Introduction to Trigonometry — Detailed Notes

Chapter 8 | CBSE Class 10 Mathematics (NCERT)

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Introduction to Trigonometry — Decision Map

Trigonometric Ratios in a Right Triangle

Standard Angles — Quick Table

8.1 Introduction

Trigonometry (Greek: tri + gon + metron) studies relations between sides and angles of triangles. NCERT motivates it with heights (Qutub Minar), river width, and balloon altitude — situations modelled by right triangles. Discussion here is restricted mainly to acute angles, with definitions extended to $0^{\circ}$ and $9 0^{\circ}$ .

8.2 Trigonometric Ratios

In right $△ A B C$ with right angle at $B$ , for acute $∠ A$ :

Ratio	Definition
$sin A$	$\frac{opposite}{hypotenuse} = \frac{B C}{A C}$
$cos A$	$\frac{adjacent}{hypotenuse} = \frac{A B}{A C}$
$tan A$	$\frac{opposite}{adjacent} = \frac{B C}{A B}$
$cosec A$	$\frac{1}{sin A} = \frac{A C}{B C}$
$sec A$	$\frac{1}{cos A} = \frac{A C}{A B}$
$cot A$	$\frac{1}{tan A} = \frac{A B}{B C}$

Also: $tan A = \frac{sin A}{cos A}, cot A = \frac{cos A}{sin A} .$

Important: $sin A$ is not “sin $\times$ $A$ ”. The name and angle form one symbol.

Independence of size

By AA similarity, any right triangle with the same acute angle $A$ gives the same trigonometric ratios. Ratios depend on the angle, not on the particular side lengths.

Notation: $sin^{2} A$ means $(sin A)^{2}$ . Note $cosec A = (sin A)^{- 1} \neq = sin^{- 1} A$ .

Finding other ratios from one

If $sin A = \frac{1}{3}$ , take opposite $= k$ , hypotenuse $= 3 k$ , then adjacent $= 8 k = 22 k$ (Pythagoras; length positive). Then write $cos A$ , $tan A$ , etc.

Remark: Since hypotenuse is longest, $sin A$ and $cos A$ are at most $1$ .

Worked ideas

Example 1: $tan A = \frac{4}{3}$ $\Rightarrow$ sides $3 k, 4 k, 5 k$ $\Rightarrow$ $sin A = \frac{4}{5}, cos A = \frac{3}{5}, cot A = \frac{3}{4}, cosec A = \frac{5}{4}, sec A = \frac{5}{3} .$

Example 2: If $sin B = sin Q$ (acute), then $∠ B = ∠ Q$ (similar right triangles via SSS on proportional sides).

Example 3: Right-angled at $C$ , $A B = 29$ , $B C = 21$ $\Rightarrow$ $A C = 20$ . For $θ = ∠ A B C$ : $sin^{2} θ + cos^{2} θ = 1, cos^{2} θ - sin^{2} θ = \frac{41}{29} .$

Example 4: $tan A = 1$ $\Rightarrow$ $sin A = cos A = \frac{1}{2}$ $\Rightarrow$ $2 sin A cos A = 1$ .

Example 5: Right-angled at $P$ , $O P = 7$ , $O Q - P Q = 1$ $\Rightarrow$ $P Q = 24$ , $O Q = 25$ $\Rightarrow$ $sin Q = \frac{7}{25}$ , $cos Q = \frac{24}{25}$ .

Better Method / Strategy: Always introduce a positive scale $k$ for sides from a given ratio; finish with Pythagoras; only then write all six ratios.

8.3 Trigonometric Ratios of Some Specific Angles

$4 5^{\circ}$

Isosceles right triangle: legs $a$ , hypotenuse $a 2$ . $sin 4 5^{\circ} = cos 4 5^{\circ} = \frac{1}{2}, tan 4 5^{\circ} = 1.$

$3 0^{\circ}$ and $6 0^{\circ}$

Equilateral of side $2 a$ ; altitude splits base into $a$ and height $a 3$ . $sin 3 0^{\circ} sin 6 0^{\circ} = \frac{1}{2}, = \frac{3}{2}, cos 3 0^{\circ} cos 6 0^{\circ} = \frac{3}{2}, = \frac{1}{2}, tan 3 0^{\circ} tan 6 0^{\circ} = \frac{1}{3}, = 3 .$

$0^{\circ}$ and $9 0^{\circ}$

Limiting definitions: $sin 0^{\circ} = 0$ , $cos 0^{\circ} = 1$ , $sin 9 0^{\circ} = 1$ , $cos 9 0^{\circ} = 0$ .

$A$	$0^{\circ}$	$3 0^{\circ}$	$4 5^{\circ}$	$6 0^{\circ}$	$9 0^{\circ}$
$sin A$	$0$	$\frac{1}{2}$	$\frac{1}{2}$	$\frac{3}{2}$	$1$
$cos A$	$1$	$\frac{3}{2}$	$\frac{1}{2}$	$\frac{1}{2}$	$0$
$tan A$	$0$	$\frac{1}{3}$	$1$	$3$	ND
$cosec A$	ND	$2$	$2$	$\frac{2}{3}$	$1$
$sec A$	$1$	$\frac{2}{3}$	$2$	$2$	ND
$cot A$	ND	$3$	$1$	$\frac{1}{3}$	$0$

(ND = not defined.) As $A$ increases from $0^{\circ}$ to $9 0^{\circ}$ , $sin A$ increases and $cos A$ decreases.

Worked ideas

Find sides (Example 6): Right-angled at $B$ , $A B = 5$ , $∠ C = 3 0^{\circ}$ $\Rightarrow$ $tan 3 0^{\circ} = \frac{5}{B C}$ $\Rightarrow$ $B C = 53$ ; $sin 3 0^{\circ} = \frac{5}{A C}$ $\Rightarrow$ $A C = 10$ .

Find angles (Example 7): $P Q = 3$ , $P R = 6$ $\Rightarrow$ $sin R = \frac{1}{2}$ $\Rightarrow$ $∠ R = 3 0^{\circ}$ , $∠ P = 6 0^{\circ}$ .

Solve for $A, B$ (Example 8): $sin (A - B) = \frac{1}{2}$ , $cos (A + B) = \frac{1}{2}$ with constraints $\Rightarrow$ $A - B = 3 0^{\circ}$ , $A + B = 6 0^{\circ}$ $\Rightarrow$ $A = 4 5^{\circ}$ , $B = 1 5^{\circ}$ .

Better Method / Strategy: Choose the ratio that links the known side to the unknown side. Prefer $tan$ when both legs appear; $sin$ / $cos$ when hypotenuse is involved.

8.4 Trigonometric Identities

An equation true for all admissible angles is an identity.

From $A B^{2} + B C^{2} = A C^{2}$ in right $△ A B C$ :

Divide by $A C^{2}$ : $sin^{2} A + cos^{2} A = 1 (0^{\circ} \leq A \leq 9 0^{\circ}) .$
Divide by $A B^{2}$ : $1 + tan^{2} A = sec^{2} A (0^{\circ} \leq A < 9 0^{\circ}) .$
Divide by $B C^{2}$ : $1 + cot^{2} A = cosec^{2} A (0^{\circ} < A \leq 9 0^{\circ}) .$

Worked ideas

Express in terms of $sin A$ (Example 9): $cos A = 1 - sin^{2} A (acute \Rightarrow positive), tan A = \frac{sin A}{1 - sin ^{2} A} .$

Prove (Example 10): $sec A (1 - sin A) (sec A + tan A) = 1$ — expand in $sin, cos$ and use $1 - sin^{2} A = cos^{2} A$ .

Prove (Example 11): $\frac{cot A - cos A}{cot A + cos A} = \frac{cosec A - 1}{cosec A + 1}$ .

Prove (Example 12): Convert LHS to $tan, sec$ by dividing by $cos$ , then use $1 + tan^{2} = sec^{2}$ .

Better Method / Strategy:

Default: write everything in $sin$ and $cos$ .

If $sec / tan$ dominate, use $1 + tan^{2} = sec^{2}$ .

Sometimes simplify LHS and RHS separately.

Standard-Angle Memory System

Write $sin$ row as: $0, \frac{1}{2}, \frac{1}{2}, \frac{3}{2}, 1$ for $0^{\circ}, 3 0^{\circ}, 4 5^{\circ}, 6 0^{\circ}, 9 0^{\circ}$ .
$cos$ is the reverse order. $tan = sin / cos$ . Reciprocals flip the fractions (and create undefined entries where $sin$ or $cos$ is $0$ ).

Complementary check: $sin (9 0^{\circ} - A) = cos A$ , $tan (9 0^{\circ} - A) = cot A$ — useful for spotting calculation errors.

Identity Proof Toolkit

Situation	Move
Fractions with $sin, cos$	Common denominator $sin cos$ or $cos^{2}$
$sec, tan$ everywhere	Replace using $1 + tan^{2} = sec^{2}$
$cosec, cot$ everywhere	Use $1 + cot^{2} = cosec^{2}$
$(1 - sin)$ factors	Multiply by $(1 + sin)$ to get $cos^{2}$
Stuck LHS	Convert both sides to $sin, cos$ separately

Classic transforms

$(1 - sin A) (1 + sin A) = cos^{2} A$ $(sec A + tan A) (sec A - tan A) = 1$ $(cosec A + cot A) (cosec A - cot A) = 1$

True/False Hotspots (Board Style)

Statement	Verdict
$tan A$ always $< 1$	False ( $tan 6 0^{\circ} = 3$ )
$sin θ = \frac{4}{3}$ possible	False
$sec A = \frac{12}{5}$ possible	True
$sin (A + B) = sin A + sin B$	False
$sin θ$ increases as $θ$ increases on $[0^{\circ}, 9 0^{\circ}]$	True
$cos θ$ increases on $[0^{\circ}, 9 0^{\circ}]$	False
$cot 0^{\circ}$ not defined	True

Linking One Ratio to All Others (Recipe)

Write sides with parameter $k$ from the given ratio.
Find the third side by Pythagoras (positive root).
Write all six ratios; simplify.
Optional check: $sin^{2} + cos^{2} = 1$ .

Exam-Oriented Analysis

Aspect	Detail
Weight	Major; often paired with Chapter 9 (combined 6–10 marks common)
Question types	Compute ratios; evaluate; MCQ on values; prove identities; true/false
1-mark	Table value / definition
2-mark	Find other ratios from one
3–4 mark	Prove identity or multi-part evaluate
Common mistakes	$sin A > 1$ ; treating $sin$ as a factor; wrong table values; using identities outside domain; $sin^{- 1}$ confusion
Scoring tips	Quote identity used; keep acute-angle positive roots; box final simplified value

Scoring tip: In 3-mark proofs, show each substitution line; end with “LHS = RHS”.

Method chooser

Evaluate? → Table → simplify
One ratio → all? → k-sides → Pythagoras
Prove identity?
  → Prefer sin/cos, OR sec/tan identity, OR cosec/cot identity
Find side in right Δ? → Choose sin/cos/tan linking known↔unknown