CBSE Class 10 · Mathematics

Class 10 Mathematics – Chapter 8: Introduction to Trigonometry

In Chapter 7 – Coordinate Geometry you measured distances on a numbered grid, and in Chapter 6 – Triangles you reasoned with ratios and similarity. This chapter takes a different route into geometry: it fixes one acute angle of a right-angled triangle and studies the unchanging ratios of its sides. Whichever size the triangle is drawn, these six ratios depend only on the angle — which is why builders, engineers, architects, navigators and computer programmers all rely on trigonometry. By the end you will be able to name all six ratios from a right triangle, recall their exact values at the standard angles \(0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ\), and simplify and evaluate expressions built from them and from the three fundamental identities.

1. Introduction to Trigonometry

Trigonometry comes from two Greek words: trigonon (triangle) and metron (measure). It began as the study of triangles — first in astronomy and map-making — and in Class 10 we study its foundation: the link between an acute angle of a right-angled triangle and the ratio of two sides of that triangle.

The key observation: in right-angled triangles of every size, if the acute angle is the same, then the ratios of corresponding sides are the same. A small triangle and a huge triangle that both contain a \(60^\circ\) angle both give the value \(\frac{1}{2}\) for \(\sin 60^\circ\). The ratio is a pure number attached to the angle, with no unit of length attached.

Right-angled triangle

A triangle with one angle of \(90^\circ\). It is the only triangle we study in this chapter.

Hypotenuse

The side opposite the right angle. It is always the longest side.

Perpendicular (P)

The side opposite the acute angle θ; also called the side opposite the angle.

Base (B)

The side adjacent to the acute angle θ that is not the hypotenuse.

Fix an acute angle θ in a right-angled triangle (draw it at vertex \(A\) of \(\triangle ABC\), right-angled at \(B\)):

  • Perpendicular (P) — the side facing the angle, opposite to it: \(BC\).
  • Base (B) — the side along which the angle rests, adjacent to it: \(AB\).
  • Hypotenuse (H) — the longest side, facing the right angle: \(AC\).
θ Base (B) — adjacent side Perpendicular (P) Hypotenuse (H) right angle at B A — acute angle θ
Watch the angle

The words "perpendicular" and "base" are always relative to the chosen acute angle. The side that is the perpendicular for one acute angle is the base for the other. Always begin by marking the acute angle you are working with.

2. Trigonometric Ratios

For a fixed acute angle θ in a right-angled triangle with perpendicular \(P\), base \(B\) and hypotenuse \(H\), we define three main ratios:

\(\sin \theta\)

\[ \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{P}{H} \]

\(\cos \theta\)

\[ \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{B}{H} \]

\(\tan \theta\)

\[ \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{P}{B} \]

Their reciprocals are the other three ratios:

\(\mathrm{cosec}\,\theta\)

\[ \mathrm{cosec}\,\theta = \frac{H}{P} = \frac{1}{\sin\theta} \]

\(\sec \theta\)

\[ \sec \theta = \frac{H}{B} = \frac{1}{\cos\theta} \]

\(\cot \theta\)

\[ \cot \theta = \frac{B}{P} = \frac{1}{\tan\theta} \]

Notice how each reciprocal simply turns the original fraction upside down:

  • \(\mathrm{cosec}\,\theta = \frac{1}{\sin \theta}\) — reciprocal of sine.
  • \(\sec \theta = \frac{1}{\cos \theta}\) — reciprocal of cosine.
  • \(\cot \theta = \frac{1}{\tan \theta}\) — reciprocal of tangent.

Two more useful consequences come straight from the definitions:

  • \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \] because \(\frac{P}{H} \div \frac{B}{H} = \frac{P}{B}\).
  • \[ \cot \theta = \frac{\cos \theta}{\sin \theta} \] because \(\frac{B}{H} \div \frac{P}{H} = \frac{B}{P}\).
B P H θ sin θ = P/H — perpendicular over hypotenuse cos θ = B/H — base over hypotenuse tan θ = P/B — perpendicular over base B P H θ sin θ = P/H ↔ cosec θ = H/P cos θ = B/H ↔ sec θ = H/B tan θ = P/B ↔ cot θ = B/P Each right-hand ratio is 1 divided by the left-hand ratio.
Reasoning tip

To decide which ratio to use, match the two sides you know (or need) to the letters in the formulas. If a question gives the perpendicular and asks for the hypotenuse, the formula that contains both \(P\) and \(H\) is \(\sin \theta = \frac{P}{H}\).

3. Trigonometric Ratios of Specific Angles

We do not need a calculator for the angles \(0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ\): their exact values come from two simple right triangles and the definitions above.

Right triangle with angles \(30^\circ, 60^\circ, 90^\circ\)

Half of an equilateral triangle of side 2 gives a right-angled triangle with sides \(1, \sqrt{3}, 2\): the shortest side \(1\) is opposite the smallest angle \(30^\circ\), the side \(\sqrt{3}\) is opposite \(60^\circ\), and the hypotenuse is \(2\).

30° √3 1 2 60°

Right triangle with angles \(45^\circ, 45^\circ, 90^\circ\)

A square divided along a diagonal gives a right-angled triangle with two equal legs \(1, 1\) and hypotenuse \(\sqrt{1^2 + 1^2} = \sqrt{2}\). Both acute angles are \(45^\circ\).

45° 45° 1 1 √2

The standard-value table

θ0°30°45°60°90°
\(\sin \theta\)\(0\)\(\frac{1}{2}\)\(\frac{1}{\sqrt{2}}\)\(\frac{\sqrt{3}}{2}\)\(1\)
\(\cos \theta\)\(1\)\(\frac{\sqrt{3}}{2}\)\(\frac{1}{\sqrt{2}}\)\(\frac{1}{2}\)\(0\)
\(\tan \theta\)\(0\)\(\frac{1}{\sqrt{3}}\)\(1\)\(\sqrt{3}\)Not defined
\(\mathrm{cosec}\,\theta\)Not defined\(2\)\(\sqrt{2}\)\(\frac{2}{\sqrt{3}}\)\(1\)
\(\sec \theta\)\(1\)\(\frac{2}{\sqrt{3}}\)\(\sqrt{2}\)\(2\)Not defined
\(\cot \theta\)Not defined\(\sqrt{3}\)\(1\)\(\frac{1}{\sqrt{3}}\)\(0\)
Undefined values are not zero

\(\tan 90^\circ\), \(\sec 90^\circ\), \(\mathrm{cosec}\,0^\circ\) and \(\cot 0^\circ\) are not defined. They are never zero and never "infinity" in a CBSE answer. Each has a denominator of zero: \(\sec 90^\circ = \frac{H}{B}\) with base \(B = 0\) in a triangle where the angle is \(90^\circ\), and \(\tan 90^\circ = \frac{P}{B}\) likewise has \(B = 0\). Write the words Not defined.

One-line memory pattern

For \(\sin \theta\) the values are \(0, \frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{\sqrt{3}}{2}, 1\) reading the columns 0° to 90°; for \(\cos \theta\) the same column reads back to front. \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) every time, and each reciprocal fills its own row.

4. Finding Trigonometric Ratios

Given a right-angled triangle, the four-step routine is: (1) mark the chosen acute angle, (2) write down \(P, B, H\) for that angle, (3) use Pythagoras if only two sides are given, (4) substitute into the required ratio.

Worked example — all six ratios from two sides

Question: In a right-angled triangle the perpendicular is 6 units and the base is 8 units. Find the hypotenuse and all six trigonometric ratios of the angle.

Approach: find the missing side by Pythagoras, then read each ratio from the definitions.

Working: \(P = 6\), \(B = 8\). Then \(H^2 = 6^2 + 8^2 = 36 + 64 = 100\), so \(H = 10\).

\[ \sin\theta = \frac{6}{10} = \frac{3}{5},\quad \cos\theta = \frac{8}{10} = \frac{4}{5},\quad \tan\theta = \frac{6}{8} = \frac{3}{4} \]

\[ \mathrm{cosec}\,\theta = \frac{5}{3},\quad \sec\theta = \frac{5}{4},\quad \cot\theta = \frac{4}{3} \]

Final answer: \(H = 10\); \(\sin\theta = \frac{3}{5}\), \(\cos\theta = \frac{4}{5}\), \(\tan\theta = \frac{3}{4}\), \(\mathrm{cosec}\,\theta = \frac{5}{3}\), \(\sec\theta = \frac{5}{4}\), \(\cot\theta = \frac{4}{3}\).

Worked example — finding ratios when a ratio is known

Question: If \(\cos \theta = \frac{8}{17}\), find \(\sin \theta\) and \(\tan \theta\).

Approach: \(\cos\theta = \frac{B}{H}\), so \(B = 8\), \(H = 17\); find \(P\) by Pythagoras.

Working: \(8^2 + P^2 = 17^2\), so \(P^2 = 289 - 64 = 225\), hence \(P = 15\). Then \(\sin\theta = \frac{P}{H} = \frac{15}{17}\) and \(\tan\theta = \frac{P}{B} = \frac{15}{8}\).

Final answer: \(\sin\theta = \frac{15}{17}\), \(\tan\theta = \frac{15}{8}\).

Pythagorean triples save time

Memorise the common triples: \(3,4,5\); \(5,12,13\); \(7,24,25\); \(8,15,17\); \(9,40,41\); and \(20,21,29\). If you recognise a triple you can skip the squaring step entirely and still write the working clearly.

5. Trigonometric Identities

The three fundamental identities follow in one line each from Pythagoras' theorem \(P^2 + B^2 = H^2\).

Identity 1

\[ \sin^2 \theta + \cos^2 \theta = 1 \]

Identity 2

\[ 1 + \tan^2 \theta = \sec^2 \theta \]

Identity 3

\[ 1 + \cot^2 \theta = \mathrm{cosec}^2\, \theta \]

Derivation

Start from Pythagoras: \(P^2 + B^2 = H^2\). This holds for every right triangle.

1
Divide \(P^2 + B^2 = H^2\) by \(H^2\): \(\frac{P^2}{H^2} + \frac{B^2}{H^2} = 1\), giving \(\sin^2\theta + \cos^2\theta = 1\).
2
Divide \(P^2 + B^2 = H^2\) by \(B^2\): \(\frac{P^2}{B^2} + 1 = \frac{H^2}{B^2}\), giving \(\tan^2\theta + 1 = \sec^2\theta\), i.e. \(1 + \tan^2\theta = \sec^2\theta\).
3
Divide \(P^2 + B^2 = H^2\) by \(P^2\): \(1 + \frac{B^2}{P^2} = \frac{H^2}{P^2}\), giving \(1 + \cot^2\theta = \mathrm{cosec}^2\,\theta\).
How identities are used
  • \(\sin^2\theta = 1 - \cos^2\theta\) and \(\cos^2\theta = 1 - \sin^2\theta\) substitute a square you do not know.
  • \(\sec^2\theta - \tan^2\theta = 1\) and \(\mathrm{cosec}^2\theta - \cot^2\theta = 1\) follow by rearranging.
  • An identity is true for every value at which both sides are defined, so it can never be proved by substituting one angle.

6. Simplification of Trigonometric Expressions

Class 10 simplification problems combine the reciprocal relationships, the three identities and ordinary algebra. Convert everything to \(\sin\theta\) and \(\cos\theta\) if you get stuck — that route always works.

Worked example — simplify a product

Question: Simplify \(\dfrac{\sin\theta \cdot \sec\theta}{\tan\theta}\).

Approach: write \(\sec\theta\) and \(\tan\theta\) in terms of sine and cosine.

Working: \(\frac{\sin\theta \cdot \frac{1}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}} = \frac{\frac{\sin\theta}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}} = 1\).

Final answer: the expression simplifies to \(1\).

Worked example — simplify with an identity

Question: Prove that \((1 + \cot^2\theta)(1 - \cos^2\theta) = 1\).

Approach: use both identities \(1 + \cot^2\theta = \mathrm{cosec}^2\theta\) and \(1 - \cos^2\theta = \sin^2\theta\).

Working: \((1 + \cot^2\theta)(1 - \cos^2\theta) = \mathrm{cosec}^2\theta \cdot \sin^2\theta = \frac{1}{\sin^2\theta}\cdot\sin^2\theta = 1\).

Final answer: \((1 + \cot^2\theta)(1 - \cos^2\theta) = 1\). ✓

Cross-check

Before finishing, test a standard angle, e.g. take \(\theta = 45^\circ\) and check the left side still equals the right side. In the second example, \(\cot 45^\circ = 1\) and \(\cos 45^\circ = \frac{1}{\sqrt{2}}\), giving \((1+1)(1-\frac12) = 2\cdot\frac12 = 1\). ✓

7. Evaluation of Trigonometric Expressions

To evaluate an expression, replace each standard angle by its exact value from the table and simplify the arithmetic exactly (keep surds, avoid decimals).

Worked example — evaluate an expression

Question: Evaluate \(\sin 30^\circ + \cos 60^\circ\).

Approach: table values: \(\sin 30^\circ = \frac12\) and \(\cos 60^\circ = \frac12\).

Working: \(\frac12 + \frac12 = 1\).

Final answer: \(\sin 30^\circ + \cos 60^\circ = 1\).

Worked example — evaluate with sec, tan and sin

Question: Evaluate \(\sec 60^\circ - \tan 45^\circ + \sin 0^\circ\).

Approach: use \(\sec 60^\circ = 2\), \(\tan 45^\circ = 1\), \(\sin 0^\circ = 0\).

Working: \(2 - 1 + 0 = 1\).

Final answer: the value is \(1\).

Exam rigour

Write the table value before doing the arithmetic: "\(\sin 30^\circ = \frac12\)" earns the substitution mark even if the final arithmetic slips.

8. Finding Unknown Ratios

When one ratio is given, the others come from a right triangle built with the given \(P, B, H\), or directly from the identities. Both routes are equally acceptable in CBSE answers.

Worked example — from sin to cos and tan

Question: If \(\sin \theta = \frac{3}{5}\), find \(\cos \theta\) and \(\tan \theta\).

Right-triangle route: \(\sin\theta = \frac{P}{H}\) gives \(P = 3\), \(H = 5\). Then \(B = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = 4\). So \(\cos\theta = \frac{B}{H} = \frac{4}{5}\) and \(\tan\theta = \frac{P}{B} = \frac{3}{4}\).

Identity route (check): \(\cos^2\theta = 1 - \sin^2\theta = 1 - \frac{9}{25} = \frac{16}{25}\), so \(\cos\theta = \frac{4}{5}\), and \(\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3/5}{4/5} = \frac{3}{4}\).

Final answer: \(\cos\theta = \frac{4}{5}\) and \(\tan\theta = \frac{3}{4}\).

Worked example — from tan to the rest

Question: If \(\tan \theta = \frac{5}{12}\), determine the other five ratios.

Approach: \(\tan\theta = \frac{P}{B}\) gives \(P = 5\), \(B = 12\), and \(H = \sqrt{5^2 + 12^2} = 13\).

Working: \(\sin\theta = \frac{5}{13}\), \(\cos\theta = \frac{12}{13}\), \(\mathrm{cosec}\,\theta = \frac{13}{5}\), \(\sec\theta = \frac{13}{12}\), \(\cot\theta = \frac{12}{5}\).

Final answer: \(\sin\theta = \frac{5}{13},\ \cos\theta = \frac{12}{13},\ \mathrm{cosec}\,\theta = \frac{13}{5},\ \sec\theta = \frac{13}{12},\ \cot\theta = \frac{12}{5}\).

Easy mistake

When a ratio is given as a fraction, its denominator is the hypotenuse only when the ratio is \(\sin\), \(\cos\), \(\sec\) or \(\mathrm{cosec}\). For \(\tan\) and \(\cot\) the denominator is one of the legs, so always build the triangle and compute the hypotenuse separately.

9. Application and Competency-Based Questions

These problems test whether you can spot the correct ratio inside a described right-angled situation. The angle stays inside the triangle — nothing is measured across open ground; height-and-distance questions with elevation angles belong to the separate chapter Some Applications of Trigonometry.

Case 1 — Choosing the ratio

A carpenter uses a right-angled template with sides 7, 24 and 25 to check a corner. For the acute angle whose perpendicular is 7 and base is 24, find \(\cot \theta\) and \(\sec \theta\).

Reasoning: for this angle \(P = 7\), \(B = 24\), \(H = 25\). \(\cot\theta = \frac{B}{P} = \frac{24}{7}\) and \(\sec\theta = \frac{H}{B} = \frac{25}{24}\).

Final answer: \(\cot\theta = \frac{24}{7}\), \(\sec\theta = \frac{25}{24}\).

Case 2 — All six ratios from a figure

In \(\triangle DEF\), right-angled at \(E\), the hypotenuse \(DF = 29\), \(DE = 21\) and \(EF = 20\). Write all six trigonometric ratios of angle \(D\).

Reasoning: for angle \(D\), the perpendicular is the opposite side \(EF = 20\), the base is the adjacent side \(DE = 21\), and the hypotenuse is \(DF = 29.\)

Working: \(\sin D = \frac{20}{29}\), \(\cos D = \frac{21}{29}\), \(\tan D = \frac{20}{21}\), and reciprocals \(\mathrm{cosec}\,D = \frac{29}{20}\), \(\sec D = \frac{29}{21}\), \(\cot D = \frac{21}{20}\).

Final answer: the six ratios above. (Check: \(\left(\frac{20}{29}\right)^2 + \left(\frac{21}{29}\right)^2 = \frac{400 + 441}{841} = 1\).)

Competency check — identify the wrong statement

Which of the following statements is false?

  1. \(\sin 45^\circ = \cos 45^\circ\)
  2. \(\sin 30^\circ = \cos 60^\circ\)
  3. \(\tan 90^\circ = 0\)
  4. \(\sec 60^\circ = 2\)

Answer: statement 3 is false. \(\tan 90^\circ\) is not defined, never 0.

10. Important Formulas

Six trigonometric ratios

\[ \sin\theta = \frac{P}{H},\qquad \cos\theta = \frac{B}{H},\qquad \tan\theta = \frac{P}{B} \]

\[ \mathrm{cosec}\,\theta = \frac{H}{P},\qquad \sec\theta = \frac{H}{B},\qquad \cot\theta = \frac{B}{P} \]

Reciprocal relationships

\[ \mathrm{cosec}\,\theta = \frac{1}{\sin\theta},\qquad \sec\theta = \frac{1}{\cos\theta},\qquad \cot\theta = \frac{1}{\tan\theta} \]

\[ \tan\theta = \frac{\sin\theta}{\cos\theta},\qquad \cot\theta = \frac{\cos\theta}{\sin\theta} \]

Three fundamental identities

\[ \sin^2\theta + \cos^2\theta = 1 \]

\[ 1 + \tan^2\theta = \sec^2\theta \]

\[ 1 + \cot^2\theta = \mathrm{cosec}^2\,\theta \]

Where each letter belongs

In the right triangle, \(P\) is opposite the acute angle, \(B\) is adjacent to it, and \(H\) is the hypotenuse (always the longest side). Every formula above is just a proud spelling-out of these three definitions.

11. Common Mistakes in Introduction to Trigonometry

MistakeWhy it happensFix
Swapping perpendicular and baseNot re-marking the sides for the chosen angleLabel the triangle first: P opposite θ, B beside θ, H opposite the right angle
Choosing the wrong ratioPicking a formula before listing the known sidesMatch the two given sides to the letters in the formulas
Confusing \(\sin\) with \(\mathrm{cosec}\)Both start with s and involve P and H\(\sin = \frac{P}{H}\), \(\mathrm{cosec} = \frac{H}{P}\); sine is never a flipped fraction
Confusing \(\cos\) with \(\sec\)Both involve B and H\(\cos = \frac{B}{H}\), \(\sec = \frac{H}{B}\)
Confusing \(\tan\) with \(\cot\)Both involve P and B\(\tan = \frac{P}{B}\), \(\cot = \frac{B}{P}\)
Wrong standard-angle valuesMisremembering the tableRebuild values from the \(30^\circ\)-\(60^\circ\)-\(90^\circ\) and \(45^\circ\)-\(45^\circ\)-\(90^\circ\) triangles
Writing "0" for an undefined ratioTreating division by zero like an answerWrite Not defined for \(\tan 90^\circ\), \(\sec 90^\circ\), \(\mathrm{cosec}\,0^\circ\), \(\cot 0^\circ\)
Wrong reciprocal pairingPairing sine with sec, tangent with cosecOnly the function with the same two letters pairs: sin↔cosec, cos↔sec, tan↔cot
Wrong identity manipulationTurning \(1 + \tan^2\theta = \sec^2\theta\) into \(1 - \tan^2\theta = \sec^2\theta\)Keep the 1 positive; rearranged forms are \(\sec^2 - \tan^2 = 1\)
Arithmetic slips in fractions and surdsRushing the simplificationWrite each line; check \(\frac{1}{\sqrt2} \cdot \frac{\sqrt2}{\sqrt2}\) style rationalising is applied
Skipping PythagorasTaking two sides as \(P\) and \(H\) without the thirdFrom any two sides always compute the missing one before writing ratios
Calculator mode errorsForgetting whether degrees or radians are onIn class 10 use exact table values; a calculator (degree mode) is only a check
Pattern to internalise

A right-angled triangle gives six ratios, but only three are independent: sine, cosine, tangent. The other three are their reciprocals, and the three identities are the same Pythagoras fact written three ways.

12. CBSE Exam Tips

Exam-ready checklist
  • Identify the required ratio first — read which sides the question mentions before writing anything.
  • Draw or reconstruct the right triangle, label \(P, B, H\), and write the angle clearly.
  • Write the formula, then substitute the values underneath it: each written step earns marks.
  • Use exact table values only (keep \(\frac{1}{\sqrt2}\), \(\frac{\sqrt3}{2}\), never rounded decimals).
  • Show sufficient steps even for "simple" evaluations.
  • Apply the identities only where they genuinely help; otherwise use the right triangle.
  • Check the final answer with a standard angle or by the identity \(\sin^2\theta + \cos^2\theta = 1\).
  • Remember the standard-angle table — reciting it from the two special triangles is safer than memory.
  • Check whether a ratio is defined before you claim a value; "not defined" is a valid CBSE answer.
Time-saving habits
  • Spot Pythagorean triples \((3,4,5)\), \((5,12,13)\), \((7,24,25)\), \((8,15,17)\) to skip arithmetic.
  • When two ratios are given, prefer the right-triangle route over repeated identities.
  • In expressions, simplify reciprocals before you combine fractions.

13. Quick Revision

ConceptKey result
Sine\(\sin\theta = \frac{P}{H}\)
Cosine\(\cos\theta = \frac{B}{H}\)
Tangent\(\tan\theta = \frac{P}{B}\)
Reciprocal pairs\(\mathrm{cosec}=\frac1{\sin}\), \(\sec=\frac1{\cos}\), \(\cot=\frac1{\tan}\)
Ratio to ratio\(\tan=\frac{\sin}{\cos}\), \(\cot=\frac{\cos}{\sin}\)
Identity 1\(\sin^2\theta + \cos^2\theta = 1\)
Identity 2\(1 + \tan^2\theta = \sec^2\theta\)
Identity 3\(1 + \cot^2\theta = \mathrm{cosec}^2\,\theta\)
θ0°30°45°60°90°
\(\sin \theta\)\(0\)\(\frac{1}{2}\)\(\frac{1}{\sqrt{2}}\)\(\frac{\sqrt{3}}{2}\)\(1\)
\(\cos \theta\)\(1\)\(\frac{\sqrt{3}}{2}\)\(\frac{1}{\sqrt{2}}\)\(\frac{1}{2}\)\(0\)
\(\tan \theta\)\(0\)\(\frac{1}{\sqrt{3}}\)\(1\)\(\sqrt{3}\)Not defined
\(\mathrm{cosec}\,\theta\)Not defined\(2\)\(\sqrt{2}\)\(\frac{2}{\sqrt{3}}\)\(1\)
\(\sec \theta\)\(1\)\(\frac{2}{\sqrt{3}}\)\(\sqrt{2}\)\(2\)Not defined
\(\cot \theta\)Not defined\(\sqrt{3}\)\(1\)\(\frac{1}{\sqrt{3}}\)\(0\)
Key problem-solving steps
  1. Mark the angle, then identify \(P, B, H\).
  2. Find the missing side with Pythagoras when only two sides are known.
  3. Substitute into the required ratio using exact values.
  4. For expressions, use reciprocals and identities to simplify fully.
  5. Check that every ratio you wrote is defined; otherwise write "Not defined".
Important exam reminders
  • The hypotenuse is always opposite the right angle and is always longest.
  • Perpendicular and base are decided by the chosen acute angle.
  • Table values are exact; never write decimals in a final CBSE answer.

14. Worked Examples — Consolidated Section

Thirty fully solved examples triangulate the whole chapter: basic concepts (1–6), finding ratios (7–14), standard-angle evaluation (15–20), identity-based simplification (21–26) and application/competency problems (27–30).

Worked Example 1
Question: In \(\triangle ABC\), right-angled at \(B\), with the acute angle at \(A\) equal to θ, name the perpendicular, base and hypotenuse for θ.
Concept: perpendicular is opposite the angle, base is beside it, hypotenuse faces the right angle.
Working: for angle \(A\): opposite side \(BC\) → perpendicular; adjacent side \(AB\) → base; side \(AC\) facing the right angle → hypotenuse.
Final Answer: \(P = BC\), \(B = AB\), \(H = AC\).
Worked Example 2
Question: A triangle has sides 8, 15 and 17 units. Confirm it is right-angled and name the hypotenuse.
Concept: Pythagoras in reverse: largest side squared equals the sum of the squares of the other two.
Working: \(8^2 + 15^2 = 64 + 225 = 289 = 17^2\).
Final Answer: it is right-angled and the hypotenuse is 17 units.
Worked Example 3
Question: For an acute angle with \(P = 3\), \(B = 4\), \(H = 5\), write \(\sin\theta\), \(\cos\theta\), \(\tan\theta\).
Concept: direct substitution into the three definitions.
Working: \(\sin\theta = \frac{3}{5}\), \(\cos\theta = \frac{4}{5}\), \(\tan\theta = \frac{3}{4}\).
Final Answer: \(\frac{3}{5},\ \frac{4}{5},\ \frac{3}{4}\).
Worked Example 4
Question: In a right triangle, the hypotenuse is 13 and the base is 5 for angle \(A\). Find the perpendicular and the three main ratios.
Concept: identify the 5-12-13 triple and locate P, B, H for angle A.
Working: \(P = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = 12\). So \(\sin A = \frac{12}{13}\), \(\cos A = \frac{5}{13}\), \(\tan A = \frac{12}{5}\).
Final Answer: \(P = 12\); \(\sin A = \frac{12}{13}\), \(\cos A = \frac{5}{13}\), \(\tan A = \frac{12}{5}\).
Worked Example 5
Question: Can \(\sin\theta\) or \(\cos\theta\) be greater than 1? Explain.
Concept: compare each ratio's numerator with its denominator.
Working: \(\sin\theta = \frac{P}{H}\), and \(P \le H\) always, so \(\frac{P}{H} \le 1\). Similarly \(B \le H\), so \(\cos\theta \le 1\).
Final Answer: No. \(\sin\theta \le 1\) and \(\cos\theta \le 1\) for every angle.
Worked Example 6
Question: Which ratio equals \(\frac{P}{B}\) and which equals \(\frac{H}{P}\)?
Concept: match letters to the six definitions.
Working: \(\frac{P}{B} = \tan\theta\) and \(\frac{H}{P} = \mathrm{cosec}\,\theta\).
Final Answer: \(\frac{P}{B} = \tan\theta\); \(\frac{H}{P} = \mathrm{cosec}\,\theta\).
Worked Example 7
Question: Given \(P = 7\) and \(B = 24\), find all six ratios of the angle.
Concept: 7-24-25 triple gives \(H = 25\).
Working: \(\sin = \frac{7}{25}\), \(\cos = \frac{24}{25}\), \(\tan = \frac{7}{24}\), \(\mathrm{cosec} = \frac{25}{7}\), \(\sec = \frac{25}{24}\), \(\cot = \frac{24}{7}\).
Final Answer: \(\frac{7}{25},\frac{24}{25},\frac{7}{24},\frac{25}{7},\frac{25}{24},\frac{24}{7}\).
Worked Example 8
Question: If \(\sin\theta = \frac{3}{5}\), find \(\cos\theta\) and \(\tan\theta\).
Concept: \(P = 3\), \(H = 5\), so \(B = 4\) by the 3-4-5 triple.
Working: \(\cos\theta = \frac{4}{5}\), \(\tan\theta = \frac{3}{4}\).
Final Answer: \(\cos\theta = \frac{4}{5}\), \(\tan\theta = \frac{3}{4}\).
Worked Example 9
Question: If \(\tan\theta = \frac{5}{12}\), determine all the other ratios.
Concept: \(P = 5\), \(B = 12\), then \(H = 13\).
Working: \(\sin = \frac{5}{13}\), \(\cos = \frac{12}{13}\), \(\mathrm{cosec} = \frac{13}{5}\), \(\sec = \frac{13}{12}\), \(\cot = \frac{12}{5}\).
Final Answer: \(\frac{5}{13},\frac{12}{13},\frac{13}{5},\frac{13}{12},\frac{12}{5}\).
Worked Example 10
Question: If \(\cos\theta = \frac{15}{17}\), find \(\sin\theta\) and \(\tan\theta\).
Concept: \(B = 15\), \(H = 17\), so \(P = 8\) (8-15-17 triple).
Working: \(\sin\theta = \frac{8}{17}\), \(\tan\theta = \frac{8}{15}\).
Final Answer: \(\sin\theta = \frac{8}{17}\), \(\tan\theta = \frac{8}{15}\).
Worked Example 11
Question: If \(\sec\theta = \frac{13}{12}\), find all the other ratios.
Concept: \(\sec = \frac{H}{B}\) gives \(H = 13\), \(B = 12\), hence \(P = 5\).
Working: \(\sin = \frac{5}{13}\), \(\cos = \frac{12}{13}\), \(\tan = \frac{5}{12}\), \(\mathrm{cosec} = \frac{13}{5}\), \(\cot = \frac{12}{5}\).
Final Answer: \(\frac{5}{13},\frac{12}{13},\frac{5}{12},\frac{13}{5},\frac{12}{5}\).
Worked Example 12
Question: If \(\cot\theta = \frac{7}{24}\), find \(\sin\theta\), \(\cos\theta\) and \(\tan\theta\).
Concept: \(\cot = \frac{B}{P}\) gives \(B = 7\), \(P = 24\), so \(H = 25\).
Working: \(\sin\theta = \frac{24}{25}\), \(\cos\theta = \frac{7}{25}\), \(\tan\theta = \frac{24}{7}\).
Final Answer: \(\frac{24}{25},\ \frac{7}{25},\ \frac{24}{7}\).
Worked Example 13
Question: A right triangle has hypotenuse 25 and base 7. Find the perpendicular and \(\mathrm{cosec}\,\theta\).
Concept: 7-24-25 triple; \(\mathrm{cosec} = \frac{H}{P}\).
Working: \(P = \sqrt{25^2 - 7^2} = 24\); \(\mathrm{cosec}\,\theta = \frac{25}{24}\). Also \(\cos\theta = \frac{7}{25}\).
Final Answer: \(P = 24\); \(\mathrm{cosec}\,\theta = \frac{25}{24}\).
Worked Example 14
Question: In a right triangle, the perpendicular is 8 and the hypotenuse is 17. Find the base and \(\tan\theta\).
Concept: 8-15-17 triple; \(\tan = \frac{P}{B}\).
Working: \(B = \sqrt{17^2 - 8^2} = 15\); \(\tan\theta = \frac{8}{15}\).
Final Answer: \(B = 15\), \(\tan\theta = \frac{8}{15}\).
Worked Example 15
Question: Evaluate \(\sin 30^\circ + \cos 60^\circ\).
Concept: table values \(\sin 30^\circ = \frac12\), \(\cos 60^\circ = \frac12\).
Working: \(\frac12 + \frac12 = 1\).
Final Answer: \(1\).
Worked Example 16
Question: Evaluate \(\tan 45^\circ + \sin 90^\circ\).
Concept: \(\tan 45^\circ = 1\), \(\sin 90^\circ = 1\).
Working: \(1 + 1 = 2\).
Final Answer: \(2\).
Worked Example 17
Question: Evaluate \(\sec 60^\circ - \tan 45^\circ\).
Concept: \(\sec 60^\circ = 2\), \(\tan 45^\circ = 1\).
Working: \(2 - 1 = 1\).
Final Answer: \(1\).
Worked Example 18
Question: Evaluate \(\sin^2 60^\circ + \cos^2 60^\circ\).
Concept: identity \(\sin^2\theta + \cos^2\theta = 1\), or table values.
Working (table): \(\left(\frac{\sqrt3}{2}\right)^2 + \left(\frac12\right)^2 = \frac34 + \frac14 = 1\).
Final Answer: \(1\).
Worked Example 19
Question: Evaluate \(\tan 30^\circ \cdot \tan 60^\circ + \sin 0^\circ\).
Concept: reciprocal-looking pair \(\frac{1}{\sqrt3}\cdot\sqrt3 = 1\), and \(\sin 0^\circ = 0\).
Working: \(1 + 0 = 1\).
Final Answer: \(1\).
Worked Example 20
Question: Evaluate \(\sec^2 60^\circ - \tan^2 60^\circ\).
Concept: \(\sec 60^\circ = 2\), \(\tan 60^\circ = \sqrt3\).
Working: \(4 - 3 = 1\).
Final Answer: \(1\) — matching the identity \(\sec^2\theta - \tan^2\theta = 1\).
Worked Example 21
Question: Prove that \(\sin^4\theta - \cos^4\theta = \sin^2\theta - \cos^2\theta\).
Concept: factorise as a difference of two squares, then apply the identity.
Working: \(\sin^4\theta - \cos^4\theta = (\sin^2\theta - \cos^2\theta)(\sin^2\theta + \cos^2\theta) = (\sin^2\theta - \cos^2\theta)(1)\).
Final Answer: \(\sin^4\theta - \cos^4\theta = \sin^2\theta - \cos^2\theta\). ✓
Worked Example 22
Question: Simplify \(\dfrac{1}{1 + \tan^2\theta}\).
Concept: \(1 + \tan^2\theta = \sec^2\theta\), then invert.
Working: \(\frac{1}{\sec^2\theta} = \cos^2\theta\).
Final Answer: \(\cos^2\theta\).
Worked Example 23
Question: Simplify \((1 - \sin^2\theta)\,\mathrm{cosec}^2\,\theta\).
Concept: \(1 - \sin^2\theta = \cos^2\theta\) and \(\mathrm{cosec}^2\theta = \frac{1}{\sin^2\theta}\).
Working: \(\cos^2\theta \cdot \frac{1}{\sin^2\theta} = \frac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta\).
Final Answer: \(\cot^2\theta\).
Worked Example 24
Question: Simplify \(\dfrac{\cos\theta}{1 - \sin^2\theta}\).
Concept: \(1 - \sin^2\theta = \cos^2\theta\).
Working: \(\frac{\cos\theta}{\cos^2\theta} = \frac{1}{\cos\theta} = \sec\theta\).
Final Answer: \(\sec\theta\).
Worked Example 25
Question: Prove that \(\tan^2\theta - \sin^2\theta = \tan^2\theta \cdot \sin^2\theta\).
Concept: write \(\tan^2\theta = \frac{\sin^2\theta}{\cos^2\theta}\) and combine over the same denominator.
Working: \(\tan^2\theta - \sin^2\theta = \frac{\sin^2\theta}{\cos^2\theta} - \sin^2\theta = \frac{\sin^2\theta - \sin^2\theta\cos^2\theta}{\cos^2\theta} = \frac{\sin^2\theta(1 - \cos^2\theta)}{\cos^2\theta} = \frac{\sin^2\theta \cdot \sin^2\theta}{\cos^2\theta} = \tan^2\theta\sin^2\theta.\)
Final Answer: \(\tan^2\theta - \sin^2\theta = \tan^2\theta\sin^2\theta\). ✓
Worked Example 26
Question: If \(\tan\theta = \frac{3}{4}\), find the value of \(\dfrac{2\sin\theta + 3\cos\theta}{\sin\theta + \cos\theta}\).
Concept: \(\tan\theta = \frac34\) gives the 3-4-5 triangle, so \(\sin\theta = \frac35\) and \(\cos\theta = \frac45\).
Working: numerator \(= 2\cdot\frac35 + 3\cdot\frac45 = \frac{6+12}{5} = \frac{18}{5}\); denominator \(= \frac{3+4}{5} = \frac75\). Ratio \(= \frac{18/5}{7/5} = \frac{18}{7}\).
Final Answer: \(\frac{18}{7}\).
Worked Example 27
Question: A 5 m ladder rests with its foot 4 m from a wall; its top touches the wall 3 m up. For the angle θ between the ladder and the floor, write \(\sin\theta\), \(\cos\theta\) and \(\tan\theta\).
Concept: the right triangle is a 3-4-5; for angle θ at the foot of the ladder: \(P = 3\) (wall), \(B = 4\) (floor), \(H = 5\) (ladder).
Working: \(\sin\theta = \frac35\), \(\cos\theta = \frac45\), \(\tan\theta = \frac34\).
Final Answer: \(\sin\theta = \frac35\), \(\cos\theta = \frac45\), \(\tan\theta = \frac34\).
Worked Example 28
Question: A ramp material runs 12 m along the ground and reaches 5 m vertically, with a slanted length of 13 m. Write the three main ratios for the angle θ the ramp makes with the ground.
Concept: identify \(P = 5\), \(B = 12\), \(H = 13\) using the 5-12-13 triple.
Working: \(\sin\theta = \frac{5}{13}\), \(\cos\theta = \frac{12}{13}\), \(\tan\theta = \frac{5}{12}\).
Final Answer: \(\frac{5}{13},\ \frac{12}{13},\ \frac{5}{12}\).
Worked Example 29
Question: In \(\triangle ABC\), right-angled at \(B\), \(AB = 5\), \(BC = 12\) and \(AC = 13\). Which of the following equals \(\frac{13}{12}\): \(\sin A\), \(\sec A\), \(\mathrm{cosec}\,A\), or \(\tan A\)?
Concept: ratio = hypotenuse over perpendicular, i.e. \(\mathrm{cosec}\).
Working: \(H = AC = 13\), perpendicular to \(A\) is \(BC = 12\), so \(\mathrm{cosec}\,A = \frac{H}{P} = \frac{13}{12}\).
Final Answer: \(\mathrm{cosec}\,A = \frac{13}{12}\).
Worked Example 30
Question: If \(\sin\theta = \frac{12}{13}\), find the value of \((1 + \cos\theta)(1 - \cos\theta)\).
Concept: \((1+\cos\theta)(1-\cos\theta) = 1 - \cos^2\theta = \sin^2\theta\) by identity.
Working: \((1+\cos\theta)(1-\cos\theta) = \sin^2\theta = \left(\frac{12}{13}\right)^2 = \frac{144}{169}\).
Final Answer: \(\frac{144}{169}\).

15. Self-Practice Questions

Thirty-two questions covering every skill in this chapter. Answer them fully before checking the answer list, which is given separately at the end.

A. Identifying ratios

  1. In \(\triangle ABC\), right-angled at \(B\), with the acute angle at \(C\) marked as θ, name the perpendicular, base and hypotenuse.
  2. The sides of a right triangle are 6, 8 and 10. Which side is the hypotenuse?
  3. Can \(\sin\theta\) be greater than 1? Give a one-line reason.
  4. Which ratio equals \(\frac{\text{Base}}{\text{Hypotenuse}}\)? Which equals \(\frac{\text{Perpendicular}}{\text{Base}}\)?

B. Calculating ratios from sides

  1. A right triangle has \(P = 9\), \(B = 12\), \(H = 15\). Write \(\sin\theta\), \(\cos\theta\), \(\tan\theta\).
  2. A right triangle has \(P = 5\), \(B = 12\), \(H = 13\). Write all six ratios of the angle.
  3. For the 8-15-17 triangle, with \(P = 8\) and \(B = 15\), find \(\mathrm{cosec}\,\theta\), \(\sec\theta\), \(\cot\theta\).
  4. If \(\sin\theta = \frac{12}{13}\), write the corresponding values of \(P\), \(B\), \(H\) and hence \(\cos\theta\).

C. Finding the missing side

  1. In a right triangle, \(B = 12\) and \(H = 20\). Find \(P\), then write \(\tan\theta\).
  2. In a right triangle, \(P = 7\) and \(H = 25\). Find \(B\), then write \(\sec\theta\).

D. Standard-angle values

  1. Write the values of \(\sin 30^\circ\), \(\cos 60^\circ\) and \(\tan 45^\circ\).
  2. Write \(\sin 0^\circ\), \(\cos 90^\circ\), \(\sin 90^\circ\), \(\cos 0^\circ\).
  3. Write \(\tan 30^\circ\), \(\tan 60^\circ\) and \(\cot 45^\circ\).
  4. Which of \(\tan 90^\circ\) and \(\sin 90^\circ\) is not defined?

E. Reciprocal ratios

  1. State the reciprocal of each of \(\sin\theta\), \(\cos\theta\), \(\tan\theta\).
  2. If \(\sin\theta = \frac{4}{5}\), find \(\mathrm{cosec}\,\theta\).
  3. If \(\sec\theta = \frac{17}{15}\), find \(\cos\theta\) and \(\sin\theta\).

F. Fundamental identities

  1. Evaluate \(\sin^2 30^\circ + \cos^2 30^\circ\).
  2. If \(\tan\theta = \sqrt3\), use \(1 + \tan^2\theta = \sec^2\theta\) to find \(\sec\theta\).
  3. If \(\cot\theta = 1\), use \(1 + \cot^2\theta = \mathrm{cosec}^2\theta\) to find \(\mathrm{cosec}\,\theta\).

G. Simplification

  1. Simplify \((1 + \tan^2\theta)\cos^2\theta\).
  2. Simplify \(\dfrac{1 - \sin^2\theta}{1 - \cos^2\theta}\).
  3. Simplify \(\dfrac{1}{1 + \cot^2\theta}\).
  4. Prove that \(\sin\theta \cdot \cot\theta = \cos\theta\).

H. Evaluation of expressions

  1. Evaluate \(\sin 45^\circ + \cos 45^\circ\).
  2. Evaluate \(\tan 45^\circ + \tan 30^\circ \cdot \tan 60^\circ\).
  3. Evaluate \(\sin^2 30^\circ + \cos^2 30^\circ\) from the table (do not use the identity).

I. Finding unknown ratios

  1. If \(\cos\theta = \frac{12}{13}\), find \(\sin\theta\) and \(\tan\theta\).
  2. If \(\tan\theta = 1\), find \(\sin\theta\) and \(\cos\theta\).

J. Application and competency

  1. A ramp rises 3 m straight up over a base of 4 m, with a strongly slanted length of 5 m. For the angle the ramp makes with the ground, write \(\sin\theta\), \(\cos\theta\), \(\tan\theta\).
  2. A right triangle has hypotenuse 10 and one perpendicular side 6. Find the remaining side and \(\tan\theta\) for that angle.
  3. If \(\sin\theta = \frac{5}{13}\), evaluate \(1 + \tan^2\theta\).

Answers

  1. \(P = AB\), \(B = BC\), \(H = AC\) (with the acute angle at \(C\)).
  2. 10 units.
  3. No: \(P \le H\), so \(P/H \le 1\).
  4. \(\cos\theta\) and \(\tan\theta\).
  5. \(\frac35\), \(\frac45\), \(\frac34\).
  6. \(\frac{5}{13}\), \(\frac{12}{13}\), \(\frac{5}{12}\), \(\frac{13}{5}\), \(\frac{13}{12}\), \(\frac{12}{5}\).
  7. \(\frac{17}{8}\), \(\frac{17}{15}\), \(\frac{15}{8}\).
  8. \(P = 12\), \(H = 13\), \(B = 5\), \(\cos\theta = \frac{5}{13}\).
  9. \(P = 16\), \(\tan\theta = \frac{16}{12} = \frac43\).
  10. \(B = 24\), \(\sec\theta = \frac{25}{24}\).
  11. \(\frac12\), \(\frac12\), \(1\).
  12. \(0\), \(0\), \(1\), \(1\).
  13. \(\frac{1}{\sqrt3}\), \(\sqrt3\), \(1\).
  14. \(\tan 90^\circ\) is not defined.
  15. \(\mathrm{cosec}\,\theta\), \(\sec\theta\), \(\cot\theta\).
  16. \(\frac54\).
  17. \(\cos\theta = \frac{15}{17}\), \(\sin\theta = \frac{8}{17}\).
  18. \(1\).
  19. \(\sec 60^\circ = 2\) (since \(\tan^2 60^\circ = 3\)).
  20. \(\mathrm{cosec}\,45^\circ = \sqrt2\) (since \(\cot^2 45^\circ = 1\)).
  21. \(\cos^2\theta \cdot \sec^2\theta = 1\).
  22. \(\frac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta\).
  23. \(\frac{1}{\mathrm{cosec}^2\theta} = \sin^2\theta\).
  24. \(\sin\theta \cdot \frac{\cos\theta}{\sin\theta} = \cos\theta\).
  25. \(\frac{1}{\sqrt2} + \frac{1}{\sqrt2} = \sqrt2\).
  26. \(1 + 1 = 2\).
  27. \(\frac14 + \frac34 = 1\).
  28. \(\sin\theta = \frac{5}{13}\), \(\tan\theta = \frac{5}{12}\).
  29. \(\sin\theta = \cos\theta = \frac{1}{\sqrt2}\).
  30. \(\sin\theta = \frac35\), \(\cos\theta = \frac45\), \(\tan\theta = \frac34\).
  31. Other side \(= 8\), \(\tan\theta = \frac68 = \frac34\).
  32. \(1 + \tan^2\theta = \sec^2\theta = \left(\frac{13}{12}\right)^2 = \frac{169}{144}\).

Next: Chapter 9 — Some Applications of Trigonometry

Apply the six ratios you have mastered here to actual heights and distances.

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