite; if θ is measured from the vertical side, the vertical side is adjacent and the base is opposite. Changing the marked angle changes both names at once, and the ratio changes with it.

The Six Trigonometric Ratios

The three primary ratios

sin θ = \frac{opposite}{hypotenuse}, cos θ = \frac{adjacent}{hypotenuse}, tan θ = \frac{opposite}{adjacent}

Their reciprocals

csc θ = \frac{hypotenuse}{opposite}, sec θ = \frac{hypotenuse}{adjacent}, cot θ = \frac{adjacent}{opposite}

Read every ratio aloud as 'opposite over hypotenuse' rather than by its symbol, because the spoken form cannot be misremembered.
cosec is written csc in some books; both mean 1 / sin.
The three reciprocals exist for every acute angle and are the three ratios written upside down.
Only sin, cos and tan are used to compute values; cosec, sec and cot are read off once sin, cos and tan are known.

Three sentences to memorise

Sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse. Tangent is opposite over adjacent. Cosecant, secant and cotangent are each of these turned upside down.

Why the Ratios Are Well Defined

Two questions must be answered before a ratio may be called a function of θ: does it give the same value for every triangle with that θ, and is it finite? The syllabus asks you to be satisfied on both counts, and the proof is short.

Draw two right triangles sharing the angle θ. Their other acute angles are 90° − θ in each, so by the AAA criterion the two triangles are similar.
Similar triangles keep one common ratio of corresponding sides, so opposite ÷ hypotenuse, adjacent ÷ hypotenuse and opposite ÷ adjacent are the same number in both triangles.
Therefore each ratio depends on θ alone and not on how large the triangle is drawn — which is what makes the ratio a property of the angle.
For an acute θ the adjacent side, the opposite side and the hypotenuse are all positive, so none of the denominators is zero and every ratio exists.
Since θ is the only free quantity, the ratio can be tabulated against θ, which is exactly what the standard values are.

Why this matters in an answer

A one-mark question may ask you to justify that a trigonometric ratio is well defined. The sentence that earns the mark is: two right triangles with the same acute angle θ are similar by AAA, so the ratio of any two corresponding sides is the same in both, so the ratio depends on θ alone.

The Values at 0° and 90°

As θ slides from 0° to 90° the triangle flattens and then stands upright, and the two end positions are worth understanding because they explain exactly which ratios break down. At 0° the opposite side vanishes; at 90° the adjacent side vanishes.

At 0°: sin 0° = 0, cos 0° = 1, tan 0° = 0, sec 0° = 1.
At 0° cosec 0° does not exist and cot 0° does not exist, because both divide by sin 0° = 0.
At 90°: sin 90° = 1, cos 90° = 0, cosec 90° = 1, cot 90° = 0.
At 90° tan 90° does not exist and sec 90° does not exist, because both divide by cos 90° = 0.
So the undefined ratios are exactly cosec and cot at 0°, and tan and sec at 90° — two ratios at each end, and no others anywhere in between.

A zero answer and an undefined answer are different

sin 0° = 0 is a number, while cosec 0° is not a number at all. Writing cosec 0° = 0 is wrong; dividing by zero is not defined, and a question on the undefined values is a standard one-mark trap.

The Values at 30°, 45° and 60°

These three angles and only these three are used in numerical questions, because their values are exact and short. Learn the three primary ratios first and get the reciprocals by inversion.

θ = 30°: sin 30° = 1 / 2, cos 30° = √3 / 2, tan 30° = 1 / √3 = √3 / 3.
θ = 30° reciprocals: cosec 30° = 2, sec 30° = 2 / √3 = 2√3 / 3, cot 30° = √3.
θ = 45°: sin 45° = 1 / √2 = √2 / 2, cos 45° = 1 / √2 = √2 / 2, tan 45° = 1.
θ = 45° reciprocals: cosec 45° = √2, sec 45° = √2, cot 45° = 1.
θ = 60°: sin 60° = √3 / 2, cos 60° = 1 / 2, tan 60° = √3.
θ = 60° reciprocals: cosec 60° = 2 / √3 = 2√3 / 3, sec 60° = 2, cot 60° = 1 / √3 = √3 / 3.

Two patterns worth memorising

The sin and cos values simply swap between 30° and 60°, so sin 30° = cos 60° = 1 / 2 and cos 30° = sin 60° = √3 / 2. And at 45° every ratio equals 1 / √2 except tan, which equals 1. That halves the amount you have to remember.

No complementary-angle shortcuts this year

Relations of the form sin(90° − A) = cos A, cos(90° − A) = sin A and tan(90° − A) = cot A were deleted in the rationalised 2024-25 syllabus and must not be used as a working step or quoted as an identity. Use the tabulated values directly, and note that a problem needing a complementary angle will not be set this year.

The Relationships Between the Ratios

The six ratios are not six independent facts. Three of them are the reciprocals of the other three, and one more follows from dividing, so six quantities collapse to three. These relationships are what let you answer a question about sec or cot when only sin and cos have been given.

The reciprocal pair relationships

sec θ = \frac{1}{cos θ}, csc θ = \frac{1}{sin θ}, cot θ = \frac{1}{tan θ}

The quotient relationships

tan θ = \frac{sin θ}{cos θ}, cot θ = \frac{cos θ}{sin θ}

Check with 45°: cos 45° = √2 / 2, so sec 45° = 1 / (√2 / 2) = 2 / √2 = √2, which agrees with the table.
Check with 30°: sin 30° = 1 / 2, so cosec 30° = 2, and tan 30° = 1 / √3, so cot 30° = √3. Both agree with the table.
The identity sin²θ + cos²θ = 1 is the next chapter's statement and the only trigonometric identity retained this year.
Nothing here needs a table: every value follows from sin, cos and tan at the three standard angles.

How to Choose the Right Ratio in a Problem

In any heights-and-distances or right-triangle question exactly one ratio is the tool for the job, and choosing it wrongly is the standard cause of a wrong answer. The choice is made from the answer, not from the drawing.

Step 1 — Write down what the question asks for. Name that length before touching the triangle.
Step 2 — If the required length is one of the two legs, the answer must be a ratio. If it is the hypotenuse or the whole line of sight, it is also a ratio but of a different kind, so keep going.
Step 3 — Compare the required length with the hypotenuse. If the required length is the hypotenuse, use sin θ = opposite / hypotenuse or cos θ = adjacent / hypotenuse, inverted.
Step 4 — Compare the required length with the adjacent side. If the required length is the adjacent side, use cos θ = adjacent / hypotenuse or cot θ = adjacent / opposite, inverted.
Step 5 — If the required length is the opposite side and the adjacent side is known, use tan θ = opposite / adjacent.
Step 6 — Take the value of that one ratio from the table at the stated angle and solve a single-step equation for the unknown.
Step 7 — Write the unit and, if the answer is a surd, rationalise the denominator so the form matches the table.

The two questions that settle it

Ask: what do I have, and what do I need? Knowing the hypotenuse and needing a leg means sine or cosine. Knowing both legs means tangent. Knowing the opposite side and needing the hypotenuse means sine. Knowing the adjacent side and needing the hypotenuse means cosine.

How the Questions Are Asked

Trigonometry carries 12 of the 80 theory marks, and this opening chapter supplies the one-mark and two-mark items that feed the later ones. The 3-hour paper has 38 questions across Sections A to E, and two internal choices each in Sections B, C and D.

State the six trigonometric ratios. (A two-mark very short answer.)
Which of the six ratios are undefined, and at which angle? (A one-mark MCQ or an assertion-reason.)
Prove that a trigonometric ratio is well defined, that is, that it depends on θ alone. (A three-mark short answer.)
Find the values of sin 30°, cos 60° and tan 45°. (One or two marks.)
If sin θ = 3 / 5, find cosec θ and cot θ. (A two or three mark question using the reciprocal relationships.)
A case-study question on a ramp or a ladder usually asks for a trigonometric ratio in one part and an area or volume calculation in the other.

Marks come from the naming

In a three-mark or five-mark trigonometry answer, the mark for the ratio, the mark for the table value and the mark for the arithmetic are separate. Writing 'by tangent, tan θ = 1, so the height is 12 m' earns all three; a bare answer earns one.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Sine

sin θ = \frac{opposite}{hypotenuse}

The hypotenuse is always the side opposite the right angle, and opposite is measured from θ.

Cosine

cos θ = \frac{adjacent}{hypotenuse}

Adjacent is the leg touching θ that is not the hypotenuse.

Tangent

tan θ = \frac{opposite}{adjacent}

The only one of the six with no hypotenuse in it, so it uses both legs.

Cosecant

csc θ = \frac{hypotenuse}{opposite} = \frac{1}{sin θ}

Undefined at 0°, because it divides by sin 0° = 0.

Secant

sec θ = \frac{hypotenuse}{adjacent} = \frac{1}{cos θ}

Undefined at 90°, because it divides by cos 90° = 0.

Cotangent

cot θ = \frac{adjacent}{opposite} = \frac{1}{tan θ}

Undefined at 0°, because it divides by sin 0° = 0.

The three standard angles

sin 3 0^{\circ} = \frac{1}{2}, cos 3 0^{\circ} = \frac{3}{2}, tan 3 0^{\circ} = \frac{1}{3} sin 4 5^{\circ} = \frac{2}{2}, cos 4 5^{\circ} = \frac{2}{2}, tan 4 5^{\circ} = 1 sin 6 0^{\circ} = \frac{3}{2}, cos 6 0^{\circ} = \frac{1}{2}, tan 6 0^{\circ} = 3

The only values a numerical question may need, since no tables are supplied.

The reciprocal values

csc 3 0^{\circ} = 2, sec 3 0^{\circ} = \frac{2 3}{3}, cot 3 0^{\circ} = 3 csc 4 5^{\circ} = 2, sec 4 5^{\circ} = 2, cot 4 5^{\circ} = 1 csc 6 0^{\circ} = \frac{2 3}{3}, sec 6 0^{\circ} = 2, cot 6 0^{\circ} = \frac{1}{3}

Each of these is 1 divided by the corresponding sin, cos or tan value.

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

Name the opposite, adjacent and hypotenuse sides from the marked angle θ, not from the triangle, or the ratio you choose will be wrong.
Only cosec 0° and cot 0° are undefined at 0°, and only tan 90° and sec 90° at 90° — every other ratio exists at both ends.
sin 0° = 0 is a value, while cosec 0° is undefined; a zero is not the same answer as a non-existent quantity.
The reciprocal relationships are sec θ = 1 / cos θ, cosec θ = 1 / sin θ and cot θ = 1 / tan θ, and tan θ = sin θ / cos θ.
No trigonometric tables are provided in the paper, so every numerical answer must come from 30°, 45° or 60°.
sin 30° = cos 60° = 1 / 2 and cos 30° = sin 60° = √3 / 2 — memorising the swap removes half the table.
Rationalise the denominator when you write 1 / √3 or 2 / √3, giving √3 / 3 and 2√3 / 3, so that your answer matches the printed table.
The complementary-angle relations sin(90° − A) = cos A and its family were deleted for 2024-25; do not use them and expect no question that needs them.
Decide which ratio you need from the quantity asked for, not from the diagram, and name the ratio in your answer because the mark for it is separate from the arithmetic.
Trigonometry carries 12 of the 80 theory marks, and internal assessment is 20 marks split 10 for the pen-paper test and multiple assessment, 5 for the portfolio and 5 for the lab practical, not chapter by chapter.

FAQ

Frequently asked questions

What are the six trigonometric ratios and how are the sides named?

For an acute angle θ in a right-angled triangle, sin θ = opposite / hypotenuse, cos θ = adjacent / hypotenuse and tan θ = opposite / adjacent, while cosec θ = hypotenuse / opposite, sec θ = hypotenuse / adjacent and cot θ = adjacent / opposite. The hypotenuse is the side opposite the right angle. The opposite and adjacent sides are named relative to θ, so the side opposite θ becomes the adjacent side if a different angle is marked.

What are the values of the six ratios at 30°, 45° and 60°?

At 30°: sin = 1 / 2, cos = √3 / 2, tan = 1 / √3, cosec = 2, sec = 2 / √3, cot = √3. At 45°: sin = cos = 1 / √2, tan = 1, cosec = sec = √2, cot = 1. At 60°: sin = √3 / 2, cos = 1 / 2, tan = √3, cosec = 2 / √3, sec = 2, cot = 1 / √3. The reciprocals are each obtained by dividing 1 by sin, cos or tan.

Which trigonometric ratios do not exist, and why?

Two ratios fail at each end of the range. At 0° the opposite side is zero, so cosec 0° = 1 / sin 0° and cot 0° = cos 0° / sin 0° both divide by zero and do not exist. At 90° the adjacent side is zero, so tan 90° = sin 90° / cos 90° and sec 90° = 1 / cos 90° both divide by zero and do not exist. All four remaining ratios exist at both angles.

How do I decide which ratio a problem needs?

Write down what is asked for, then compare it with what is given. Hypotenuse and a leg means sine or cosine; both legs means tangent; opposite and hypotenuse means sine; adjacent and hypotenuse means cosine. Then take that one ratio's value at 30°, 45° or 60° from the table and solve one step. Naming the ratio in the answer earns the mark for the ratio itself, separately from the arithmetic.

Why is a trigonometric ratio a function of the angle alone?

Draw two right triangles with the same acute angle θ. Each has a third angle of 90° − θ, so the two triangles are similar by the AAA criterion. Similar triangles keep one common ratio for corresponding sides, so opposite over hypotenuse, adjacent over hypotenuse and opposite over adjacent are numerically the same in both triangles. Hence the ratio depends on θ alone, and for an acute θ all three denominators are non-zero, so each ratio exists.

Next Chapters

Continue learning

Class 10 Maths Notes Trigonometric Identities Class 10 Maths Notes Some Applications of Trigonometry Class 10 Maths Notes Triangles Class 10 Maths Notes All Class 11 & 12 Notes

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