Introduction to Trigonometry starts the study of how the sides and angles of a right triangle are linked. It defines the six trigonometric ratios of an acute angle, shows that they depend only on the angle, finds every ratio from one known ratio, derives the values for 0°, 30°, 45°, 60° and 90°, and proves the three identities built on the Pythagoras theorem. Board papers test it through quick value questions, ratio calculations and step-by-step identity proofs, and Chapter 9 on heights and distances rests on it.
Key Concepts
1. What Trigonometry Studies
The word trigonometry comes from three Greek words: tri (three), gon (sides) and metron (measure). Trigonometry is the study of the relationships between the sides and the angles of a triangle.
The NCERT chapter opens with situations where a right triangle can be imagined: a student looking at the top of the Qutub Minar, a girl on a balcony looking at a flower pot across a river, and a hot air balloon moving across the sky. In each, a height or distance is wanted without measuring it directly.
In this chapter we work only with acute angles of a right triangle, and we also define the ratios for 0° and 90°. The ratios can be extended to other angles, which you will meet in higher classes.
Naming the sides with respect to an angle
Take triangle ABC, right-angled at B. Look at the acute angle A (that is, angle CAB).
- BC faces angle A. It is the side opposite to angle A.
- AC faces the right angle. It is the hypotenuse, the longest side.
- AB is one of the two arms of angle A and is not the hypotenuse. It is the side adjacent to angle A.
Now switch to angle C. The side opposite to C is AB, the side adjacent to C is BC, and the hypotenuse is still AC. So the labels “opposite” and “adjacent” depend on which angle you are looking at. The hypotenuse is the only side whose name stays the same.
| Side of triangle ABC (right angle at B) | With respect to angle A | With respect to angle C |
|---|---|---|
| BC | Opposite | Adjacent |
| AB | Adjacent | Opposite |
| AC | Hypotenuse | Hypotenuse |
2. The Six Trigonometric Ratios
A trigonometric ratio of an acute angle is a ratio of two sides of a right triangle containing that angle. In triangle ABC, right-angled at B, the six ratios of angle A are:
| Ratio | Short form | Definition | In triangle ABC |
|---|---|---|---|
| sine of angle A | sin A | side opposite to A / hypotenuse | BC/AC |
| cosine of angle A | cos A | side adjacent to A / hypotenuse | AB/AC |
| tangent of angle A | tan A | side opposite to A / side adjacent to A | BC/AB |
| cosecant of angle A | cosec A | hypotenuse / side opposite to A = 1/sin A | AC/BC |
| secant of angle A | sec A | hypotenuse / side adjacent to A = 1/cos A | AC/AB |
| cotangent of angle A | cot A | side adjacent to A / side opposite to A = 1/tan A | AB/BC |
The last three are the reciprocals of the first three: cosec A = 1/sin A, sec A = 1/cos A and cot A = 1/tan A.
Two quotient relations
Dividing top and bottom of BC/AB by AC: tan A = (BC/AC) / (AB/AC) = sin A / cos A, and so cot A = cos A / sin A.
These two relations let you rewrite any expression in terms of sin and cos alone, which is the safest first move in most proofs.
Ratios of angle C in the same triangle
For angle C in the same triangle, sin C = AB/AC, cos C = BC/AC and tan C = AB/BC. Compare with angle A: sin C = cos A and cos C = sin A. This happens because the opposite and adjacent sides swap when you move from A to C.
What the symbol means
sin A is one symbol for “the sine of the angle A”. It is never the product of “sin” and A. (sin A)2 is written sin2A. cosec A = (sin A)−1, which differs from sin−1A (“sine inverse A”, studied in higher classes). The Greek letter θ is often used to name an angle.
3. Ratios Depend Only on the Angle
Take triangle ABC right-angled at B. Choose a point P on the hypotenuse AC and drop a perpendicular PM to AB. Choose a point Q on AC extended and drop a perpendicular QN to AB extended. Now there are three right triangles that all contain angle A: triangle PAM, triangle CAB and triangle QAN.
Triangles PAM and CAB share angle A and each has a right angle, so by the AA similarity criterion (Chapter 6) they are similar. Corresponding sides of similar triangles are proportional:
AM/AB = AP/AC = MP/BC
Rearranging, MP/AP = BC/AC = sin A, AM/AP = AB/AC = cos A, and MP/AM = BC/AB = tan A. The same argument works for triangle QAN.
Conclusion: the values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same.
This is why, given sin A = 1/3, you may call the sides k and 3k for any positive k.
Bounds on the ratios
The hypotenuse is the longest side, so the top of sin A and cos A is never larger than the bottom: 0 ≤ sin A ≤ 1 and 0 ≤ cos A ≤ 1 for 0° ≤ A ≤ 90°.
Their reciprocals sec A (for 0° ≤ A < 90°) and cosec A (for 0° < A ≤ 90°) are always greater than or equal to 1.
tan A has no upper bound: tan 60° = √3 is already more than 1.
4. Finding All Six Ratios from One Known Ratio
If one trigonometric ratio of an acute angle is known, the other five can be found. The method uses the definitions, one positive multiplier k and the Pythagoras theorem.
- Write the given ratio as a ratio of two named sides.
- Give those two sides lengths in terms of k, where k is a positive number.
- Find the third side by the Pythagoras theorem. Take the positive square root, because a length is positive.
- Write the remaining ratios straight from their definitions.
NCERT illustration. In right triangle ABC (right angle at B), sin A = 1/3. Then BC/AC = 1/3, so let BC = k and AC = 3k.
AB2 = AC2 − BC2 = (3k)2 − k2 = 8k2 = (2√2 k)2, so AB = 2√2 k.
The negative value −2√2 k is rejected because AB is a length. Now cos A = AB/AC = 2√2 k / 3k = 2√2/3, tan A = BC/AB = 1/(2√2), and the reciprocals follow: cosec A = 3, sec A = 3/(2√2), cot A = 2√2.
Pythagorean triples that appear again and again
Many questions are built on whole-number right triangles. Spotting them saves time.
| Legs | Hypotenuse | Where it appears in the chapter |
|---|---|---|
| 3, 4 | 5 | tan A = 4/3 (NCERT Example 1); 3 cot A = 4 (Exercise 8.1) |
| 5, 12 | 13 | sec θ = 13/12; PR + QR = 25, PQ = 5 (Exercise 8.1) |
| 7, 24 | 25 | AB = 24 cm, BC = 7 cm (Exercise 8.1); OP = 7 cm (NCERT Example 5) |
| 8, 15 | 17 | 15 cot A = 8 (Exercise 8.1) |
| 20, 21 | 29 | AB = 29, BC = 21 (NCERT Example 3) |
The identities of Concept 7 give a second route: from tan A = 1/√3, sec2A = 1 + 1/3 = 4/3, so sec A = 2/√3, cos A = √3/2 and sin A = √(1 − 3/4) = 1/2.
5. Trigonometric Ratios of 45°, 30° and 60°
These values are derived from two simple triangles. Knowing the derivation means you can rebuild the table in the exam hall if memory fails.
45°: the isosceles right triangle
In triangle ABC right-angled at B, let angle A = 45°. Then angle C = 45° too, because the angles add up to 180°. Sides opposite equal angles are equal, so BC = AB. Let BC = AB = a.
By the Pythagoras theorem, AC2 = a2 + a2 = 2a2, so AC = a√2.
- sin 45° = BC/AC = a/(a√2) = 1/√2
- cos 45° = AB/AC = 1/√2
- tan 45° = BC/AB = a/a = 1
- cosec 45° = √2, sec 45° = √2, cot 45° = 1
30° and 60°: half of an equilateral triangle
Take an equilateral triangle ABC, so each angle is 60°. Draw the perpendicular AD from A to BC. Triangles ABD and ACD are congruent (RHS: hypotenuse AB = AC, common side AD, right angles at D). So BD = DC and angle BAD = angle CAD (CPCT). This makes triangle ABD a right triangle, right-angled at D, with angle BAD = 30° and angle ABD = 60°.
Let AB = 2a. Then BD = half of BC = a, and AD2 = AB2 − BD2 = 4a2 − a2 = 3a2, so AD = a√3.
For 30° (angle BAD): opposite side is BD = a, adjacent side is AD = a√3, hypotenuse AB = 2a.
- sin 30° = BD/AB = 1/2, cos 30° = AD/AB = √3/2, tan 30° = BD/AD = 1/√3
- cosec 30° = 2, sec 30° = 2/√3, cot 30° = √3
For 60° (angle ABD): opposite side is AD = a√3, adjacent side is BD = a, hypotenuse AB = 2a.
- sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3
- cosec 60° = 2/√3, sec 60° = 2, cot 60° = 1/√3
Notice that the values for 30° and 60° are the same numbers with the roles of sin and cos swapped. 30° and 60° are the two acute angles of the same right triangle, so the side opposite one is adjacent to the other.
6. Ratios of 0° and 90° and the Complete Table
In triangle ABC right-angled at B, let angle A shrink towards 0°. BC shrinks towards 0 and AC becomes almost equal to AB, so sin A = BC/AC approaches 0 and cos A = AB/AC approaches 1. So we define sin 0° = 0 and cos 0° = 1. Then tan 0° = 0 and sec 0° = 1, while cot 0° = 1/0 and cosec 0° = 1/0 are not defined.
Now let A grow towards 90°. Angle C shrinks towards 0°, AB shrinks, and AC almost lies along BC, so sin A approaches 1 and cos A approaches 0. So we define sin 90° = 1 and cos 90° = 0. Then cot 90° = 0 and cosec 90° = 1, while tan 90° and sec 90° need division by 0 and are not defined.
Table 8.1: values for the standard angles
| Ratio | 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 |
How the table behaves. As A increases from 0° to 90°, sin A increases from 0 to 1 and cos A decreases from 1 to 0. tan A also increases, from 0 at 0° through 1 at 45° to √3 at 60°.
A memory aid for the first row
Write the numbers 0, 1, 2, 3, 4 under 0°, 30°, 45°, 60°, 90°. Divide each by 4 and take the square root:
√(0/4) = 0, √(1/4) = 1/2, √(2/4) = 1/√2, √(3/4) = √3/2, √(4/4) = 1.
That is the sin row. Write it backwards to get the cos row. Divide sin by cos to get the tan row. Take reciprocals to get the last three rows.
Finding sides and angles. If one side and one other part (an acute angle or another side) of a right triangle are known, the remaining sides and angles can be found. Pick the ratio that links what you know with what you want, and read angles from the table: opposite/hypotenuse = 1/2 means the angle is 30°.
7. Trigonometric Identities
An equation is an identity when it is true for all values of the variables involved. An equation involving trigonometric ratios of an angle is a trigonometric identity if it is true for all values of the angle(s) involved. An ordinary equation, such as sin A = 1/2, is true only for particular angles.
Proving the first identity
In triangle ABC right-angled at B, the Pythagoras theorem gives
AB2 + BC2 = AC2 (1)
Divide every term of (1) by AC2: (AB/AC)2 + (BC/AC)2 = (AC/AC)2, that is (cos A)2 + (sin A)2 = 1:
cos2A + sin2A = 1 (2)
This holds for all A with 0° ≤ A ≤ 90°. At 0° it reads 1 + 0 = 1 and at 90° it reads 0 + 1 = 1.
The second identity
Divide (1) by AB2: (AB/AB)2 + (BC/AB)2 = (AC/AB)2, which gives
1 + tan2A = sec2A (3)
At A = 0° it reads 1 + 0 = 1, which is true. At A = 90°, tan A and sec A are not defined. So (3) is true for all A with 0° ≤ A < 90°.
The third identity
Divide (1) by BC2: (AB/BC)2 + (BC/BC)2 = (AC/BC)2, which gives
cot2A + 1 = cosec2A (4)
cosec A and cot A are not defined at A = 0°, so (4) is true for all A with 0° < A ≤ 90°.
Useful rearranged forms
| Identity | Rearranged forms | Factorised form |
|---|---|---|
| sin2A + cos2A = 1 | sin2A = 1 − cos2A; cos2A = 1 − sin2A | (1 − sin A)(1 + sin A) = cos2A; (1 − cos A)(1 + cos A) = sin2A |
| 1 + tan2A = sec2A | sec2A − tan2A = 1 | (sec A − tan A)(sec A + tan A) = 1 |
| cot2A + 1 = cosec2A | cosec2A − cot2A = 1 | (cosec A − cot A)(cosec A + cot A) = 1 |
The factorised forms are the hidden key to many proofs. For example, (sec A − tan A)(sec A + tan A) = 1 tells you at once that sec A + tan A = 1/(sec A − tan A).
Expressing one ratio in terms of another
The identities let you write every ratio in terms of any single one. NCERT Example 9 writes cos A, tan A and sec A in terms of sin A:
- cos2A = 1 − sin2A, so cos A = √(1 − sin2A). The positive root is taken because A is acute, so cos A is positive.
- tan A = sin A / cos A = sin A / √(1 − sin2A)
- sec A = 1 / cos A = 1 / √(1 − sin2A)
A method for proving identities
Start from the more complicated side and work only on it (unless told to simplify both sides separately). Rewrite mixed ratios in sin and cos, take an LCM, and look for sin2A + cos2A = 1 or a factorised form. If the question names an identity, divide by the ratio that brings those two functions in and replace 1 by, for example, sec2θ − tan2θ. End with “= RHS”.
Formula and Theorem Sheet
| Result | Statement | Valid for |
|---|---|---|
| Reciprocals | cosec A = 1/sin A, sec A = 1/cos A, cot A = 1/tan A | Wherever the denominator is not zero |
| Quotients | tan A = sin A / cos A, cot A = cos A / sin A | Wherever the denominator is not zero |
| 0° | sin 0° = 0, cos 0° = 1, tan 0° = 0; cot 0° and cosec 0° not defined | By definition |
| 30° | sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3 | Half equilateral triangle |
| 45° | sin 45° = cos 45° = 1/√2, tan 45° = 1 | Isosceles right triangle |
| 60° | sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3 | Half equilateral triangle |
| 90° | sin 90° = 1, cos 90° = 0, cot 90° = 0; tan 90° and sec 90° not defined | By definition |
| Identity 1 | sin2A + cos2A = 1 | 0° ≤ A ≤ 90° |
| Identity 2 | 1 + tan2A = sec2A | 0° ≤ A < 90° |
| Identity 3 | cot2A + 1 = cosec2A | 0° < A ≤ 90° |
Read the rest of the chapter →Hide the rest ↑
Important Definitions
- Trigonometry: the study of relationships between the sides and angles of a triangle. The name comes from the Greek tri (three), gon (sides) and metron (measure).
- Hypotenuse: the side opposite the right angle in a right triangle; the longest side.
- Side opposite to an angle: the side of the right triangle that faces the given acute angle.
- Side adjacent to an angle: the side that forms the given acute angle along with the hypotenuse.
- sin A, cos A, tan A: opposite/hypotenuse, adjacent/hypotenuse and opposite/adjacent for angle A; cosec, sec and cot are their reciprocals.
- Not defined: a ratio whose calculation would need division by zero, such as tan 90°, sec 90°, cot 0° and cosec 0°.
- Identity: an equation that is true for all values of the variables involved.
- Trigonometric identity: an equation involving trigonometric ratios of an angle that is true for all values of the angle(s) involved.
Solved Examples (NCERT-Based)
Example 1: All ratios from tan A (NCERT Example 1)
Question: Given tan A = 4/3, find the other trigonometric ratios of the angle A.
Solution: Draw triangle ABC right-angled at B. tan A = BC/AB = 4/3, so let BC = 4k and AB = 3k, where k is a positive number.
By the Pythagoras theorem, AC2 = AB2 + BC2 = (3k)2 + (4k)2 = 25k2, so AC = 5k.
sin A = BC/AC = 4k/5k = 4/5 and cos A = AB/AC = 3k/5k = 3/5.
cot A = 1/tan A = 3/4, cosec A = 1/sin A = 5/4, sec A = 1/cos A = 5/3.
Example 2: Two triangles, 24 and 7 (NCERT Exercise 8.1, Q1)
Question: In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A (ii) sin C, cos C
Solution: AC2 = AB2 + BC2 = 576 + 49 = 625, so AC = 25 cm.
(i) For angle A, the opposite side is BC and the adjacent side is AB. sin A = BC/AC = 7/25 and cos A = AB/AC = 24/25.
(ii) For angle C, the opposite side is AB and the adjacent side is BC. sin C = AB/AC = 24/25 and cos C = BC/AC = 7/25.
Example 3: From cot A to sin A and sec A (NCERT Exercise 8.1, Q4)
Question: Given 15 cot A = 8, find sin A and sec A.
Solution: cot A = 8/15 = adjacent/opposite. In triangle ABC right-angled at B, let AB = 8k and BC = 15k.
AC2 = (8k)2 + (15k)2 = 64k2 + 225k2 = 289k2, so AC = 17k.
sin A = BC/AC = 15/17 and sec A = AC/AB = 17/8.
Example 4: A sum of two sides (NCERT Exercise 8.1, Q10)
Question: In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Solution: Let QR = x cm. Then PR = (25 − x) cm. PR is the hypotenuse, so
PR2 = PQ2 + QR2
(25 − x)2 = 25 + x2
625 − 50x + x2 = 25 + x2
50x = 600, so x = 12.
So QR = 12 cm and PR = 13 cm. For angle P the opposite side is QR and the adjacent side is PQ.
sin P = QR/PR = 12/13, cos P = PQ/PR = 5/13, tan P = QR/PQ = 12/5.
Example 5: Using both acute angles (NCERT Exercise 8.1, Q9)
Question: In triangle ABC, right-angled at B, if tan A = 1/√3, find the value of: (i) sin A cos C + cos A sin C (ii) cos A cos C − sin A sin C
Solution: tan A = 1/√3 = BC/AB. Let BC = k and AB = √3 k. Then AC = √(3k2 + k2) = 2k.
sin A = BC/AC = 1/2, cos A = AB/AC = √3/2, sin C = AB/AC = √3/2, cos C = BC/AC = 1/2.
(i) sin A cos C + cos A sin C = (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1
(ii) cos A cos C − sin A sin C = (√3/2)(1/2) − (1/2)(√3/2) = √3/4 − √3/4 = 0
Example 6: Finding sides from an angle (NCERT Example 6)
Question: In Δ ABC, right-angled at B, AB = 5 cm and ∠ACB = 30°. Determine the lengths of the sides BC and AC.
Solution: With respect to angle C, AB is the opposite side and BC is the adjacent side. So
AB/BC = tan C, that is 5/BC = tan 30° = 1/√3, which gives BC = 5√3 cm.
For AC, use sin C = AB/AC: 1/2 = 5/AC, so AC = 10 cm.
Check by the Pythagoras theorem: AC = √(52 + (5√3)2) = √(25 + 75) = √100 = 10 cm.
Example 7: Finding angles from sides (NCERT Example 7)
Question: In Δ PQR, right-angled at Q, PQ = 3 cm and PR = 6 cm. Determine ∠QPR and ∠PRQ.
Solution: PQ is opposite to angle R and PR is the hypotenuse. So sin R = PQ/PR = 3/6 = 1/2, which gives ∠PRQ = 30°.
The angles of the triangle add up to 180° and ∠Q = 90°, so ∠QPR = 180° − 90° − 30° = 60°.
Example 8: Evaluating with standard values (NCERT Exercise 8.2, Q1)
Question: Evaluate the following: (ii) 2 tan2 45° + cos2 30° − sin2 60° (v) (5 cos2 60° + 4 sec2 30° − tan2 45°)/(sin2 30° + cos2 30°)
Solution:
(ii) 2(1)2 + (√3/2)2 − (√3/2)2 = 2 + 3/4 − 3/4 = 2
(v) Numerator = 5(1/2)2 + 4(2/√3)2 − 12 = 5/4 + 16/3 − 1 = (15 + 64 − 12)/12 = 67/12.
Denominator = (1/2)2 + (√3/2)2 = 1/4 + 3/4 = 1.
So the value is 67/12.
Example 9: Two angles from two equations (NCERT Exercise 8.2, Q3)
Question: If tan (A + B) = √3 and tan (A − B) = 1/√3; 0° < A + B ≤ 90°; A > B, find A and B.
Solution: tan (A + B) = √3 = tan 60°, so A + B = 60° (1)
tan (A − B) = 1/√3 = tan 30°, so A − B = 30° (2)
Adding (1) and (2): 2A = 90°, so A = 45°. Then B = 60° − 45° = 15°.
Example 10: A short identity proof (NCERT Example 10)
Question: Prove that sec A (1 − sin A)(sec A + tan A) = 1.
Solution:
LHS = (1/cos A)(1 − sin A)(1/cos A + sin A/cos A)
= (1 − sin A)(1 + sin A) / cos2A
= (1 − sin2A) / cos2A
= cos2A / cos2A = 1 = RHS
Example 11: Identity with a square (NCERT Exercise 8.3, Q4(i))
Question: Prove that (cosec θ − cot θ)2 = (1 − cos θ)/(1 + cos θ).
Solution:
LHS = (1/sin θ − cos θ/sin θ)2 = (1 − cos θ)2/sin2θ
= (1 − cos θ)2/(1 − cos2θ)
= (1 − cos θ)(1 − cos θ) / [(1 − cos θ)(1 + cos θ)]
= (1 − cos θ)/(1 + cos θ) = RHS
Example 12: Using a named identity (NCERT Exercise 8.3, Q4(v))
Question: Prove that (cos A − sin A + 1)/(cos A + sin A − 1) = cosec A + cot A, using the identity cosec2A = 1 + cot2A.
Solution: Divide the numerator and the denominator of the LHS by sin A:
LHS = (cot A − 1 + cosec A)/(cot A + 1 − cosec A)
In the numerator, replace 1 by cosec2A − cot2A:
Numerator = (cosec A + cot A) − (cosec2A − cot2A)
= (cosec A + cot A) − (cosec A + cot A)(cosec A − cot A)
= (cosec A + cot A)(1 − cosec A + cot A)
The bracket (1 − cosec A + cot A) is exactly the denominator. So
LHS = (cosec A + cot A)(1 − cosec A + cot A)/(cot A + 1 − cosec A) = cosec A + cot A = RHS
Competency-Based Questions (with answers)
1. Case-based: The wheelchair ramp
A school builds a ramp to its library door. The ramp surface is 13 m long and it rises 5 m from the ground to the door level. The ramp surface, the vertical rise and the horizontal ground form a right triangle with the right angle between the rise and the ground. Let θ be the angle the ramp makes with the ground.
(i) Find the horizontal distance covered by the ramp. (ii) Find sin θ, cos θ and tan θ. (iii) A second ramp has the same angle θ but a vertical rise of only 2 m. Find its length.
Answer: (i) Horizontal distance = √(132 − 52) = √(169 − 25) = √144 = 12 m.
(ii) The rise is opposite to θ and the ground is adjacent to it. sin θ = 5/13, cos θ = 12/13, tan θ = 5/12.
(iii) The ratios depend only on the angle, so sin θ is still 5/13. Length = rise / sin θ = 2 × 13/5 = 26/5 = 5.2 m.
2. Case-based: The kite string
Riya flies a kite. The taut string from her hand to the kite makes an angle of 60° with the horizontal line through her hand, and the kite is 30 m directly above a point on that horizontal line. Treat the string as a straight line.
(i) Name the side of the right triangle that the string represents. (ii) Find the length of the string. (iii) Find the horizontal distance from her hand to the point below the kite.
Answer: (i) The string is the hypotenuse, as it faces the right angle between the vertical and the horizontal.
(ii) sin 60° = 30/string, so √3/2 = 30/L and L = 60/√3 = 20√3 m.
(iii) tan 60° = 30/d, so √3 = 30/d and d = 30/√3 = 10√3 m.
3. Source-based: From jya to sine
Read the passage: “The first use of the idea of sine in the way we use it today was in the work Aryabhatiyam by Aryabhata, in A.D. 500. Aryabhata used the word ardha-jya for the half-chord, which was shortened to jya or jiva in due course. An English Professor of astronomy Edmund Gunter (1581 to 1626), first used the abbreviated notation ‘sin’. The cosine function arose from the need to compute the sine of the complementary angle. Aryabhatta called it kotijya.”
(i) Who first used the abbreviation “sin”? (ii) In triangle ABC right-angled at B, angles A and C are complementary. Show that cos A = sin C. (iii) If sin C = 0.6, find cos A.
Answer: (i) Edmund Gunter, an English professor of astronomy, first used “sin”.
(ii) cos A = AB/AC (adjacent to A over hypotenuse). AB is opposite to angle C, so sin C = AB/AC. Hence cos A = sin C.
(iii) cos A = sin C = 0.6.
4. Assertion-Reason
Assertion (A): For an acute angle A, the value of sin A can be 4/3.
Reason (R): The hypotenuse is the longest side of a right triangle.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (d). sin A = opposite/hypotenuse, and the hypotenuse is the longest side, so sin A can never exceed 1. 4/3 is more than 1, so A is false. R is a true fact.
5. Assertion-Reason
Assertion (A): sin 30° = cos 60°.
Reason (R): In a right triangle, the side opposite one acute angle is the side adjacent to the other acute angle.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (a). Both values equal 1/2. In the half equilateral triangle, 30° and 60° are the two acute angles; the side BD is opposite to 30° and adjacent to 60°, so BD/AB is at once sin 30° and cos 60°. R explains A.
6. Assertion-Reason
Assertion (A): 1 + tan2A = sec2A is true for A = 90°.
Reason (R): tan 90° and sec 90° are not defined.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (d). tan 90° = sin 90°/cos 90° = 1/0 and sec 90° = 1/0 are not defined, so the identity cannot be applied at 90°. It holds only for 0° ≤ A < 90°. A is false and R is true.
7. Error analysis: The wrong expansion
A student writes: “sin (30° + 60°) = sin 30° + sin 60° = 1/2 + √3/2.”
(i) Find the correct value of sin (30° + 60°). (ii) Is sin (A + B) = sin A + sin B an identity?
Answer: (i) sin (30° + 60°) = sin 90° = 1. The student’s value is (1 + √3)/2, which is about 1.37 and is more than 1, which a sine can never be.
(ii) No. A single counter-example, such as A = 30°, B = 60°, is enough to show the statement is false. This is NCERT Exercise 8.2, Q4(i).
8. Competency MCQ: Which value is possible
For an acute angle A, which one of these can be true?
(a) cos A = 1.2 (b) cosec A = 0.8 (c) sec A = 12/5 (d) sin A = √5/2
Answer: (c). cos A and sin A cannot exceed 1: 1.2 is too big, and √5/2 is about 1.118, also too big. cosec A cannot be less than 1, so 0.8 is impossible. sec A = 12/5 means hypotenuse 12 and adjacent side 5, which is a real right triangle with opposite side √(144 − 25) = √119. This matches NCERT Exercise 8.1, Q11(ii).
Important Questions for Board Exams
1-Mark Questions
Q1. Choose the correct option: 9 sec2A − 9 tan2A = (A) 1 (B) 9 (C) 8 (D) 0
Answer: (B) 9, because 9(sec2A − tan2A) = 9 × 1 = 9.
Q2. Choose the correct option: (1 − tan2 45°)/(1 + tan2 45°) = (A) tan 90° (B) 1 (C) sin 45° (D) 0
Answer: (D) 0, because tan 45° = 1 makes the numerator 1 − 1 = 0 and the denominator 2.
Q3. Choose the correct option: sin 2A = 2 sin A is true when A = (A) 0° (B) 30° (C) 45° (D) 60°
Answer: (A) 0°. Both sides equal 0. At 30°, sin 60° = √3/2 while 2 sin 30° = 1, so 30° fails; 45° and 60° fail the same way.
Q4. Choose the correct option: (1 + tan2A)/(1 + cot2A) = (A) sec2A (B) −1 (C) cot2A (D) tan2A
Answer: (D) tan2A, because sec2A/cosec2A = (1/cos2A)(sin2A) = tan2A.
Q5. State whether true or false and justify: cot A is not defined for A = 0°.
Answer: True. cot 0° = cos 0°/sin 0° = 1/0, which is not defined.
2-Mark Questions
Q6. If sin A = 3/4, calculate cos A and tan A.
Answer: Let the opposite side be 3k and the hypotenuse 4k. Adjacent side = √(16k2 − 9k2) = √7 k. So cos A = √7/4 and tan A = 3/√7.
Q7. Evaluate: sin 60° cos 30° + sin 30° cos 60°
Answer: (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1.
Q8. If cot θ = 7/8, evaluate: (i) (1 + sin θ)(1 − sin θ)/[(1 + cos θ)(1 − cos θ)] (ii) cot2θ
Answer: (i) The expression is (1 − sin2θ)/(1 − cos2θ) = cos2θ/sin2θ = cot2θ = 49/64. (ii) cot2θ = (7/8)2 = 49/64.
Q9. Choose the correct option and justify: 2 tan 30°/(1 + tan2 30°) = (A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
Answer: 2 tan 30° = 2/√3 and 1 + tan2 30° = 1 + 1/3 = 4/3. The value is (2/√3) × (3/4) = 3/(2√3) = √3/2 = sin 60°. Option (A).
3-Mark Questions
Q10. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Answer: A and B are any two acute angles, so place each in its own right triangle (the method of NCERT Example 2). Let triangle ACD be right-angled at C with acute angle A, and triangle BEF be right-angled at E with acute angle B.
cos A = AC/AD and cos B = BE/BF. Since cos A = cos B, AC/AD = BE/BF, so AC/BE = AD/BF = k, say. (1)
By the Pythagoras theorem, CD = √(AD2 − AC2) = √(k2BF2 − k2BE2) = k√(BF2 − BE2) = k × EF, so CD/EF = k. (2)
From (1) and (2), AC/BE = AD/BF = CD/EF. So triangle ACD ~ triangle BEF by the SSS similarity criterion, and therefore ∠A = ∠B.
Q11. Evaluate: cos 45°/(sec 30° + cosec 30°)
Answer: sec 30° + cosec 30° = 2/√3 + 2 = (2 + 2√3)/√3. So the value is (1/√2) × √3/(2 + 2√3) = √3/[2√2(1 + √3)]. Multiply top and bottom by (√3 − 1): √3(√3 − 1)/[2√2 × 2] = (3 − √3)/(4√2) = (3√2 − √6)/8.
Q12. Prove that (1 + sec A)/sec A = sin2A/(1 − cos A).
Answer: LHS = (1 + 1/cos A)/(1/cos A) = cos A + 1. RHS = (1 − cos2A)/(1 − cos A) = (1 − cos A)(1 + cos A)/(1 − cos A) = 1 + cos A. LHS = RHS.
Q13. Prove that √[(1 + sin A)/(1 − sin A)] = sec A + tan A.
Answer: Multiply inside the root by (1 + sin A)/(1 + sin A): LHS = √[(1 + sin A)2/(1 − sin2A)] = √[(1 + sin A)2/cos2A] = (1 + sin A)/cos A, taking the positive root since A is acute. This equals 1/cos A + sin A/cos A = sec A + tan A = RHS.
5-Mark Questions
Q14. Prove that tan θ/(1 − cot θ) + cot θ/(1 − tan θ) = 1 + sec θ cosec θ. [Write the expression in terms of sin θ and cos θ.]
Answer: Write s = sin θ and c = cos θ.
First term = (s/c) / [(s − c)/s] = s2/[c(s − c)].
Second term = (c/s) / [(c − s)/c] = c2/[s(c − s)] = −c2/[s(s − c)].
LHS = [s3 − c3]/[sc(s − c)] = (s − c)(s2 + sc + c2)/[sc(s − c)] = (1 + sc)/(sc) = 1/(sc) + 1.
1/(sin θ cos θ) = sec θ cosec θ, so LHS = 1 + sec θ cosec θ = RHS.
Q15. Prove that (sin θ − cos θ + 1)/(sin θ + cos θ − 1) = 1/(sec θ − tan θ), using the identity sec2θ = 1 + tan2θ.
Answer: Divide the numerator and denominator by cos θ:
LHS = (tan θ − 1 + sec θ)/(tan θ + 1 − sec θ)
Multiply top and bottom by (tan θ − sec θ):
= [(tan θ + sec θ) − 1](tan θ − sec θ) / {[(tan θ − sec θ) + 1](tan θ − sec θ)}
The numerator is (tan2θ − sec2θ) − (tan θ − sec θ) = −1 − tan θ + sec θ.
So LHS = −(1 + tan θ − sec θ) / [(tan θ − sec θ + 1)(tan θ − sec θ)] = −1/(tan θ − sec θ) = 1/(sec θ − tan θ) = RHS. (This is NCERT Example 12.)
Q16. Prove that (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2A + cot2A.
Answer: Expand each square:
(sin A + cosec A)2 = sin2A + cosec2A + 2 sin A cosec A = sin2A + cosec2A + 2
(cos A + sec A)2 = cos2A + sec2A + 2
LHS = (sin2A + cos2A) + cosec2A + sec2A + 4 = 1 + (1 + cot2A) + (1 + tan2A) + 4 = 7 + tan2A + cot2A = RHS.
Common Mistakes and Examiner Tips
- Mixing up opposite and adjacent. Students often label sides once and use the same labels for the other angle. Fix: before writing any ratio, say which angle you are working with and mark its opposite side. For angle C in triangle ABC (right angle at B), AB is the opposite side and BC is the adjacent side.
- Using the hypotenuse as the adjacent side. The hypotenuse also touches the angle. Fix: the adjacent side is the arm of the angle that is not the hypotenuse.
- Treating sin A as sin × A. Writing sin 2A = 2 sin A, or cancelling “sin” from top and bottom, is wrong. Fix: sin A is one symbol for one number. Check with 30°: sin 60° = √3/2 while 2 sin 30° = 1.
- Splitting sin (A + B). sin (A + B) = sin A + sin B is false (try A = 30°, B = 60°). Fix: find the angle A + B first, then take its sine.
- Reading sin−1A as cosec A. cosec A = (sin A)−1 = 1/sin A. The symbol sin−1A means something else (sine inverse). Fix: write 1/sin A or cosec A in answers.
- Keeping the negative root. When the Pythagoras theorem gives AB2 = 8k2, some students write AB = ±2√2 k. Fix: a side is a length, so take the positive root. For acute angles all six ratios are positive.
- Swapping the sin and cos rows. Writing sin 30° = √3/2 is a frequent slip. Fix: sin increases from 0 to 1, so the smaller angle has the smaller sine: sin 30° = 1/2 is less than sin 60° = √3/2.
- Calling tan 90° “infinity” or 0. Fix: write “not defined”, and give the reason, division by cos 90° = 0. The same wording is needed for sec 90°, cot 0° and cosec 0°.
- Ignoring the range of an identity. 1 + tan2A = sec2A fails to make sense at 90°, and cot2A + 1 = cosec2A fails at 0°. Fix: learn each identity together with its range.
- Working on both sides of an identity at once. Moving terms across the equals sign assumes what you are trying to prove. Fix: start from one side and reach the other, unless the question says “simplify LHS and RHS separately”, as in NCERT Exercise 8.3 Q4(iv) and (ix).
Quick Revision Points
- Trigonometry studies the relationships between the sides and angles of a triangle; the word means “three sides measure”.
- Opposite and adjacent depend on the angle chosen. The hypotenuse is always the side facing the right angle.
- sin A = opposite/hypotenuse, cos A = adjacent/hypotenuse, tan A = opposite/adjacent.
- cosec A = 1/sin A, sec A = 1/cos A, cot A = 1/tan A.
- tan A = sin A/cos A and cot A = cos A/sin A.
- In a right triangle, sin of one acute angle equals cos of the other.
- sin2A means (sin A)2; sin−1A is a different idea from cosec A.
- Ratios depend only on the angle, by AA similarity; the size of the triangle does not matter.
- Given one ratio, call the two sides mk and nk, find the third by Pythagoras, then write the rest.
- sin A and cos A lie between 0 and 1; sec A and cosec A are at least 1; tan A can be any non-negative value.
- 45° comes from an isosceles right triangle with sides a, a, a√2.
- 30° and 60° come from half an equilateral triangle with sides a, a√3, 2a.
- sin row: 0, 1/2, 1/√2, √3/2, 1. cos row: the same in reverse.
- tan row: 0, 1/√3, 1, √3, not defined.
- Not defined: tan 90°, sec 90°, cot 0°, cosec 0°.
- sin2A + cos2A = 1 for 0° ≤ A ≤ 90°.
- 1 + tan2A = sec2A for 0° ≤ A < 90°.
- cot2A + 1 = cosec2A for 0° < A ≤ 90°.
- (sec A − tan A)(sec A + tan A) = 1 and (cosec A − cot A)(cosec A + cot A) = 1.
- To prove an identity, convert to sin and cos, simplify one side, and end with “= RHS”.
Weightage in Board Exams
Introduction to Trigonometry belongs to the Trigonometry unit of Class 10 Maths, together with Chapter 9, Some Applications of Trigonometry. Check the current CBSE course structure for the marks given to the unit. Within it, this chapter is assessed in a predictable way.
| Question type | What is usually asked |
|---|---|
| MCQ and Assertion-Reason | Standard values; “not defined” cases; possible and impossible values of a ratio; ranges of the identities; one-step simplifications such as 9 sec2A − 9 tan2A |
| Short answers | All ratios from one given ratio; evaluating an expression with standard values; finding two angles from two equations |
| Long answers | Proving an identity, often with a hint to write it in sin and cos or to use a named identity |
| Case-based | A real right-triangle situation (ramp, ladder, string) split into parts: name the sides, find a ratio, find a missing length |
Preparation plan: know Table 8.1 without error, practise every NCERT exercise item including all ten parts of Exercise 8.3 Q4, and revise this chapter together with Chapter 9.
Class 10 Maths · Chapter 8 – swipe through all 9 cards to understand the whole chapter.
Opposite, adjacent, hypotenuse
In triangle ABC right-angled at B, the names of the legs depend on the angle you look at.
Only the hypotenuse keeps its name for both acute angles.
- For angle A: BC is opposite, AB is adjacent
- For angle C: AB is opposite, BC is adjacent
- AC faces the right angle, so it is the hypotenuse
sin, cos and tan
Three basic ratios of an acute angle A in a right triangle.
tan A = sin A / cos A
- sin A = BC/AC
- cos A = AB/AC
- tan A = BC/AB
cosec, sec and cot
The other three ratios are the reciprocals of the first three.
cot A = cos A / sin A
- cosec pairs with sin
- sec pairs with cos
- cot pairs with tan
Ratios depend only on the angle
Right triangles with the same acute angle are similar (AA), so their side ratios match.
sin A, cos A are between 0 and 1; sec A, cosec A are 1 or more.
- Triangle size never changes sin, cos or tan
- This is why sides can be taken as k and 3k
- sin θ = 4/3 is impossible
All ratios from one ratio
Write the given ratio as two sides in k, find the third side by Pythagoras, then use the definitions.
Take the positive square root: a side is a length.
- sin A = 4/5, cos A = 3/5
- cosec A = 5/4, sec A = 5/3, cot A = 3/4
- Common triples: 3-4-5, 5-12-13, 7-24-25, 8-15-17
30°, 45° and 60°
45° comes from an isosceles right triangle; 30° and 60° from half an equilateral triangle.
Sides a, a, a√2 for 45°; a, a√3, 2a for 30° and 60°.
- cos 30° = √3/2, cos 45° = 1/√2, cos 60° = 1/2
- tan 30° = 1/√3, tan 45° = 1, tan 60° = √3
- sin of one acute angle = cos of the other
0° and 90°
Defined by shrinking or growing angle A inside a right triangle.
As A goes from 0° to 90°, sin A rises and cos A falls.
- tan 0° = 0, sec 0° = 1, cot 90° = 0, cosec 90° = 1
- Not defined: tan 90°, sec 90°
- Not defined: cot 0°, cosec 0°
The three identities
Divide AB2 + BC2 = AC2 by AC2, AB2 and BC2 in turn.
1 + tan2A = sec2A fails at 90°; cot2A + 1 = cosec2A fails at 0°.
- sin2A + cos2A = 1 holds for 0° ≤ A ≤ 90°
- (sec A − tan A)(sec A + tan A) = 1
- (cosec A − cot A)(cosec A + cot A) = 1
Proving an identity
Work from one side to the other, never both at once.
Proof: (1 − sin A)(1 + sin A)/cos2A = cos2A/cos2A = 1
- Convert to sin and cos first
- Take an LCM and use sin2A + cos2A = 1
- End with the line = RHS
📝 Practice Introduction to Trigonometry - 10 board questions
CBSE previous-year and competency-based · with answers & explanations
Start →Close ✕
Why C: Step 1 (take a right triangle): opposite side = a, hypotenuse = b, so adjacent side = √(b² − a²) by Pythagoras. Step 2 (cos = adjacent/hypotenuse): cos θ = √(b² − a²)/b.
Why not A: b/√(b² − a²) is hypotenuse over adjacent, which is sec θ.
Why not B: b/a is hypotenuse over opposite, which is cosec θ.
Why not D: a/√(b² − a²) is opposite over adjacent, which is tan θ.
Remember: sin θ = a/b means opposite a, hypotenuse b, adjacent √(b² − a²).
Why B: Step 1 (sin²A + cos²A = 1): cos²A = 1/4, so sin²A = 3/4. Step 2 (substitute): 3/4 + 2 × 1/4 = 3/4 + 2/4 = 5/4.
Why not A: 3/2 uses cos A = 1/2 instead of cos²A = 1/4 in the second term.
Why not C: −1 is impossible, since both terms are squares and cannot add up to a negative number.
Why not D: 1/2 is cos A itself, not the value of the expression.
Remember: sin²A + 2 cos²A = 1 + cos²A.
Why C: Step 1 (1 + tan²A = sec²A): √(sec²A − 1) = tan A for an acute angle. Step 2 (write in sin and cos): sec A/tan A = (1/cos A) × (cos A/sin A) = 1/sin A = cosec A.
Why not A: sin A is the reciprocal of the correct answer.
Why not B: tan A is only the denominator √(sec²A − 1), not the whole fraction.
Why not D: cos A is 1/sec A, which ignores the tan A in the denominator.
Remember: sec²A − 1 = tan²A, and sec A/tan A = cosec A.
Why B: Step 1 (Pythagoras, sides 1 and 3): opposite = 1, hypotenuse = 3, adjacent = √(9 − 1) = √8 = 2√2. Step 2 (cot = adjacent/opposite): cot A = 2√2/1 = 2√2.
Why not A: 2√2/3 is adjacent/hypotenuse, which is cos A.
Why not C: 1/(2√2) is opposite/adjacent, which is tan A.
Why not D: 3 is hypotenuse/opposite, which is cosec A.
Remember: cot A = cos A/sin A = adjacent/opposite.
Why D: Step 1 (cosec θ = 1/sin θ): cosec θ = 9, so 9 cosec θ = 81. Step 2 (substitute): (81 + 1)/(81 − 1) = 82/80.
Why not A: 0 would need the numerator to be zero, but 9 cosec θ + 1 = 82.
Why not B: 80/81 is (81 − 1)/81, which divides by 9 cosec θ instead of by 9 cosec θ − 1.
Why not C: 1 comes from cancelling 9 cosec θ between numerator and denominator, which is not allowed across a sum or difference.
Remember: cosec θ is the reciprocal of sin θ, so sin θ = 1/9 gives cosec θ = 9.
Why A: Step 1 (standard table): sin 30° = 1/2, tan 45° = 1, sec 60° = 2. Step 2 (solve): 1/2 × 1 = 2/k, so k = 4.
Why not B: k = 3 gives 2/3 on the right, but the left side is 1/2.
Why not C: 2 is sec 60° itself, not k.
Why not D: k = 1 would need sec 60° = 1/2, which is cos 60°, not sec 60°.
Remember: sec 60° = 1/cos 60° = 2.
Why C: Step 1 (split 7 cos²θ as 3 cos²θ + 4 cos²θ): 3(sin²θ + cos²θ) + 4 cos²θ = 4. Step 2 (sin²θ + cos²θ = 1): 3 + 4 cos²θ = 4, so cos²θ = 1/4 and cos θ = 1/2. Step 3 (standard table): cos 60° = 1/2, so θ = 60°.
Why not A: at 30°, cos²θ = 3/4 and sin²θ = 1/4, giving 21/4 + 3/4 = 6, not 4.
Why not B: at 45°, both squares are 1/2, giving 7/2 + 3/2 = 5, not 4.
Why not D: at 90°, cos²θ = 0 and sin²θ = 1, giving 3, not 4.
Remember: pull out a sin²θ + cos²θ pair and replace it with 1.
Why D: Step 1 (standard table): tan 60° = √3, so 3θ = 60°. Step 2 (divide): θ = 20°, so θ/2 = 10°.
Why not A: 60° is 3θ, not θ/2.
Why not B: 30° takes tan 30° = √3, but tan 30° = 1/√3.
Why not C: 20° is θ; the question asks for half of it.
Remember: tan 60° = √3 and tan 30° = 1/√3.
Why C: Step 1 (1/cos θ = sec θ): (1/cos θ) × sec θ = sec²θ. Step 2 (1 + tan²θ = sec²θ): tan²θ − sec²θ = −(sec²θ − tan²θ) = −1.
Why not A: 1 is sec²θ − tan²θ, the expression in the other order.
Why not B: 0 would need tan²θ = sec²θ, but sec²θ is always 1 more than tan²θ.
Why not D: 2 comes from adding 1 + 1 instead of subtracting using the identity.
Remember: sec²θ − tan²θ = 1, so tan²θ − sec²θ = −1.
Why A: Step 1 (divide numerator and denominator by cos θ): (sin θ + cos θ)/(sin θ − cos θ) = (tan θ + 1)/(tan θ − 1). Step 2 (substitute tan θ = 5/12): (5/12 + 1)/(5/12 − 1) = (17/12)/(−7/12) = −17/7.
Why not B: 17/7 drops the minus sign; sin θ is smaller than cos θ here, so the denominator is negative.
Why not C: 17/13 uses the hypotenuse 13 as the denominator instead of sin θ − cos θ.
Why not D: −7/13 is sin θ − cos θ = (5 − 12)/13 on its own, only the denominator.
Remember: an expression in sin θ and cos θ of the same degree turns into tan θ when you divide by cos θ.
Chapter Navigation
Previous: Coordinate Geometry Class 10 Notes
Next: Some Applications of Trigonometry Class 10 Notes
Related Chapters in Class 10 Maths
- Real Numbers Class 10 Notes
- Polynomials Class 10 Notes
- Pair of Linear Equations in Two Variables Class 10 Notes
- Quadratic Equations Class 10 Notes
- Arithmetic Progressions Class 10 Notes
- Triangles Class 10 Notes
Explore More
Frequently Asked Questions
First fix the acute angle you are working with. The side facing that angle is the opposite side. The side facing the right angle is the hypotenuse. The remaining side, which forms the angle along with the hypotenuse, is the adjacent side. When you switch to the other acute angle, the opposite and adjacent sides swap, while the hypotenuse stays the same.
Two right triangles with the same acute angle are similar by the AA criterion, so their corresponding sides are proportional. A ratio of two sides is therefore the same in both. That is why sin 30° is 1/2 in every right triangle with a 30° angle, and why you may take sides as k and 3k for any positive k.
No. sin A and cos A each have the hypotenuse in the denominator, and the hypotenuse is the longest side of a right triangle, so their values lie from 0 to 1. For the same reason sec A and cosec A, their reciprocals, are always 1 or more where defined. tan A has no such limit: tan 60° is √3.
At 0°, sin 0° = 0, so cosec 0° = 1/0 and cot 0° = cos 0°/sin 0° = 1/0 are not defined. At 90°, cos 90° = 0, so sec 90° = 1/0 and tan 90° = sin 90°/cos 90° = 1/0 are not defined. Write the words ‘not defined’ in answers, never infinity.
Write 0, 1, 2, 3, 4 under 0°, 30°, 45°, 60°, 90°, divide each by 4 and take the square root. This gives the sin row: 0, 1/2, 1/√2, √3/2, 1. Reverse it for the cos row, divide sin by cos for the tan row, and take reciprocals for cosec, sec and cot.
sin²A + cos²A = 1 holds for 0° ≤ A ≤ 90°. 1 + tan²A = sec²A holds for 0° ≤ A < 90°, since tan and sec are not defined at 90°. cot²A + 1 = cosec²A holds for 0° < A ≤ 90°, since cot and cosec are not defined at 0°. All three come from dividing AB² + BC² = AC² by a squared side.
Begin with the more complicated side, usually the LHS, and work only on that side. Rewrite everything in sin and cos, take an LCM, and look for places to use sin²A + cos²A = 1 or a factorised form like (sec A − tan A)(sec A + tan A) = 1. Show each step on its own line and end with ‘= RHS’.
