Chapter 8 — Introduction to Trigonometry

Every idea in this chapter lives inside a right triangle. Consider ΔABC right-angled at B. The side AC opposite the right angle is the hypotenuse, the longest side. The other two sides are named relative to whichever acute angle we are studying. For ∠A, the side BC, which lies opposite A, is the side opposite to A (or the perpendicular), and the side AB, which together with the hypotenuse forms the angle A, is the side adjacent to A (or the base). If we turn to ∠C instead, the roles exchange: AB becomes the opposite side and BC the adjacent side. The hypotenuse stays the same for both angles.

Get this naming right before anything else, because every trigonometric ratio is defined by it. A quick check: the opposite side is the one that does not touch the angle; the adjacent side is the one that touches the angle and is not the hypotenuse.

Why does a right triangle deserve special treatment? Because its shape is fixed by one acute angle. The angles of a triangle add up to 180°; with one angle 90°, the other two add to 90°, so fixing ∠A fixes ∠C = 90° − A. Two right triangles with the same acute angle A are therefore equiangular and hence similar (AA criterion), so all their corresponding sides are in the same ratio. This is the fact that makes trigonometry possible: the ratio of any two sides of a right triangle depends only on the acute angle, not on how big the triangle is drawn.

Consider a student looking at the top of a tall building. If she stands at different distances, the line from her eye to the top makes different angles with the horizontal, and the height of the building compared with the distance changes accordingly. The ratio height ÷ distance is a fixed function of the angle. To compute heights from angles we must therefore tabulate such ratios for every angle, and that is what the trigonometric ratios do.

Notation: the angle is written ∠A or simply A, and when a Greek letter is used it is usually θ (theta). Angles are measured in degrees in this chapter; the sixtieth part of a degree is a minute, written 1′, but the examination uses whole degrees.

Two facts from the previous chapter are used constantly: the Pythagoras theorem, hypotenuse² = opposite² + adjacent², and the similarity of equiangular triangles.

Let ΔABC be right-angled at B and let A be the acute angle under study. The trigonometric ratios of ∠A are the following six ratios of sides.

The last three are the reciprocals of the first three: cosec A = 1/sin A, sec A = 1/cos A, cot A = 1/tan A. Also, dividing the definitions, tan A = sin A/cos A and cot A = cos A/sin A. So in practice only sin and cos need to be remembered from the triangle; the rest follow.

These ratios are numbers, not lengths, since the units cancel. For the same angle A drawn in a smaller right triangle AB′C′ (with C′ on AC and B′ on AB, B′C′ ∥ BC), similarity gives B′C′/AC′ = BC/AC, so sin A is the same. The ratios depend only on the angle. This is why we may speak of sin 30° without mentioning any triangle.

Because the hypotenuse is the longest side, sin A and cos A are always less than or equal to 1 and cosec A and sec A are always greater than or equal to 1. tan A and cot A can take any positive value.

The symbol sin A stands for one number; it is not sin multiplied by A, and sin A cannot be separated as sin × A. The square of sin A is written sin²A, which means (sin A)² and never sin(A²).

A memory aid used by many students: for sin, cos, tan think of the syllables OH, AH, OA (Opposite over Hypotenuse, Adjacent over Hypotenuse, Opposite over Adjacent). Another is the phrase in which the first letters give the pattern: Some People Have, Curly Brown Hair, Through Proper Brushing, where P is perpendicular (opposite), B is base (adjacent) and H is hypotenuse.

The names come from history. Aryabhata used the Sanskrit word jya (half-chord), which passed through Arabic as jiba and was read by Latin translators as jaib, meaning bay or bosom, translated as sinus, hence sine. The cosine is the sine of the complement.

If a single trigonometric ratio of an acute angle is known, the shape of the right triangle is fixed and all the other ratios can be found. The method has three steps: draw a right triangle in which the given ratio appears as a ratio of two sides, use the Pythagoras theorem to find the third side, and then read off the remaining ratios.

Worked example. Given tan A = 4/3, find the other ratios. Since tan A = opposite/adjacent, take BC = 4k and AB = 3k for some positive number k (we cannot assume the sides are exactly 4 and 3; only their ratio is known). Then AC = √(16k² + 9k²) = 5k. Hence sin A = 4k/5k = 4/5, cos A = 3/5, cosec A = 5/4, sec A = 5/3, cot A = 3/4. The k cancels in every ratio, as it must.

Worked example. Given sin θ = 8/17, find cos θ and tan θ. Opposite = 8k, hypotenuse = 17k, adjacent = √(289 − 64)k = 15k. So cos θ = 15/17 and tan θ = 8/15.

Worked example. If cot θ = 7/8, evaluate (1 + sin θ)(1 − sin θ)/[(1 + cos θ)(1 − cos θ)]. The expression equals (1 − sin²θ)/(1 − cos²θ) = cos²θ/sin²θ = cot²θ = 49/64. Here no triangle is needed once the identity is spotted; but a triangle with adjacent 7k, opposite 8k, hypotenuse √113 k gives the same answer.

Worked example. If 3 cot A = 4, check whether (1 − tan²A)/(1 + tan²A) = cos²A − sin²A. cot A = 4/3 so tan A = 3/4; the left side is (1 − 9/16)/(1 + 9/16) = (7/16)/(25/16) = 7/25. With adjacent 4k, opposite 3k, hypotenuse 5k: cos²A − sin²A = 16/25 − 9/25 = 7/25. Equal, so the statement is true.

Worked example. If sin A = cos A for an acute angle A, then opposite = adjacent, so the triangle is isosceles and A = 45°. And if ∠B and ∠Q are acute with sin B = sin Q, then the two right triangles containing them have opposite/hypotenuse equal; by similarity ∠B = ∠Q. A trigonometric ratio determines the acute angle uniquely.

A caution: the equality sin A = 4/5 does not mean the opposite side is 4 units; it means the opposite side is 4/5 of the hypotenuse. Always introduce the multiplier k.

For a few special angles the trigonometric ratios can be found exactly, without tables, from simple geometry. The first is 45°.

In ΔABC right-angled at B, if ∠A = 45°, then ∠C = 90° − 45° = 45° too. So the triangle is isosceles with AB = BC (sides opposite equal angles are equal). Let AB = BC = a. By the Pythagoras theorem, AC² = a² + a² = 2a², so AC = a√2.

Now read off the ratios for ∠A:
sin 45° = BC/AC = a/(a√2) = 1/√2,
cos 45° = AB/AC = a/(a√2) = 1/√2,
tan 45° = BC/AB = a/a = 1,
cosec 45° = √2, sec 45° = √2, cot 45° = 1.

These values are exact; 1/√2 is approximately 0.7071. In examination answers leave 1/√2 as it is, or write it as √2/2, unless a decimal is demanded.

The value tan 45° = 1 has a geometric meaning: a line making 45° with the horizontal rises one unit for each unit it advances. A ramp inclined at 45° is as high as it is long along the ground. Similarly sin 45° = cos 45° reflects the symmetry of the isosceles right triangle: for this angle the opposite and adjacent sides are the same.

The 45° triangle is the diagonal-half of a square. Cutting a square of side a along a diagonal gives two right triangles each with angles 45°, 45°, 90° and hypotenuse a√2; so the diagonal of a square is √2 times its side, a fact you already used in the chapter on triangles.

An immediate application is verifying identities numerically. For A = 45°, sin²A + cos²A = 1/2 + 1/2 = 1, and 1 + tan²A = 2 = sec²A, consistent with the identities proved later in the chapter. Whenever you derive a general identity, testing it at 45° is a quick sanity check.

Worked example: evaluate 2 tan²45° + cos²30° − sin²60°. tan 45° = 1; cos 30° and sin 60° are both √3/2 (found in the next topic), so their squares cancel and the value is 2 × 1 + 0 = 2.

The angles 30° and 60° come from the equilateral triangle. Take an equilateral triangle ABC of side 2a. Each angle is 60°. Draw the perpendicular AD from A to BC. In an equilateral triangle this perpendicular bisects the base and the vertical angle, so BD = DC = a and ∠BAD = ∠DAC = 30°. ΔABD is now a right triangle with ∠ABD = 60°, ∠BAD = 30°, hypotenuse AB = 2a, BD = a, and by the Pythagoras theorem AD = √(4a² − a²) = a√3.

Ratios of 30° (angle at A in ΔABD; opposite side BD = a, adjacent side AD = a√3, hypotenuse 2a):
sin 30° = a/2a = 1/2, cos 30° = a√3/2a = √3/2, tan 30° = a/(a√3) = 1/√3,
cosec 30° = 2, sec 30° = 2/√3, cot 30° = √3.

Ratios of 60° (angle at B in ΔABD; opposite side AD = a√3, adjacent side BD = a, hypotenuse 2a):
sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3,
cosec 60° = 2/√3, sec 60° = 2, cot 60° = 1/√3.

Observe that sin 30° = cos 60° and cos 30° = sin 60°, and tan 30° = cot 60°. This is no coincidence: 30° and 60° are complementary angles, and the ratios of complementary angles are related in exactly this way, as a later topic proves in general.

The numbers are worth memorising as a pattern. Writing sin 0°, sin 30°, sin 45°, sin 60°, sin 90° as √0/2, √1/2, √2/2, √3/2, √4/2 gives 0, 1/2, 1/√2, √3/2, 1; the cosines are the same list reversed. The tangents are 0, 1/√3, 1, √3 and undefined at 90°.

Worked example: in ΔABC right-angled at B, AB = 5 cm and ∠ACB = 30°. Find BC and AC. tan 30° = AB/BC gives 1/√3 = 5/BC, so BC = 5√3 cm. sin 30° = AB/AC gives 1/2 = 5/AC, so AC = 10 cm.

Worked example: evaluate sin 60° cos 30° + sin 30° cos 60° = (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1. In the Intermediate course you will recognise this as sin(60° + 30°) = sin 90°.

Worked example: the value of tan 30°/cot 60° is (1/√3)/(1/√3) = 1, since cot 60° = tan 30°.

The definitions of the ratios use a right triangle with an acute angle, so 0° and 90° cannot be handled by drawing a triangle directly. Instead we watch what happens to the ratios as the angle shrinks towards 0° or grows towards 90°.

Angle approaching 0°. In ΔABC right-angled at B, keep the hypotenuse AC fixed and let ∠A become smaller and smaller. The point C moves down towards B; the opposite side BC shrinks towards 0 and the adjacent side AB grows until it almost equals AC. Therefore sin A = BC/AC approaches 0, cos A = AB/AC approaches 1, and tan A = BC/AB approaches 0. We define
sin 0° = 0, cos 0° = 1, tan 0° = 0.
Then cot 0° = 1/tan 0° = 1/0 is not defined, cosec 0° = 1/sin 0° is not defined, and sec 0° = 1/cos 0° = 1.

Angle approaching 90°. Now let ∠A grow towards 90° with AC fixed. The point A moves close to B; the adjacent side AB shrinks to 0 and the opposite side BC grows to almost AC. So sin A approaches 1, cos A approaches 0, and tan A = BC/AB grows without bound. We define
sin 90° = 1, cos 90° = 0, and tan 90° is not defined.
Then cot 90° = 0, cosec 90° = 1 and sec 90° is not defined.

The complete table of values is now:

Reading the table across, sin A increases from 0 to 1 as A increases from 0° to 90°, while cos A decreases from 1 to 0, and tan A increases from 0 without limit. These trends are asked as true-or-false questions: the statement that sin A increases as A increases is true; the statement that cos A increases as A increases is false.

Worked example: evaluate 2 sin 90° − 3 cos 0° + tan 45° = 2 − 3 + 1 = 0. And sin 0° × cos 90° + sin 90° × cos 0° = 0 + 1 = 1.

A regular board question gives an expression in the standard angles and asks for its exact value. The method is mechanical: substitute the values from the table, simplify the surds and fractions with care, and present the result in simplest form. The difficulty lies only in arithmetic discipline.

Worked example 1. Evaluate sin 60° cos 30° + sin 30° cos 60°.
= (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1.

Worked example 2. Evaluate 2 tan²45° + cos²30° − sin²60°.
= 2(1)² + (√3/2)² − (√3/2)² = 2 + 3/4 − 3/4 = 2.

Worked example 3. Evaluate cos 45°/(sec 30° + cosec 30°).
Numerator 1/√2; denominator 2/√3 + 2 = (2 + 2√3)/√3. So the value is (1/√2) × √3/(2 + 2√3) = √3/[√2 × 2(1 + √3)] = √3/[2√2(1 + √3)]. Rationalise by multiplying numerator and denominator by (√3 − 1): numerator √3(√3 − 1) = 3 − √3; denominator 2√2(3 − 1) = 4√2. The value is (3 − √3)/(4√2), which can be written as (3√2 − √6)/8 after multiplying top and bottom by √2.

Worked example 4. Evaluate (sin 30° + tan 45° − cosec 60°)/(sec 30° + cos 60° + cot 45°).
Numerator: 1/2 + 1 − 2/√3 = 3/2 − 2/√3 = (3√3 − 4)/(2√3). Denominator: 2/√3 + 1/2 + 1 = 2/√3 + 3/2 = (4 + 3√3)/(2√3). The ratio is (3√3 − 4)/(3√3 + 4). Rationalising with (3√3 − 4): [(3√3 − 4)²]/(27 − 16) = (27 − 24√3 + 16)/11 = (43 − 24√3)/11.

Worked example 5. Evaluate (5 cos²60° + 4 sec²30° − tan²45°)/(sin²30° + cos²30°).
Numerator: 5(1/4) + 4(4/3) − 1 = 5/4 + 16/3 − 1 = (15 + 64 − 12)/12 = 67/12. Denominator: 1/4 + 3/4 = 1. Value 67/12.

Finding an angle. If tan 2A = cot(A − 18°), where 2A is acute, then since tan 2A = cot(90° − 2A), we get 90° − 2A = A − 18°, so 3A = 108° and A = 36°. And if sin 2A = 2 sin A, the only solution with 0° ≤ A ≤ 90° is A = 0°, which can be seen by testing: at A = 30°, sin 60° = √3/2 while 2 sin 30° = 1, and for every acute A the two sides differ.

Present the working line by line, keep surds unsimplified until the final step, and rationalise the denominator at the end.

Two angles are complementary if their sum is 90°. In ΔABC right-angled at B, the acute angles A and C are complementary: C = 90° − A. Let us compare the ratios of A with those of C.

For ∠A: opposite side BC, adjacent side AB, hypotenuse AC. For ∠C: opposite side AB, adjacent side BC, hypotenuse AC. So
sin C = AB/AC = cos A, cos C = BC/AC = sin A, tan C = AB/BC = cot A,
cosec C = AC/AB = sec A, sec C = AC/BC = cosec A, cot C = BC/AB = tan A.

Writing C = 90° − A gives the complementary angle relations:

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.

Each ratio changes to its co-ratio: sine to cosine, tangent to cotangent, secant to cosecant, and back. This is where the prefix co comes from: cosine is the sine of the complement. The relations hold for every A between 0° and 90° inclusive, which is why sin 90° = cos 0° = 1 and tan 0° = cot 90° = 0 are consistent with the table.

Worked example. Evaluate tan 65°/cot 25°. Since cot 25° = cot(90° − 65°) = tan 65°, the ratio is 1.

Worked example. Evaluate sin 25° cos 65° + cos 25° sin 65°. Write cos 65° = sin 25° and sin 65° = cos 25°; the expression becomes sin²25° + cos²25° = 1 by the fundamental identity.

Worked example. Evaluate cos 48° − sin 42°. Since sin 42° = cos(90° − 42°) = cos 48°, the value is 0.

Worked example. Evaluate cosec 31° − sec 59°. sec 59° = cosec(90° − 59°) = cosec 31°, so the value is 0.

Worked example. If tan A = cot B, prove A + B = 90°. cot B = tan(90° − B), so tan A = tan(90° − B); since a tangent value fixes an acute angle uniquely, A = 90° − B.

Worked example. If sec 4A = cosec(A − 20°) with 4A acute, then cosec(A − 20°) = sec(90° − (A − 20°)) = sec(110° − A), so 4A = 110° − A, 5A = 110°, A = 22°.

The strategy in every such problem is to convert each ratio of an angle above 45° into a co-ratio of its complement below 45°, or the other way round, so that the same angle appears throughout.

An identity is an equation that is true for every value of the variable for which both sides are defined. The identity (a + b)² = a² + 2ab + b² holds for all numbers a and b; a trigonometric identity holds for all angles. The three fundamental trigonometric identities all come from the Pythagoras theorem.

Theorem. For every angle A with 0° ≤ A ≤ 90°, sin²A + cos²A = 1.

Proof. Take ΔABC right-angled at B. By the Pythagoras theorem, AB² + BC² = AC². Divide every term by AC²:
AB²/AC² + BC²/AC² = AC²/AC², that is, (AB/AC)² + (BC/AC)² = 1.
But AB/AC = cos A and BC/AC = sin A. Hence cos²A + sin²A = 1. This holds for every acute A, and by direct substitution it holds at 0° (0 + 1 = 1) and 90° (1 + 0 = 1).

The identity is used in the forms sin²A = 1 − cos²A and cos²A = 1 − sin²A, and, after factorising, 1 − sin²A = (1 − sin A)(1 + sin A) = cos²A and 1 − cos²A = (1 − cos A)(1 + cos A) = sin²A. These factorised forms are the key to many rationalisation steps in proofs.

Worked example. Express sin A, tan A and sec A in terms of cos A. sin A = √(1 − cos²A) (positive root, since sin A is positive for an acute angle); tan A = sin A/cos A = √(1 − cos²A)/cos A; sec A = 1/cos A.

Worked example. Express the other ratios in terms of cot A. Let cot A = c. Then tan A = 1/c; cosec²A = 1 + c² (from the third identity), so sin A = 1/√(1 + c²); cos A = cot A sin A = c/√(1 + c²); sec A = √(1 + c²)/c.

Worked example. Prove (1 − sin θ)/(1 + sin θ) = (sec θ − tan θ)². Right side: (1/cos θ − sin θ/cos θ)² = (1 − sin θ)²/cos²θ = (1 − sin θ)²/(1 − sin²θ) = (1 − sin θ)²/[(1 − sin θ)(1 + sin θ)] = (1 − sin θ)/(1 + sin θ) = left side.

A numerical check at 30°: (1/2)² + (√3/2)² = 1/4 + 3/4 = 1. Such a check does not prove the identity but catches algebra errors.

The other two fundamental identities come from the same Pythagoras relation AB² + BC² = AC² in ΔABC right-angled at B, divided by a different side.

Second identity. Divide by AB² (the side adjacent to A):
AB²/AB² + BC²/AB² = AC²/AB², that is, 1 + (BC/AB)² = (AC/AB)².
Since BC/AB = tan A and AC/AB = sec A, we get 1 + tan²A = sec²A, valid for 0° ≤ A < 90° (at 90° tan and sec are not defined).

Third identity. Divide by BC² (the side opposite to A):
AB²/BC² + BC²/BC² = AC²/BC², that is, (AB/BC)² + 1 = (AC/BC)².
Since AB/BC = cot A and AC/BC = cosec A, we get 1 + cot²A = cosec²A, valid for 0° < A ≤ 90°.

Both are used in rearranged forms: sec²A − tan²A = 1, tan²A = sec²A − 1, cosec²A − cot²A = 1, cot²A = cosec²A − 1. The difference of squares gives further useful factorisations: sec²A − tan²A = (sec A − tan A)(sec A + tan A) = 1, so sec A + tan A and sec A − tan A are reciprocals of each other; similarly cosec A + cot A and cosec A − cot A are reciprocals.

Worked example. Simplify (sec A + tan A)(1 − sin A). Write in sines and cosines: (1 + sin A)/cos A × (1 − sin A) = (1 − sin²A)/cos A = cos²A/cos A = cos A.

Worked example. Prove (cosec θ − cot θ)² = (1 − cos θ)/(1 + cos θ). Left side = [(1 − cos θ)/sin θ]² = (1 − cos θ)²/sin²θ = (1 − cos θ)²/[(1 − cos θ)(1 + cos θ)] = (1 − cos θ)/(1 + cos θ).

Worked example. If sec A + tan A = p, find sec A − tan A. Since (sec A + tan A)(sec A − tan A) = sec²A − tan²A = 1, sec A − tan A = 1/p. Adding and subtracting, sec A = (p + 1/p)/2 = (p² + 1)/(2p) and tan A = (p² − 1)/(2p), so sin A = tan A/sec A = (p² − 1)/(p² + 1).

Worked example. Which of the following is correct: 9 sec²A − 9 tan²A equals 1, 9, 8 or 0? It is 9(sec²A − tan²A) = 9 × 1 = 9. Such one-mark questions test whether the identity is at your fingertips.

Numerical check at 60°: 1 + tan²60° = 1 + 3 = 4 = sec²60° = 2²; and 1 + cot²60° = 1 + 1/3 = 4/3 = cosec²60° = (2/√3)².

Proving identities is the highest-scoring skill in this chapter, and it rewards method more than cleverness. An identity is proved by transforming one side, usually the more complicated one, step by step until it becomes the other side; or by reducing both sides independently to the same expression. You must never cross-multiply or treat the identity as an equation to be solved, because that assumes what is to be proved.

Strategies, in the order to try them.

• Convert to sine and cosine. Replace tan, cot, sec and cosec by sin/cos, cos/sin, 1/cos and 1/sin. Most identities then reduce to algebra with sin and cos and the identity sin² + cos² = 1.
• Combine fractions over a common denominator and simplify the numerator.
• Use the factorised forms 1 − sin²A = (1 − sin A)(1 + sin A), sec²A − tan²A = (sec A − tan A)(sec A + tan A) to cancel factors.
• Rationalise. When the target has 1 + sin A in the denominator and you have 1 − sin A, multiply numerator and denominator by 1 + sin A (or by sec A + tan A, and so on).
• Substitute the identity into a 1. Replace a stray 1 by sin²A + cos²A or by sec²A − tan²A when a factorisation is needed.
• Algebraic identities. a³ ± b³ = (a ± b)(a² ∓ ab + b²) turns sin³A + cos³A into (sin A + cos A)(1 − sin A cos A).

Worked example. Prove (cos A − sin A + 1)/(cos A + sin A − 1) = cosec A + cot A. Divide numerator and denominator by sin A: (cot A − 1 + cosec A)/(cot A + 1 − cosec A). In the numerator write 1 as cosec²A − cot²A: cot A + cosec A − (cosec²A − cot²A) = (cot A + cosec A) − (cosec A − cot A)(cosec A + cot A) = (cosec A + cot A)(1 − cosec A + cot A). The denominator is (cot A + 1 − cosec A), the same as the second factor. Cancelling, the result is cosec A + cot A.

Worked example. Prove √[(1 + sin A)/(1 − sin A)] = sec A + tan A. Multiply inside the root by (1 + sin A)/(1 + sin A): √[(1 + sin A)²/(1 − sin²A)] = √[(1 + sin A)²/cos²A] = (1 + sin A)/cos A = sec A + tan A.

Worked example. Prove (sin θ − 2 sin³θ)/(2 cos³θ − cos θ) = tan θ. Factor: sin θ(1 − 2 sin²θ)/[cos θ(2 cos²θ − 1)]. Now 1 − 2 sin²θ = 1 − 2(1 − cos²θ) = 2 cos²θ − 1, so the brackets cancel and the result is sin θ/cos θ = tan θ.

Write LHS and RHS clearly, transform one at a time, and end with the line LHS = RHS.

A second family of examination identities mixes the reciprocal ratios in a way that becomes simple once the right common denominator is found. Here are the standard ones with complete working, since each appears in the board papers year after year.

Identity 1. tan θ/(1 − cot θ) + cot θ/(1 − tan θ) = 1 + sec θ cosec θ.
Write cot θ = 1/tan θ. First term: tan θ/(1 − 1/tan θ) = tan²θ/(tan θ − 1). Second term: (1/tan θ)/(1 − tan θ) = 1/[tan θ(1 − tan θ)] = −1/[tan θ(tan θ − 1)]. Sum = [tan³θ − 1]/[tan θ(tan θ − 1)] = [(tan θ − 1)(tan²θ + tan θ + 1)]/[tan θ(tan θ − 1)] = (tan²θ + tan θ + 1)/tan θ = tan θ + 1 + cot θ. Now tan θ + cot θ = sin θ/cos θ + cos θ/sin θ = (sin²θ + cos²θ)/(sin θ cos θ) = sec θ cosec θ. Hence the sum is 1 + sec θ cosec θ.

Identity 2. (cosec A − sin A)(sec A − cos A) = 1/(tan A + cot A).
Left side: (1/sin A − sin A)(1/cos A − cos A) = [(1 − sin²A)/sin A][(1 − cos²A)/cos A] = (cos²A/sin A)(sin²A/cos A) = sin A cos A. Right side: 1/(sin A/cos A + cos A/sin A) = sin A cos A/(sin²A + cos²A) = sin A cos A. Both sides equal sin A cos A.

Identity 3. (1 + sec A)/sec A = sin²A/(1 − cos A).
Left side: (1 + 1/cos A)/(1/cos A) = (cos A + 1)/cos A × cos A = 1 + cos A. Right side: (1 − cos²A)/(1 − cos A) = (1 − cos A)(1 + cos A)/(1 − cos A) = 1 + cos A.

Identity 4. (1 + cot²A) tan A/sec²A = cot A.
Left side: cosec²A × tan A × cos²A = (1/sin²A)(sin A/cos A)(cos²A) = cos A/sin A = cot A.

Identity 5. (sin A − cos A + 1)/(sin A + cos A − 1) = 1/(sec A − tan A).
Divide numerator and denominator by cos A: (tan A − 1 + sec A)/(tan A + 1 − sec A). Write 1 in the numerator as sec²A − tan²A: (tan A + sec A) − (sec A − tan A)(sec A + tan A) = (sec A + tan A)(1 − sec A + tan A). The denominator is (tan A + 1 − sec A), the same bracket, so the quotient is sec A + tan A = 1/(sec A − tan A).

Common to all: reciprocal ratios are best turned into sin and cos, and any lone 1 may be swapped for sec² − tan² or cosec² − cot² to create a factor that cancels.

To solve a right triangle means to find all its unknown sides and angles from the given ones. Since one angle is 90°, it is enough to know one side and one acute angle, or two sides. Trigonometric ratios turn the problem into a single multiplication or division, and this is the skill that the next chapter on heights and distances uses on every question.

Given one side and one acute angle. Choose the ratio that links the known side, the wanted side and the known angle. If the known side is the hypotenuse and you want the opposite side, use sine; the adjacent side, use cosine. If the known side is a leg and you want the other leg, use tangent; if you want the hypotenuse from a leg, use sine or cosine as appropriate. The other acute angle is 90° minus the given one.

Worked example. In ΔABC right-angled at B, AC = 10 cm and ∠A = 30°. Then BC = AC sin A = 10 × 1/2 = 5 cm, AB = AC cos A = 10 × √3/2 = 5√3 cm ≈ 8.66 cm, and ∠C = 60°.

Worked example. In ΔPQR right-angled at Q, PQ = 3 cm and PR = 6 cm. Find ∠QPR and ∠PRQ. sin R = PQ/PR = 3/6 = 1/2, so ∠R = 30° and ∠P = 60°. Here two sides were given and the angle was recognised from the table.

Worked example. In ΔABC right-angled at B, AB = 5 cm and ∠ACB = 30°. tan 30° = AB/BC gives BC = 5√3 cm; sin 30° = AB/AC gives AC = 10 cm.

Worked example. In ΔPQR right-angled at Q, PR + QR = 25 cm and PQ = 5 cm; find sin P, cos P and tan P. Let QR = x cm, so PR = 25 − x. Pythagoras: (25 − x)² = x² + 25, so 625 − 50x = 25 and x = 12. Then QR = 12, PR = 13; sin P = QR/PR = 12/13, cos P = PQ/PR = 5/13, tan P = 12/5.

Worked example. A right triangle has hypotenuse 8 cm and one angle 45°. Both legs equal 8 sin 45° = 8/√2 = 4√2 cm, and the area is ½ × (4√2)² = 16 cm².

When the angle is not one of the standard ones, the examination gives the needed ratio value in the question, for instance take tan 35° = 0.7, and you use it directly. Present the answer with units and, where a surd occurs, both exact and approximate values if a decimal is asked. Draw the triangle, mark the right angle, the given angle and the given side before writing any ratio; in a well-labelled figure the right ratio is obvious.