If trigonometry has been sitting on your desk looking scary, take a breath. It is one of the friendliest chapters in Class 10 Maths once you see what it really is: a way of describing the shape of a right-angled triangle using nothing but division. No new machinery, no magic — just six ratios of sides, and a handful of angle values you will end up knowing by heart. We will build it from zero together, slowly, and every idea gets worked examples before you are asked to do anything on your own.
Here is the promise: by the end of this page you will be able to look at any right triangle, name its sides correctly, write all six ratios, recall the standard angle values without panic, and prove a board-level identity. Take it in sittings — two or three sections a day is plenty. There is no prize for rushing.
- What Trigonometry Actually Means (and How to Name the Sides)
- The Six Trigonometric Ratios
- Relationships Between the Ratios
- Finding Every Ratio When You Are Given Just One
- Trigonometric Ratios of 30°, 45° and 60°
- What Happens at 0° and 90°
- Evaluating Expressions With Standard Angles
- Trigonometric Identities and Where They Come From
- Proving Simple Identities Step by Step
- Practice Worksheet (with Answers)
Your Game Plan for This Chapter
- Read the first two sections carefully and learn to name the sides of a right triangle for a chosen angle. Nothing else works until this is automatic.
- Write the six ratios for three different triangles of your own before moving on.
- Derive the 30°, 45° and 60° values once with pencil and paper — do not start by memorising the table.
- Practise ten evaluation sums until substitution feels boring. Boring is the goal.
- Only then start identities. Do the worked proofs with your book closed, then open it to check.
- Finish with the worksheet at the bottom and mark yourself honestly.
Study Notes
1. What Trigonometry Actually Means (and How to Name the Sides)
The word itself is a giveaway: tri-gon-metron is Greek for “three-angle measurement”. Long before anyone had a measuring tape long enough, people wanted to know how tall a tower was, or how far a ship sat from the shore. You cannot climb the tower with a tape, but you can stand a known distance away and measure the angle you tilt your head. Trigonometry is the toolkit that turns that angle into a height.
In Class 10 the whole subject lives inside one shape: the right-angled triangle. So the first job is to learn the vocabulary for its sides, and here is the part most students trip over — two of the three names change depending on which angle you are talking about.
- Hypotenuse: the side opposite the right angle. This one never changes. It is always the longest side.
- Opposite side (perpendicular): the side facing the angle you have chosen.
- Adjacent side (base): the remaining side, the one touching your chosen angle (and not the hypotenuse).
Now switch to angle C. The hypotenuse is still AC. But now the opposite side is AB, and the adjacent side is BC — the two labels have swapped. Why it works: the triangle did not move; only your point of view did.
Then for θ: hypotenuse = 10 cm, adjacent = 8 cm (it touches θ), opposite = 6 cm (it faces θ). If instead you looked at the other acute angle, the 6 cm side would become the adjacent one and the 8 cm side the opposite one.
2. The Six Trigonometric Ratios
Take a right triangle and fix an acute angle A. There are three sides, so there are six ways to divide one side by another. Each of those six fractions has a name. That is the entire idea — nothing is being invented, we are just giving names to divisions.
| Ratio | Definition | Reciprocal of |
|---|---|---|
| sin A | opposite ÷ hypotenuse | cosec A |
| cos A | adjacent ÷ hypotenuse | sec A |
| tan A | opposite ÷ adjacent | cot A |
| cosec A | hypotenuse ÷ opposite | sin A |
| sec A | hypotenuse ÷ adjacent | cos A |
| cot A | adjacent ÷ opposite | tan A |
The three on top are the ones you will use constantly. A lot of students remember them as SOH – CAH – TOA: Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent. The bottom three are simply those three flipped upside down.
sin A = 3/5, cos A = 4/5, tan A = 3/4,
cosec A = 5/3, sec A = 5/4, cot A = 4/3.
Notice each bottom-row value is just the matching top-row value turned over. You never need to compute the last three separately.
So cos θ = 24/25 and tan θ = 7/24. Why it works: the ratio fixes the shape of the triangle, and shape is all a ratio cares about.
In triangle 1: sin = 3/5 = 0.6. In triangle 2: sin = 6/10 = 0.6. Identical.
This is the whole reason trigonometric ratios are well defined: any two right triangles with the same acute angle are similar, and similar triangles have proportional sides, so the fraction never changes. Double the triangle and both numbers double — the fraction stays put.
3. Relationships Between the Ratios
The six ratios are not six separate facts to memorise. They are tangled together, and once you see the connections you effectively only need to remember two of them: sine and cosine.
Also sin A × cosec A = 1, cos A × sec A = 1, tan A × cot A = 1.
Where does the first one come from? Write the definitions out and watch the hypotenuse cancel: sin A ÷ cos A = (opposite/hypotenuse) ÷ (adjacent/hypotenuse) = opposite/adjacent = tan A. That is the whole proof, and it is worth doing once on paper yourself.
Now test the rule: sin A ÷ cos A = (8/17) ÷ (15/17) = (8/17) × (17/15) = 8/15. Same answer. The rule is not a new fact — it is the same triangle read a different way.
Step 1: rewrite everything in sine and cosine. cot θ = cos θ / sin θ and sec θ = 1 / cos θ.
Step 2: multiply. (cos θ / sin θ) × (1 / cos θ) = 1 / sin θ.
Step 3: name it. 1 / sin θ = cosec θ.
Answer: cosec θ.
In sine and cosine: (sin θ / cos θ) × (1 / sin θ) × cos θ. The sin θ cancels with 1/sin θ, and cos θ cancels with 1/cos θ, leaving 1.
Why it works: “convert to sine and cosine, then cancel” solves the large majority of simplification questions in this chapter. When you are stuck, that is the default move.
4. Finding Every Ratio When You Are Given Just One
This is the single most common question type in the chapter, and it has a fixed recipe. You are handed one ratio and asked for something else. Do not try to be clever — just draw the triangle.
- Read the given ratio and write down which two sides it names.
- Sketch a right triangle and mark those two sides.
- Use Pythagoras to get the third side.
- Read off whatever the question asked for.
Step 1: tan is opposite over adjacent, so opposite = 5, adjacent = 12.
Step 2: hypotenuse = √(5² + 12²) = √(25 + 144) = √169 = 13.
Step 3: sin A = 5/13, cos A = 12/13.
Step 4: numerator = 5/13 + 12/13 = 17/13. Denominator = 5/13 − 12/13 = −7/13.
Step 5: divide. (17/13) ÷ (−7/13) = −17/7.
Answer: −17/7.
cot is adjacent over opposite, so adjacent = 9 and opposite = 40. Hypotenuse = √(81 + 1600) = √1681 = 41.
So sin θ = 40/41 and cos θ = 9/41, giving sin θ + cos θ = 49/41.
Why it works: the numbers 9, 40, 41 form a Pythagorean triple, which is why the square root came out whole. Examiners choose these deliberately — if your square root is ugly, re-check your arithmetic before assuming the question is hard.
Here tan A = 4/3. Instead of building the triangle, divide every term of the fraction by cos A:
numerator becomes 3 tan A − 2 = 4 − 2 = 2,
denominator becomes 3 tan A + 2 = 4 + 2 = 6.
Answer: 2/6 = 1/3.
Both methods give the same result — use whichever you trust. The dividing trick only works when every term has exactly one sine or cosine factor.
5. Trigonometric Ratios of 30°, 45° and 60°
Three angles show up again and again, so their ratios are worth knowing cold. But please do not begin by memorising the table — derive it once. The derivation takes five minutes and it means that if your memory ever wobbles in the exam hall, you can rebuild the values from scratch.
Then the hypotenuse = √(1² + 1²) = √2.
For the 45° angle: opposite = 1, adjacent = 1, hypotenuse = √2. So
sin 45° = 1/√2, cos 45° = 1/√2, tan 45° = 1/1 = 1.
Notice sine and cosine are equal here, which makes sense: the triangle is symmetric, so “opposite” and “adjacent” are the same length.
You now have a right triangle with hypotenuse 2, base 1, and height √(2² − 1²) = √3.
Looking at the 30° angle (at the top): opposite = 1, adjacent = √3, hypotenuse = 2, so sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3.
Looking at the 60° angle (at the base): opposite = √3, adjacent = 1, hypotenuse = 2, so sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3.
Why it works: 30° and 60° sit in the same triangle looking at each other, which is exactly why their sines and cosines are swapped.
Substitute: 2 × (1/2) × (√3/2) = √3/2.
And √3/2 is exactly sin 60°. Little coincidences like this are a good sign your substitution was correct — but always finish by simplifying to a single clean value.
6. What Happens at 0° and 90°
Zero and ninety degrees are not really triangles — a triangle with a 0° angle has collapsed flat. So we reason our way to the values instead of drawing them. Picture a right triangle where you slowly squash the angle A down towards 0°. The side opposite A shrinks towards nothing, while the adjacent side and the hypotenuse become almost the same line.
So sin A = opposite/hypotenuse heads towards 0, and cos A = adjacent/hypotenuse heads towards 1. Push the angle the other way, up towards 90°, and the roles swap. Here is the complete table — the one you will use for the rest of the year.
| Angle | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan | 0 | 1/√3 | 1 | √3 | not defined |
| cosec | not defined | 2 | √2 | 2/√3 | 1 |
| sec | 1 | 2/√3 | √2 | 2 | not defined |
| cot | not defined | √3 | 1 | 1/√3 | 0 |
The “not defined” entries are not laziness. cot 0° = cos 0° / sin 0° = 1/0, and division by zero has no meaning. Same story for tan 90° and sec 90°.
= 0 + 1 + 0 = 1. Short, but it is exactly the kind of one-mark question that appears in Section A.
sin 30° = 1/2, so sin²30° = 1/4.
cos 90° = 0, so cos²90° = 0.
tan 45° = 1, so 3 tan²45° = 3.
Total = 1/4 + 0 + 3 = 13/4.
Remember that sin²30° means (sin 30°)² — square the value, not the angle.
7. Evaluating Expressions With Standard Angles
These questions look intimidating because they are long, but they ask nothing of you except careful substitution. Work in three steps every time: substitute the table values, simplify the top and bottom separately, then tidy the final surd. Rushing straight to the answer is where marks leak away.
Numerator: tan 45° = 1 so 2 tan²45° = 2. cos 30° = √3/2 so cos²30° = 3/4. sin 60° = √3/2 so sin²60° = 3/4. Numerator = 2 + 3/4 − 3/4 = 2.
Denominator: cot 30° = √3 so cot²30° = 3. sin 90° = 1 so 3 sin²90° = 3. Denominator = 3 + 3 = 6.
Answer: 2/6 = 1/3.
Why it works: cos 30° and sin 60° are the same number, so those two terms wiped each other out. Spotting that early saves you a line of work.
cot 45° = 1, so 4 cot²45° = 4.
sec 60° = 2, so sec²60° = 4.
sin 30° = 1/2, so sin²30° = 1/4.
cos 90° = 0, so cos²90° = 0.
Total = 4 − 4 + 1/4 + 0 = 1/4.
Step 1: divide both sides by √3. tan 2θ = 3/√3 = √3.
Step 2: from the table, the acute angle whose tangent is √3 is 60°. So 2θ = 60°.
Step 3: θ = 30°.
Always check the final angle is sensible: 2θ = 60° really is acute, so the answer stands.
Numerator: 1/2 + 1/2 = 1.
Denominator: √3 − 1/√3. Put over a common denominator: (3 − 1)/√3 = 2/√3.
So the expression = 1 ÷ (2/√3) = √3/2.
Answer: √3/2. Leaving it as 1 ÷ (2/√3) would likely cost you the final mark — always finish the simplification.
8. Trigonometric Identities and Where They Come From
An identity is an equation that is true for every allowed value of the angle — unlike an ordinary equation, which is only true for particular values. Your syllabus builds everything on one identity, and the other two are its children.
Dividing that by cos²A gives 1 + tan²A = sec²A.
Dividing it by sin²A gives 1 + cot²A = cosec²A.
The proof is Pythagoras wearing a disguise. In a right triangle with angle A, let the opposite side be p, the adjacent side be b and the hypotenuse be h. Pythagoras says p² + b² = h². Divide every term by h²:
p²/h² + b²/h² = 1, which is (p/h)² + (b/h)² = 1, which is sin²A + cos²A = 1. That is the entire derivation — one division.
sin 30° = 1/2 so sin²30° = 1/4. cos 30° = √3/2 so cos²30° = 3/4.
Sum = 1/4 + 3/4 = 1. ✓
Try it at 45° too: 1/2 + 1/2 = 1. It will work every single time — that is what makes it an identity rather than an equation.
sin²A/cos²A + cos²A/cos²A = 1/cos²A
tan²A + 1 = sec²A.
Do the same with sin²A on the bottom and you get 1 + cot²A = cosec²A. Rearranged forms are just as useful: sec²A − tan²A = 1 and cosec²A − cot²A = 1.
Use tan²A = sec²A − 1 = (25/7)² − 1 = 625/49 − 49/49 = 576/49.
So tan A = √(576/49) = 24/7.
Cross-check with a triangle: sec = hypotenuse/adjacent = 25/7, so the opposite side is √(625 − 49) = 24, giving tan A = 24/7. Same answer, less writing.
9. Proving Simple Identities Step by Step
Proving an identity feels different from solving an equation, and that difference is where most of the anxiety comes from. You are not finding anything. You are taking one side of the statement and rewriting it, legally, until it looks like the other side. Here is the strategy that works nearly every time.
- Start with the messier side. There is more to simplify there.
- Convert everything to sine and cosine if you cannot see a route.
- Look for sin² + cos² = 1 or its two children hiding in the expression.
- If you see 1 − sin θ or 1 + cos θ in a denominator, try multiplying top and bottom by its conjugate.
- Write “= RHS” only when you have genuinely arrived there.
LHS = (1 − cos²θ)(1 + cot²θ)
= sin²θ × cosec²θ [using 1 − cos²θ = sin²θ and 1 + cot²θ = cosec²θ]
= sin²θ × (1/sin²θ)
= 1 = RHS. ✓
Two substitutions and it collapsed. That is typical — the hard part is recognising the identities, not the algebra.
LHS is a difference of squares in disguise:
= sec²θ − tan²θ
= 1 [rearranging 1 + tan²θ = sec²θ]
= RHS. ✓
Why it works: whenever you see (something − something)(something + something), expand it as a² − b² first and see whether an identity appears.
Take LHS over a common denominator (1 + sin θ)(1 − sin θ):
= [cos θ(1 − sin θ) + cos θ(1 + sin θ)] ÷ (1 − sin²θ)
= [cos θ − cos θ sin θ + cos θ + cos θ sin θ] ÷ cos²θ
= 2 cos θ ÷ cos²θ
= 2/cos θ = 2 sec θ = RHS. ✓
Notice the two middle terms cancelled, and 1 − sin²θ quietly became cos²θ. Those are the two moves being tested.
Multiply inside the root, top and bottom, by (1 − cos θ):
= √[(1 − cos θ)² ÷ ((1 + cos θ)(1 − cos θ))]
= √[(1 − cos θ)² ÷ (1 − cos²θ)]
= √[(1 − cos θ)² ÷ sin²θ]
= (1 − cos θ)/sin θ [both quantities are positive for an acute θ]
= 1/sin θ − cos θ/sin θ
= cosec θ − cot θ = RHS. ✓
The conjugate trick in the first line is the whole question. Practise it until it is a reflex.
Practice Worksheet
Ten questions, roughly in increasing order of difficulty. Do them on paper first — reading a solution feels like learning, but it is not the same thing. Reveal each answer only after you have committed to one.
Show Answer
For angle P: opposite = QR = 7, adjacent = PQ = 24, hypotenuse = PR = 25.
sin P = 7/25 and cos P = 24/25.
Show Answer
Opposite = √(289 − 64) = √225 = 15.
sin A = 15/17, cos A = 8/17.
sin A + cos A = 23/17.
Show Answer
Total = 1/2 + 1/2 + 1 = 2.
(The first two terms had to add to 1 — that is the identity at work.)
Show Answer
Numerator = √3 − 1/√3 = (3 − 1)/√3 = 2/√3.
Denominator = 1 + √3 × (1/√3) = 1 + 1 = 2.
Value = (2/√3) ÷ 2 = 1/√3 (equivalently √3/3).
Show Answer
cosec A = 5/4, cot A = 3/4, so cosec A − cot A = 2/4 = 1/2.
sec A = 5/3, tan A = 4/3, so sec A − tan A = 1/3.
Value = (1/2) ÷ (1/3) = 3/2.
Show Answer
= (1/cos²A) ÷ (1/sin²A)
= sin²A / cos²A
= tan²A = RHS. ✓
Show Answer
From the table, θ = 45°.
Check: sin 45° = cos 45° = 1/√2. ✓
Show Answer
sec 45° = √2 so sec²45° = 2; tan²45° = 1; second bracket = 1 and 3 × 1 = 3.
Value = 2 − 3 = −1.
(The second bracket is just sec² − tan² = 1, so you could have written 3 straight away.)
Show Answer
= (1/cos A − 1) ÷ (1/cos A + 1)
Multiply numerator and denominator by cos A:
= (1 − cos A) ÷ (1 + cos A) = RHS. ✓
One substitution and one tidy multiplication — that is the whole proof.
Show Answer
Since cosec θ + cot θ = 3, we get cosec θ − cot θ = 1/3.
Add the two equations: 2 cosec θ = 3 + 1/3 = 10/3, so cosec θ = 5/3 and sin θ = 3/5.
Subtract them: 2 cot θ = 3 − 1/3 = 8/3, so cot θ = 4/3.
Then cos θ = cot θ × sin θ = (4/3) × (3/5) = 4/5.
Check: (3/5)² + (4/5)² = 9/25 + 16/25 = 1. ✓
If a question fought back, do not mark it wrong and move on — redo it tomorrow morning from a blank page. Trigonometry rewards repetition more than cleverness, and you only need to be one more correct question than yesterday. That is the whole method.