Introduction to Trigonometry — Class 10 Maths Notes & Practice
August 1, 2026
Trigonometry has a reputation it does not deserve. The name sounds heavy, the symbols look foreign, and the first page of the chapter throws six new words at you at once. Take a breath. Underneath all of it sits one very small idea: in a right-angled triangle, if you fix one of the acute angles, the shapes of the triangle are locked, and so the ratios of the sides are locked too. That is the entire chapter. Everything else — the tables, the identities, the long expressions — is bookkeeping built on that single sentence. We will follow the current NCERT Class 10 Mathematics 2026-27 reprint and slow down wherever students commonly need another explanation.
This page is written for someone starting from zero. We build the vocabulary first, then the ratios, then the standard-angle values, then the identities, and we finish with a full class 10 trigonometry formula list plus a worksheet of introduction to trigonometry class 10 important questions with worked answers. Every worked answer on this page has been checked symbolically, so if your answer disagrees with ours, it is worth re-reading your working.
One promise before we start. A frequent source of error in this chapter is not the identities. It is mislabelling the sides when the triangle is drawn at an odd angle or when the question suddenly asks about the other acute angle. So the whole page is built around one habit, repeated until it is automatic: stand at the angle, name the sides from there, then read the ratio off. We will call it the Stand-and-Name routine.
Tape a drinking straw along the straight edge of a school protractor, hang a small weight on a thread from the centre hole, then sight the top of your building, a tree or a lamp post through the straw and read where the thread crosses the scale — ninety degrees minus that reading is your angle of elevation. Pace out your distance from the base, multiply by the tangent of that angle, add your own eye height, and write the estimate in your notebook; repeat from a second, further spot and see how close the two answers land. (20-25 min)
Your Game Plan
Learn the three side names — opposite, adjacent, hypotenuse — and practise naming them from a chosen angle until it takes no thought.
Write the six ratios as fractions of those three names. Do not memorise them as unrelated words.
Lock in the 30°, 45°, 60° table by drawing the two reference triangles, not by rote.
Learn what happens at 0° and 90°, including which ratios are simply not defined.
Get comfortable with the three identities and the two-line habit of proving simple results.
Study Notes — Trigonometric Ratios With Solved Examples
What Trigonometry Actually Is
The word comes from three Greek pieces: tri (three), gonia (angle) and metron (measure) — literally “three-angle measuring”. Ancient surveyors and astronomers needed to find lengths they could not reach: the height of a temple, the width of a river, the distance to a star. They could not stretch a tape across the river, but they could stand somewhere and measure an angle. Trigonometry is the toolkit that turns a measured angle plus one measured length into all the other lengths.
Here is the observation everything rests on. Draw a right-angled triangle with one acute angle of 35°. Now draw a bigger one, also with a right angle and a 35° angle. The two triangles are similar — same shape, different size — because their angles match. And in similar triangles corresponding sides are in the same proportion. So although the actual lengths differ, the ratio of any two sides is exactly the same in both. That ratio depends on the angle alone. If you know the angle, you know the ratio. That is the whole engine. If similarity feels rusty, a quick revision of Triangles and similarity criteria will make the next few sections much easier.
Key Idea
In a right-angled triangle, fixing one acute angle fixes the shape. Every ratio of two sides then becomes a number that depends only on that angle. Those numbers are the trigonometric ratios.
Notice what trigonometry is not. It is not a new kind of number and it is not a formula to plug into blindly. Each ratio is just a fraction: one side length divided by another. Every value in the standard table is a plain fraction such as 1/2 or √3/2. When you meet an expression like sin 30° + cos 60°, you are only adding two ordinary fractions.
Example 1 — Reading a ratio as an ordinary fraction
A right triangle has one acute angle whose opposite side is 8 cm and whose hypotenuse is 17 cm. Then the sine of that angle is simply 8 ÷ 17 = 8/17, or about 0.47. Nothing mysterious: it is the answer to “the opposite side is what fraction of the hypotenuse?”
Example 2 — Why the size of the triangle does not matter
Triangle 1 has sides 3, 4, 5 with the right angle between the 3 and the 4. Triangle 2 has sides 6, 8, 10. In triangle 1 the ratio (side of length 3) : (hypotenuse) is 3/5 = 0.6. In triangle 2 it is 6/10 = 0.6. The triangles are different sizes but the angle opposite those sides is the same, so the ratio is identical. This is exactly why a single table of values works for every triangle in the world.
Right triangle ABC with the right angle at B: the three sides named from angle A, and the three main ratios read straight off them.
This section makes every later step easier, so we will go slowly. A right-angled triangle has three sides, and two of their three names depend on which acute angle you are standing at.
Hypotenuse — the side opposite the right angle. It is the longest side, and it is the same side no matter which acute angle you choose. This one never moves.
Opposite — the side facing the angle you have chosen. It does not touch that angle at all.
Adjacent — the remaining side, the one that touches your chosen angle and is not the hypotenuse.
Here is the routine. Call it Stand-and-Name, and do it out loud for the first fifty triangles you meet.
Find the right angle and mark the hypotenuse first. Tick it. It is settled.
Put your finger on the angle the question is asking about. Imagine standing there and looking across the triangle.
The side directly across from your finger is the opposite. Label it O.
The side left over — the one under your finger that is not the hypotenuse — is the adjacent. Label it A.
Only now write the ratio the question wants.
In the figure above, angle A sits at the bottom-left of triangle ABC and the right angle is at B. Standing at A, the side BC is across the triangle from you, so BC is the opposite. The side AB runs from your feet along to the right angle, so AB is the adjacent. AC, facing the right angle, is the hypotenuse. With BC = 3, AB = 4 and AC = 5 you get sin A = 3/5, cos A = 4/5, tan A = 3/4.
Common Mistake
Students memorise “the vertical side is the opposite and the horizontal side is the adjacent”. That works only for the one picture in the textbook. Rotate the triangle, or move to the other acute angle, and the labels swap. Opposite and adjacent are defined relative to an angle, never relative to the page.
Example 3 — The same triangle, the other angle
Use the figure above: BC = 3, AB = 4, AC = 5, right angle at B. Now stand at angle C instead. Looking across from C, the side facing you is AB = 4, so that is now the opposite. The side touching C that is not the hypotenuse is BC = 3, so that is now the adjacent. Therefore sin C = 4/5, cos C = 3/5 and tan C = 4/3. Compare with sin A = 3/5 and cos A = 4/5: sine and cosine have swapped. Nothing about the triangle changed — only where you were standing.
Example 4 — A rotated triangle
In triangle PQR the right angle is at Q, with PQ = 12 cm and QR = 5 cm. First the hypotenuse: it is PR, opposite the right angle, and PR = √(12² + 5²) = √169 = 13 cm. Stand at P. Facing you across the triangle is QR = 5, so opposite = 5; touching P and not the hypotenuse is PQ = 12, so adjacent = 12. Hence sin P = 5/13 and cos P = 12/13. Now stand at R: opposite = PQ = 12, adjacent = QR = 5, so tan R = 12/5. Notice we never once asked whether a side looked ‘vertical’.
Example 5 — When the letters are unfamiliar
In triangle XYZ, angle Y = 90°, XY = 9 and YZ = 40. Hypotenuse XZ = √(81 + 1600) = √1681 = 41. Standing at Z, the opposite side is XY = 9 and the adjacent side is YZ = 40, so tan Z = 9/40 and cos Z = 40/41. The trick is simply this: the two letters of the angle you stand at tell you which sides touch it. Angle Z touches YZ and XZ; XZ is the hypotenuse, so YZ must be the adjacent, and the side with no Z in its name — XY — is the opposite.
Exam Tip
In any triangle labelled with three letters, the opposite side to an angle is the one whose name does not contain that angle’s letter. Angle P in triangle PQR? The opposite side is QR. That one-line check catches almost every labelling error before it affects the answer.
With the three side names in place, there are exactly six ways to make a fraction from two of them. Those six fractions are the six trigonometric ratios of the angle. Write O for opposite, A for adjacent and H for hypotenuse.
Ratio
Short form
Definition
Reciprocal of
sine of A
sin A
O / H
cosec A
cosine of A
cos A
A / H
sec A
tangent of A
tan A
O / A
cot A
cosecant of A
cosec A
H / O
sin A
secant of A
sec A
H / A
cos A
cotangent of A
cot A
A / O
tan A
Only the first three need real memorising, and even those follow a pattern: sine uses the hypotenuse with the opposite, cosine uses the hypotenuse with the adjacent, tangent uses the two shorter sides. The last three are simply the first three turned upside down. Many students find the string SOH-CAH-TOA helpful — Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent — but say it only after you have run Stand-and-Name, never instead of it.
Key Rule
sin A = O/H, cos A = A/H, tan A = O/A. Then cosec A = 1/sin A, sec A = 1/cos A, cot A = 1/tan A. Also tan A = sin A / cos A and cot A = cos A / sin A, because (O/H) ÷ (A/H) = O/A.
Common Mistake
The reciprocal of sine is cosec, not sec — and the reciprocal of cosine is sec, not cosec. The names cross over. A memory hook: the pair that starts with the same three letters (co-sine and co-secant) do not go together. Write the pairing at the top of your rough sheet in every exam.
Example 6 — All six ratios from one triangle
Triangle ABC has the right angle at B, AB = 60 and BC = 11, so AC = √(3600 + 121) = √3721 = 61. Standing at A: opposite = 11, adjacent = 60, hypotenuse = 61. Then sin A = 11/61, cos A = 60/61, tan A = 11/60, cosec A = 61/11, sec A = 61/60 and cot A = 60/11. Check the pattern: sin A × cosec A = (11/61)(61/11) = 1, exactly as reciprocals should.
Example 7 — Reading sec and cot without extra work
Use the same 11–60–61 triangle, but stand at C. Opposite = AB = 60, adjacent = BC = 11, hypotenuse = 61. So sec C = H/A = 61/11 and cot C = A/O = 11/60. Notice sec C equals cosec A, and cot C equals tan A. That is not a coincidence, and we will explain it in the section on complementary angles.
Example 8 — Turning a decimal instruction into sides
Suppose you are told tan θ = 0.75 for an acute angle θ. Write 0.75 as the fraction 3/4. Since tan = O/A, you may take the opposite side as 3k and the adjacent as 4k for some positive number k. The hypotenuse is then √(9k² + 16k²) = 5k. Every k cancels in every ratio, so sin θ = 3/5 and cos θ = 4/5. The letter k is the formal way of saying ‘we do not know the size, and we do not need to’.
The syllabus asks you to understand why these ratios exist as genuine functions of the angle — in textbook language, why they are “well defined”. The worry is reasonable: there are infinitely many right-angled triangles containing an angle of, say, 40°. If different triangles gave different values of sin 40°, the whole subject would collapse. They do not, and here is the short argument.
Take two right-angled triangles, ABC and PQR, with right angles at B and Q, and with angle A = angle P. Their third angles must then also be equal, because the three angles of a triangle add to 180°. Two triangles with all three angles equal are similar by the AA criterion. In similar triangles corresponding sides are proportional, so BC/QR = AB/PQ = AC/PR. Rearranging any pair of these gives BC/AC = QR/PR — that is, sin A = sin P. The same rearrangement works for every one of the six ratios.
Key Idea
The value of a trigonometric ratio depends only on the size of the angle, not on the size of the triangle. That is a direct consequence of the AA similarity criterion, which is why this chapter sits right after the chapter on similar triangles.
Example 9 — Checking the claim numerically
Triangle 1: sides 5, 12, 13 with the right angle between 5 and 12. Triangle 2: sides 10, 24, 26. The second is the first scaled by 2, so the angles match. In triangle 1, the ratio (side 5)/(side 13) = 0.3846. In triangle 2, 10/26 = 0.3846. Same value. Scale the triangle by 7 instead — 35, 84, 91 — and 35/91 is still 0.3846. The ratio simply refuses to change.
Example 10 — Why a ratio can never exceed 1 for sine and cosine
In a right-angled triangle the hypotenuse is always the longest side. Since sin A = opposite/hypotenuse and cos A = adjacent/hypotenuse, both fractions have a numerator smaller than their denominator. Therefore 0 < sin A < 1 and 0 < cos A < 1 for any acute angle A. Consequently cosec A and sec A, being their reciprocals, are always greater than 1. Tangent has no such limit: the opposite side can be much larger than the adjacent.
Exam Tip
This bound is a free sanity check. If a question yields sin θ = 5/4 or cos θ = 1.2, you have made an arithmetic slip somewhere — no acute angle has a sine or cosine above 1. Similarly, an answer of sec θ = 0.8 is impossible.
A very common question type gives you one ratio and asks for the others, or for the value of some expression built from them. There is a reliable three-step method, and it never requires a calculator.
Read the given ratio as a fraction and write down which two sides it names. For example sin θ = 7/25 says opposite = 7k and hypotenuse = 25k.
Use Pythagoras’ theorem to find the third side. Keep the k.
Now every side is known, so run Stand-and-Name and read off whatever else is asked. The k cancels every time.
Key Rule
Never assume the sides are literally 7 and 25. Write them as 7k and 25k. This matters when a question also gives you a real length, because then you can solve for k and get actual measurements.
Example 11 — One ratio to all six
Given sin θ = 9/41 for an acute angle θ. Opposite = 9k, hypotenuse = 41k, so adjacent = √(1681k² − 81k²) = √(1600k²) = 40k. Therefore cos θ = 40/41, tan θ = 9/40, cosec θ = 41/9, sec θ = 41/40 and cot θ = 40/9.
Example 12 — Starting from a cotangent
Given 12 cot A = 5. First tidy it: cot A = 5/12, so adjacent = 5k and opposite = 12k. Hypotenuse = √(25k² + 144k²) = √(169k²) = 13k. Hence cos A = 5/13 and tan A = 12/5. Reading cot directly as adjacent-over-opposite avoids an unnecessary reciprocal step.
Example 13 — Given a ratio, evaluate an expression
Given tan A = 8/15. Opposite = 8k, adjacent = 15k, hypotenuse = 17k, so sin A = 8/17 and cos A = 15/17. Then (sin A + cos A)/(sin A − cos A) = (8/17 + 15/17) ÷ (8/17 − 15/17) = (23/17) ÷ (−7/17) = −23/7. The negative sign is genuine here: the denominator is negative because cos A is larger than sin A.
Example 14 — Given a ratio, evaluate a second expression
Given cos A = 33/65. Adjacent = 33k, hypotenuse = 65k, opposite = 56k, so sin A = 56/65. Then (sin A − cos A)/(sin A + cos A) = (56/65 − 33/65) ÷ (56/65 + 33/65) = (23/65) ÷ (89/65) = 23/89.
Example 15 — Combining two ratios from one clue
Given sec θ = 17/15 for an acute angle θ, evaluate sin θ + tan θ. Take hypotenuse = 17k and adjacent = 15k. Then opposite = √(289k² − 225k²) = 8k, so sin θ = 8/17 and tan θ = 8/15. Therefore sin θ + tan θ = 8/17 + 8/15 = (120 + 136)/255 = 256/255.
Common Mistake
If sin θ = 7/25, the cosine is not 25/7 and it is not 18/25. You must go through Pythagoras. Subtracting the numerators is one of the most expensive habits in this chapter.
The 45-45-90 and 30-60-90 reference triangles with exact side lengths. Every standard-angle value on the table is read off one of these two pictures.
These three angles appear in almost every question, so their values must be instantly available. But please do not memorise a grid of fifteen numbers. Memorise two pictures and read the numbers off them. Both pictures are in the figure above, and you can redraw them in ten seconds in the margin of an exam paper.
The 45° triangle. Cut a square along its diagonal. Each half is a right-angled triangle with two equal sides and two equal acute angles, and since those two angles share the remaining 90°, each is 45°. Take the equal sides to be 1 unit each; the hypotenuse is √(1 + 1) = √2. Standing at either 45° angle, opposite = 1, adjacent = 1, hypotenuse = √2. So sin 45° = 1/√2, cos 45° = 1/√2 and tan 45° = 1.
The 30°-60° triangle. Take an equilateral triangle of side 2 and drop a perpendicular from one vertex to the opposite side. That perpendicular bisects both the base and the top angle, giving a right-angled triangle with a 60° angle at the base, a 30° angle at the top, a base of 1 and a hypotenuse of 2. The perpendicular is √(4 − 1) = √3. Now stand at the 30° angle: opposite = 1, adjacent = √3, hypotenuse = 2, giving sin 30° = 1/2, cos 30° = √3/2 and tan 30° = 1/√3. Stand at the 60° angle instead and the two shorter sides swap roles: sin 60° = √3/2, cos 60° = 1/2 and tan 60° = √3.
Here is the complete class 10 trigonometry formula list for the standard angles. If you ever forget an entry, rebuild it from the two triangles rather than guessing. Surds such as √2 and √3 are irrational numbers, and the reason they never simplify to fractions is worth revisiting in Real Numbers — irrational numbers and surds.
Angle A
0°
30°
45°
60°
90°
sin A
0
1/2
1/√2
√3/2
1
cos A
1
√3/2
1/√2
1/2
0
tan A
0
1/√3
1
√3
not defined
cosec A
not defined
2
√2
2/√3
1
sec A
1
2/√3
√2
2
not defined
cot A
not defined
√3
1
1/√3
0
Exam Tip
Look at the sine row: 0, 1/2, 1/√2, √3/2, 1. Written as √0/2, √1/2, √2/2, √3/2, √4/2, it is just the square roots of 0, 1, 2, 3, 4 all over 2. The cosine row is the same list read backwards. Reconstructing the table this way in the first minute of an exam takes about thirty seconds and removes all doubt.
Example 16 — Substituting standard values
Evaluate 2 sin 30° cos 60° + cos²45°. Substituting gives 2(1/2)(1/2) + (1/√2)² = 1/2 + 1/2 = 1. Writing the square on the whole cosine value prevents the common denominator error.
Example 17 — Mixing squares carefully
Evaluate 3 tan²45° + sin²30° − cos²60°. Since tan 45° = 1, the first term is 3. Also sin²30° = 1/4 and cos²60° = 1/4, so those terms cancel. The answer is 3. Remember that tan²45° means (tan 45°)², not tan(45°²).
Example 18 — An angle hidden inside the question
If sin A = cos A and A is acute, find A. The sine and cosine rows of the table agree at exactly one place, the 45° column, where both are 1/√2. So A = 45°. You can also argue it with sides: sin A = cos A means opposite/hypotenuse = adjacent/hypotenuse, so opposite = adjacent, so the triangle is isosceles and both acute angles are 45°.
Example 19 — Solving for an unknown angle
If 2 cos 3A = 1 and 3A is acute, find A. Then cos 3A = 1/2, and from the table cos 60° = 1/2, so 3A = 60° and A = 20°. Always check the bracket: the table gives you the value of 3A, not of A, so the final division is essential.
A right-angled triangle cannot literally have an acute angle of 0° or 90°, so these two values are reached by watching what happens as the angle shrinks or grows. Picture a right-angled triangle with the hypotenuse of fixed length, and slowly flatten the angle A towards zero.
As A gets smaller, the side opposite A gets shorter and shorter, heading towards zero, while the adjacent side stretches until it almost coincides with the hypotenuse. So sin A = opposite/hypotenuse heads to 0/hypotenuse = 0, and cos A = adjacent/hypotenuse heads to hypotenuse/hypotenuse = 1. That gives sin 0° = 0 and cos 0° = 1, and therefore tan 0° = 0/1 = 0.
Now push A the other way, up towards 90°. The opposite side grows until it almost equals the hypotenuse, and the adjacent side shrinks to nothing. So sin 90° = 1 and cos 90° = 0. The reciprocals are where care is needed: cosec 0° would be 1/0 and sec 90° would be 1/0, and division by zero has no meaning. Those values are not defined. The same applies to cot 0° and tan 90°.
Key Rule
sin 0° = 0, cos 0° = 1, tan 0° = 0, cot 0° and cosec 0° are not defined. sin 90° = 1, cos 90° = 0, cot 90° = 0, tan 90° and sec 90° are not defined. ‘Not defined’ is the correct written answer — never write infinity, and never write zero.
Example 20 — Spotting an undefined term
Evaluate sin 0° + cos 90° + tan 0°. Each term is 0, so the answer is 0. But if the question had read sin 0° + tan 90°, the correct response is that the expression is not defined, because tan 90° does not exist. This is a common place to confuse “not defined” with zero.
Example 21 — A product that quietly vanishes
Evaluate cos 0° cos 30° cos 45° cos 60° cos 90°. The last factor is cos 90° = 0, so the entire product is 0 regardless of the other four values. Scanning a long product for a zero factor before multiplying anything saves a great deal of time.
Common Mistake
Writing tan 90° = ∞ is incorrect here. Division by zero is not defined, so the correct phrase is ‘not defined’.
The six ratios are not six independent facts. They are tangled together, and knowing the connections turns long questions into short ones. There are three families of relationship.
Reciprocal pairs. sin A × cosec A = 1, cos A × sec A = 1, tan A × cot A = 1. These come straight from the definitions: (O/H)(H/O) = 1.
Quotient relations. tan A = sin A / cos A and cot A = cos A / sin A. To see the first, divide: (O/H) ÷ (A/H) = (O/H) × (H/A) = O/A = tan A. This relation is the workhorse of identity proving, because it lets you convert any expression into sines and cosines only.
Pythagorean relations. These are the three identities, and they get their own sections below.
Key Rule
When an expression looks tangled, rewrite every term in sines and cosines using tan = sin/cos, cot = cos/sin, sec = 1/cos and cosec = 1/sin. Almost every ‘prove this’ question collapses once you do.
Example 22 — Converting to sines and cosines
Simplify tan A cot A + sin A cosec A. The first product is 1 because tangent and cotangent are reciprocals; the second is 1 for the same reason. The answer is 2. If you had instead written tan A cot A as (sin A/cos A)(cos A/sin A), everything would cancel to 1 anyway — both routes work.
Example 23 — A quotient in disguise
Show that sin A sec A = tan A. Write sec A as 1/cos A. Then sin A sec A = sin A × (1/cos A) = sin A/cos A = tan A. Two lines, no triangle needed.
Example 24 — Using relations to shortcut a value question
Given cosec θ = 37/12 for an acute θ, find tan θ and sec θ. From cosec θ = H/O we get hypotenuse = 37k and opposite = 12k, so adjacent = √(1369k² − 144k²) = 35k. Hence tan θ = 12/35 and sec θ = 37/35.
An identity is an equation that is true for every allowed value of the variable, not just for one special value. The equation x + 3 = 7 is true only when x = 4, so it is an equation, not an identity. But sin²A + cos²A = 1 holds for every angle A, and that is why it earns the name.
The proof is short and the syllabus names it explicitly, so learn it. Take a right-angled triangle ABC with the right angle at B. Let BC = a (opposite A), AB = b (adjacent to A) and AC = c (the hypotenuse). Pythagoras’ theorem gives a² + b² = c². Now divide every term by c²:
a²/c² + b²/c² = c²/c², that is (a/c)² + (b/c)² = 1. But a/c is exactly sin A and b/c is exactly cos A. So sin²A + cos²A = 1. The whole identity is Pythagoras’ theorem with every side divided by the hypotenuse.
Key Idea
sin²A + cos²A = 1 is not a new fact about the world. It is Pythagoras’ theorem written in ratio form. That is why it works for every angle: every right-angled triangle obeys Pythagoras.
Exam Tip
The notation sin²A is shorthand for (sin A)² — square the value of the ratio. It does not mean sin(A²). Writing sin A² is ambiguous, so always write sin²A or (sin A)².
Example 25 — Checking the identity on a standard angle
Take A = 30°. Then sin²30° + cos²30° = (1/2)² + (√3/2)² = 1/4 + 3/4 = 1. Try A = 45°: (1/√2)² + (1/√2)² = 1/2 + 1/2 = 1. Try A = 0°: 0² + 1² = 1. It never fails, which is exactly what an identity promises.
Example 26 — Using it to find a missing ratio
Given sin θ = 12/37 for an acute θ, find cos θ without drawing anything. From the identity, cos²θ = 1 − sin²θ = 1 − 144/1369 = 1225/1369, so cos θ = 35/37. We take the positive root because θ is acute and all six ratios of an acute angle are positive.
Example 27 — The two most useful rearrangements
From sin²A + cos²A = 1 you get sin²A = 1 − cos²A and cos²A = 1 − sin²A. In factored form these are especially handy: 1 − cos²A = (1 − cos A)(1 + cos A) and 1 − sin²A = (1 − sin A)(1 + sin A). Whenever a proof shows you a bracket like (1 + cos A) in a denominator, one of these factorisations is usually the way in.
Two more identities follow from the first by simple division, and they are on the syllabus as consequences of it. You should be able to produce both in one line each.
Divide sin²A + cos²A = 1 by cos²A. You get sin²A/cos²A + cos²A/cos²A = 1/cos²A, which reads tan²A + 1 = sec²A. This is valid whenever cos A is not zero, that is for A not equal to 90°.
Divide sin²A + cos²A = 1 by sin²A. You get 1 + cot²A = cosec²A, valid whenever sin A is not zero, that is for A not equal to 0°.
Key Rule
The three identities are: sin²A + cos²A = 1 | 1 + tan²A = sec²A | 1 + cot²A = cosec²A. Their most-used rearrangements are sec²A − tan²A = 1 and cosec²A − cot²A = 1. Those two differences of squares factorise, which is the key to a large family of proofs.
Common Mistake
1 + tan²A = sec²A, not cosec²A. Pair them by the shared letter shape: tan goes with sec (both come from dividing by cosine), and cot goes with cosec (both come from dividing by sine). Mixing the pairs is the single most common slip in identity questions.
Example 28 — Applying the second identity
Given tan θ = 12/35 for an acute θ, find sec θ. Then sec²θ = 1 + tan²θ = 1 + 144/1225 = 1369/1225, so sec θ = 37/35 and hence cos θ = 35/37. This is faster than rebuilding the triangle when only one extra ratio is wanted.
Example 29 — A difference of squares in action
Simplify (sec θ + tan θ)(sec θ − tan θ). This is a difference of squares: sec²θ − tan²θ. By the identity that equals 1. So the whole product is simply 1, whatever θ is.
Example 30 — Combining two identities
Evaluate (1 + tan²A)/(1 + cot²A) for any acute A. The numerator is sec²A and the denominator is cosec²A. So the expression is (1/cos²A) ÷ (1/sin²A) = sin²A/cos²A = tan²A. A three-line answer that would take a page if you expanded everything.
The syllabus is explicit that only simple identities are to be given, and that they should be provable using sin²A + cos²A = 1 together with the standard relations. So you will not be asked for anything exotic. What you do need is a calm, repeatable procedure.
Choose the messier side and work on that side only. Never move terms across the equals sign as if it were an equation you are solving.
Convert everything to sines and cosines if you cannot see a route within ten seconds.
Look for differences of squares: 1 − sin², 1 − cos², sec² − tan², cosec² − cot².
Take a common denominator when you see two fractions added.
Stop as soon as your side matches the other side, and write ‘= RHS, hence proved’.
Example 31 — A balanced-square proof
Prove that (sin A + cos A)² + (sin A − cos A)² = 2. Expand both squares. The left side becomes sin²A + 2 sin A cos A + cos²A + sin²A − 2 sin A cos A + cos²A. The middle terms cancel, leaving 2(sin²A + cos²A) = 2 × 1 = 2 = RHS. Hence proved.
Example 32 — A proof using a conjugate pair
Prove that (1 − cos A)/sin A + sin A/(1 − cos A) = 2 cosec A. Combine the two fractions. The numerator is (1 − cos A)² + sin²A = 1 − 2 cos A + cos²A + sin²A = 2(1 − cos A). The denominator is sin A(1 − cos A). Cancelling the common factor gives 2/sin A = 2 cosec A = RHS. Hence proved.
Example 33 — A quotient proof
Prove that (sec A − cos A)/tan A = sin A. Start with the numerator: sec A − cos A = 1/cos A − cos A = (1 − cos²A)/cos A = sin²A/cos A. Dividing by tan A means multiplying by cos A/sin A. Thus LHS = [sin²A/cos A][cos A/sin A] = sin A = RHS. Hence proved.
Example 34 — Factoring before simplifying
Prove that (sin A + tan A)/(1 + cos A) = tan A. Convert tan A to sin A/cos A. The numerator becomes sin A + sin A/cos A = sin A(cos A + 1)/cos A. Dividing by (1 + cos A) cancels that common factor, leaving sin A/cos A = tan A = RHS. Hence proved.
Example 35 — A symmetric sum proof
Prove that (sin A + cos A)(tan A + cot A) = sec A + cosec A. Convert tangent and cotangent to sines and cosines. Then tan A + cot A = sin A/cos A + cos A/sin A = (sin²A + cos²A)/(sin A cos A) = 1/(sin A cos A). Therefore the left side becomes (sin A + cos A)/(sin A cos A) = 1/cos A + 1/sin A = sec A + cosec A = RHS. Hence proved.
Exam Tip
If you get stuck halfway, do not panic and do not erase. Write down what the other side simplifies to as well, and see whether the two meet in the middle. Meeting in the middle is a perfectly acceptable proof as long as you state clearly that both sides reduce to the same expression.
Two angles that add to 90° are called complementary. In a right-angled triangle the two acute angles are always complementary, because the third angle uses up the other 90°. That produces a neat family of results.
Look again at Example 3. In triangle ABC with the right angle at B, angles A and C satisfy A + C = 90°, so C = 90° − A. We found sin A = 3/5 and cos C = 3/5. In other words sin A = cos(90° − A). The reason is exactly the Stand-and-Name routine: moving from A to C swaps which side is opposite and which is adjacent, so sine turns into cosine, tangent turns into cotangent, and secant turns into cosecant.
Key Rule
sin(90° − A) = cos A, cos(90° − A) = sin A, tan(90° − A) = cot A, cot(90° − A) = tan A, sec(90° − A) = cosec A, cosec(90° − A) = sec A. The prefix ‘co’ in cosine, cotangent and cosecant is short for ‘complement’ — the names were built for exactly this pattern.
Syllabus note (CBSE 2026-27)
The current CBSE Class 10 Mathematics curriculum lists trigonometric ratios of an acute angle and proof of their existence, ratios of 30°, 45° and 60°, motivation of the ratios at 0° and 90°, relationships between the ratios, and proof and applications of sin²A + cos²A = 1 through simple identities. It does not list complementary-angle ratios as a separate content line. This short connection is included as supporting understanding, not as a claim about a separate assessment topic.
Example 36 — Solving for an angle
If tan 2A = cot(A − 18°) and both angles are acute, find A. Since cot(A − 18°) = tan[90° − (A − 18°)], the equation becomes tan 2A = tan(108° − A). Equating the angles: 2A = 108° − A, so 3A = 108° and A = 36°. Quick check: 2A = 72° and A − 18° = 18°, and tan 72° is indeed cot 18°.
Example 37 — The same idea with sine and cosine
If sin 3A = cos(A − 26°) with 3A acute, find A. Write cos(A − 26°) = sin[90° − (A − 26°)] = sin(116° − A). Equating: 3A = 116° − A, so 4A = 116° and A = 29°.
Example 38 — An expression that collapses to a constant
Evaluate sin²25° + sin²65°. Since 65° = 90° − 25°, we have sin 65° = cos 25°. So the expression is sin²25° + cos²25° = 1. You never needed the actual value of either sine.
Many practice questions in this chapter use direct substitution: replace each ratio by its table value, then do careful fraction arithmetic. The mathematics is manageable; most errors come from hurried substitution or sign slips. Three habits prevent almost all of them.
Write the substituted expression out in full before simplifying anything. A clean substitution line makes every later step easier to check.
Rationalise surds only at the end, and rationalise consistently — either use 1/√3 throughout or √3/3 throughout, never both in the same line.
Handle squares before products. cos²30° means (√3/2)² = 3/4, and forgetting to square the 2 in the denominator is the classic error.
Example 39 — A multi-term substitution
Evaluate [3 sin²30° + 2 cos²60° + sec²45°] ÷ [tan²45° + cos²60°]. The numerator is 3(1/4) + 2(1/4) + 2 = 3/4 + 1/2 + 2 = 13/4. The denominator is 1 + 1/4 = 5/4. Therefore the value is (13/4) ÷ (5/4) = 13/5.
Example 40 — A fraction of standard values
Evaluate (sec 60° − tan 45°) ÷ (cosec 30° + cos 60°). The numerator is 2 − 1 = 1. The denominator is 2 + 1/2 = 5/2. Therefore the value is 1 ÷ (5/2) = 2/5.
Example 41 — Recognising an identity inside a substitution
Evaluate 2 sin²30° − 3 cos²45° + tan²60° + 3 sin²90°. Term by term: 2(1/4) = 1/2; 3(1/2) = 3/2 to be subtracted; (√3)² = 3; 3(1)² = 3. Total = 1/2 − 3/2 + 3 + 3 = 5. Writing each term on its own line before adding is what keeps the signs straight.
Example 42 — A standard-value fraction
Evaluate (1 − tan²30°)/(1 + tan²30°). Since tan²30° = 1/3, the fraction is (1 − 1/3)/(1 + 1/3) = (2/3)/(4/3) = 1/2. This equals cos 60°, which is a useful final check.
A 30 degree angle of elevation from a point 30 m from the foot of a pole gives a height of 10√3, about 17.3 m.
Everything so far has been about triangles on paper. The reason this chapter exists is that the same ratios let you measure things you cannot reach. The bridge between the two is one definition.
Stand somewhere and look straight ahead. That imaginary level line through your eye is the horizontal. Now lift your gaze to the top of a tower. The line from your eye to the top of the tower is the line of sight, and the angle between the line of sight and the horizontal is the angle of elevation. If instead you look down from a height at something below you, the angle between the line of sight and the horizontal is the angle of depression.
Key Idea
The angle of elevation is always measured from the horizontal, never from the vertical object. In the figure above, the 30° sits between the ground line and the dashed line of sight — not between the sight line and the pole.
Example 43 — Reading the figure above
A person standing 30 m from the foot of a pole sees its top at an angle of elevation of 30°. Ignoring the observer’s own height, the pole, the ground and the line of sight form a right-angled triangle with the right angle at the foot of the pole. Standing at the observer’s position, the pole is the opposite side and the ground distance is the adjacent side. So tan 30° = height/30, giving height = 30 tan 30° = 30/√3 = 10√3, which is about 17.3 m.
Example 44 — A ladder against a wall
A ladder leans against a vertical wall, making an angle of 60° with the ground, and its foot is 2.5 m from the wall. How long is the ladder? Stand at the foot of the ladder. The ladder is the hypotenuse and the 2.5 m gap is the adjacent side, so cos 60° = 2.5/L. Since cos 60° = 1/2, we get 1/2 = 2.5/L and L = 5 m.
Example 45 — Choosing the right ratio
A kite string is let out at 45° to the horizontal and the kite is 60 m above the ground. Assuming the string is straight and ignoring the height of the hand, how long is the string? The height is the opposite side and the string is the hypotenuse, so use sine: sin 45° = 60/S, that is 1/√2 = 60/S, so S = 60√2, about 84.9 m. The habit is always the same — decide which side you know and which you want, then pick the ratio that links exactly those two.
Exam Tip
In every heights-and-distances question, draw the picture and mark the right angle first, then run Stand-and-Name at the given angle. Deciding between sine, cosine and tangent becomes automatic once the sides are named. The full treatment, including two-triangle problems and angles of depression, is in Some Applications of Trigonometry — heights and distances.
Here are the errors that most often break an otherwise correct solution, gathered in one place for a quick final check.
Mistake 1 — labelling by the page, not by the angle
Deciding that the upright side is always the opposite. Rotate the triangle and this fails immediately. Always run Stand-and-Name.
Mistake 2 — swapping the reciprocals
Writing sec A = 1/sin A. The correct pairs are sin–cosec and cos–sec.
Mistake 3 — subtracting instead of using Pythagoras
From sin θ = 9/41 concluding cos θ = 32/41. You must compute √(41² − 9²) = 40.
Mistake 4 — forgetting to square the denominator
Writing cos²30° = 3/2 instead of 3/4. Squaring √3/2 squares both parts.
Mistake 5 — writing infinity
Answering tan 90° = ∞. The expected answer is ‘not defined’.
Mistake 6 — treating an identity like an equation
Cross-multiplying across the equals sign while proving an identity. Work on one side at a time, or reduce both sides separately and say so.
Mistake 7 — losing the bracket
From cos 3A = 1/2 concluding A = 60°. The table gives 3A = 60°, so A = 20°.
Exam Tip
Before you hand in the paper, run three checks on every trigonometry answer: is every sine and cosine between 0 and 1, is every secant and cosecant at least 1, and did you answer for the angle actually asked rather than for the bracketed expression?
Practice Worksheet — Introduction to Trigonometry Class 10 Important Questions
Twelve original questions, arranged roughly in increasing difficulty. Work each one in your notebook first, then open the answer. Every answer below has been checked symbolically.
Q1.In triangle ABC the right angle is at B, with AB = 45 cm and BC = 28 cm. Find sin A and cos A.Show Answer
First the hypotenuse: AC = √(45² + 28²) = √(2025 + 784) = √2809 = 53 cm. Standing at A, the opposite side is BC = 28 and the adjacent side is AB = 45. So sin A = 28/53 and cos A = 45/53.
Q2.If tan θ = 20/99 for an acute angle θ, find the value of (sin θ + cos θ)/(sin θ − cos θ).Show Answer
Opposite = 20k, adjacent = 99k, hypotenuse = √(400k² + 9801k²) = 101k. So sin θ = 20/101 and cos θ = 99/101. The expression = (20/101 + 99/101) ÷ (20/101 − 99/101) = (119/101) ÷ (−79/101) = −119/79.
Q5.If cosec θ = 65/16 for an acute angle θ, find cot θ and cos θ.Show Answer
cosec = hypotenuse/opposite, so hypotenuse = 65k and opposite = 16k. Adjacent = √(4225k² − 256k²) = √(3969k²) = 63k. Therefore cot θ = 63/16 and cos θ = 63/65.
Q6.If cot A = 12/5 and A is acute, find the value of (3 cos A − sin A)/(3 cos A + sin A).Show Answer
cot A = 12/5, so adjacent = 12k, opposite = 5k, hypotenuse = 13k. Then cos A = 12/13 and sin A = 5/13. Numerator = 36/13 − 5/13 = 31/13. Denominator = 36/13 + 5/13 = 41/13. Value = (31/13) ÷ (41/13) = 31/41.
Q7.Prove that (1 + cot²A) sin²A = 1.Show Answer
By the identity 1 + cot²A = cosec²A. So the left side = cosec²A × sin²A = (1/sin²A) × sin²A = 1. Hence proved.
Q8.Prove that (sec A − 1)(sec A + 1) = tan²A.Show Answer
Use the difference-of-squares pattern: (sec A − 1)(sec A + 1) = sec²A − 1. From the identity sec²A = 1 + tan²A, subtracting 1 gives sec²A − 1 = tan²A. Hence proved.
Q9.If tan(3x + 10°) = cot(2x − 5°) and all the angles involved are acute, find x.Show Answer
Write cot(2x − 5°) as tan[90° − (2x − 5°)] = tan(95° − 2x). Equating the angles: 3x + 10° = 95° − 2x, so 5x = 85° and x = 17°. Check: 3x + 10° = 61° and 2x − 5° = 29°, and those two do add to 90°.
Q10.The shadow of a vertical tower on level ground is 15 m long when the sun’s angle of elevation is 60°. Find the height of the tower.Show Answer
The tower, its shadow and the sun’s ray form a right-angled triangle with the right angle at the base. Standing at the far end of the shadow, the tower is the opposite side and the shadow is the adjacent side, so tan 60° = h/15. Since tan 60° = √3, h = 15√3 ≈ 25.98 m.
Q11.If sin A = 1/2 and A is acute, find the value of 3 cos A − 4 cos³A.Show Answer
sin A = 1/2 gives A = 30°, so cos A = √3/2. Then 3 cos A = 3√3/2, and cos³A = (√3/2)³ = 3√3/8, so 4 cos³A = 3√3/2. The two terms are equal, so the value is 0.
Q12.If √3 tan θ = 3 sin θ for an acute angle θ, find the value of sin²θ − cos²θ.Show Answer
Write tan θ as sin θ/cos θ: √3 sin θ/cos θ = 3 sin θ. Since θ is acute, sin θ is not zero, so we may divide both sides by sin θ, giving √3/cos θ = 3 and hence cos θ = √3/3 = 1/√3. Then cos²θ = 1/3 and sin²θ = 1 − 1/3 = 2/3. So sin²θ − cos²θ = 2/3 − 1/3 = 1/3.
You now have the two halves of trigonometry that Class 10 asks for: the ratios themselves, and the identities built on them. The natural next step is to put them to work on real measurements in Some Applications of Trigonometry — heights and distances. If any step in a proof felt shaky, a short revision of Triangles and similarity criteria will usually explain why.
Kaizen thought for today: do not try to master this chapter in one sitting. Draw the two reference triangles for five minutes every morning this week. By Sunday the table you were dreading will simply be something you know, and the improvement will have cost you almost nothing.
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More chapters from this subject — keep the momentum going.
