Trigonometry

ICSE Class 10 Mathematics • Chapter 13

1. Trigonometric Ratios

[Diagram: Right triangle ABC with right angle at B, angle θ at A, opposite side BC, adjacent side AB, hypotenuse AC]

In a right-angled triangle, for angle θ:

Ratio	Formula	Reciprocal
sin θ	$\frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{P}{H}$	$\csc\theta = \frac{H}{P}$
cos θ	$\frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{B}{H}$	$\sec\theta = \frac{H}{B}$
tan θ	$\frac{\text{Opposite}}{\text{Adjacent}} = \frac{P}{B}$	$\cot\theta = \frac{B}{P}$

Memory Tricks:

SOH-CAH-TOA: Sin=O/H, Cos=A/H, Tan=O/A
"Pandit Badri Prasad Har Har Bole" → P/H, B/H, P/B for sin, cos, tan

2. Standard Angle Values

θ	0°	30°	45°	60°	90°
sin θ	0	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	1
cos θ	1	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	0
tan θ	0	$\frac{1}{\sqrt{3}}$	1	$\sqrt{3}$	∞
cot θ	∞	$\sqrt{3}$	1	$\frac{1}{\sqrt{3}}$	0
sec θ	1	$\frac{2}{\sqrt{3}}$	$\sqrt{2}$	2	∞
cosec θ	∞	2	$\sqrt{2}$	$\frac{2}{\sqrt{3}}$	1

Quick Pattern for sin θ: $\sin 0° = \frac{\sqrt{0}}{2}$, $\sin 30° = \frac{\sqrt{1}}{2}$, $\sin 45° = \frac{\sqrt{2}}{2}$, $\sin 60° = \frac{\sqrt{3}}{2}$, $\sin 90° = \frac{\sqrt{4}}{2}$

cos θ = sin θ in reverse order!

3. Fundamental Trigonometric Identities

Pythagorean Identities:

$\sin^2\theta + \cos^2\theta = 1$
$1 + \tan^2\theta = \sec^2\theta$
$1 + \cot^2\theta = \csc^2\theta$

Derived Forms (Rearrangements)

From Identity	Derived Forms
$\sin^2\theta + \cos^2\theta = 1$	$\sin^2\theta = 1 - \cos^2\theta$
$\sin^2\theta + \cos^2\theta = 1$	$\cos^2\theta = 1 - \sin^2\theta$
$1 + \tan^2\theta = \sec^2\theta$	$\tan^2\theta = \sec^2\theta - 1$
$1 + \tan^2\theta = \sec^2\theta$	$\sec^2\theta - \tan^2\theta = 1$
$1 + \cot^2\theta = \csc^2\theta$	$\cot^2\theta = \csc^2\theta - 1$
$1 + \cot^2\theta = \csc^2\theta$	$\csc^2\theta - \cot^2\theta = 1$

Quotient Identities:

$\tan\theta = \frac{\sin\theta}{\cos\theta}$ $\qquad$ $\cot\theta = \frac{\cos\theta}{\sin\theta}$

4. Complementary Angles (90° − θ)

Ratio	Complementary Relation
$\sin(90° - \theta)$	$= \cos\theta$
$\cos(90° - \theta)$	$= \sin\theta$
$\tan(90° - \theta)$	$= \cot\theta$
$\cot(90° - \theta)$	$= \tan\theta$
$\sec(90° - \theta)$	$= \csc\theta$
$\csc(90° - \theta)$	$= \sec\theta$

Co-function Rule: "Co" goes with "Co" → sin↔cos, tan↔cot, sec↔csc

5. Proving Trigonometric Identities

Strategy for Proving LHS = RHS:

Choose the more complex side (usually LHS)
Express everything in terms of sin θ and cos θ
Use algebraic operations (factoring, taking LCM, etc.)
Apply fundamental identities
Simplify to match the other side

Example 1: Prove: $\frac{1 + \tan^2\theta}{1 + \cot^2\theta} = \tan^2\theta$

Proof:

LHS = $\frac{1 + \tan^2\theta}{1 + \cot^2\theta}$

= $\frac{\sec^2\theta}{\csc^2\theta}$ (using identities)

= $\frac{1/\cos^2\theta}{1/\sin^2\theta}$

= $\frac{\sin^2\theta}{\cos^2\theta}$

= $\tan^2\theta$ = RHS ✓

Example 2: Prove: $\frac{\sin\theta}{1 + \cos\theta} + \frac{1 + \cos\theta}{\sin\theta} = 2\csc\theta$

Proof:

LHS = $\frac{\sin^2\theta + (1 + \cos\theta)^2}{\sin\theta(1 + \cos\theta)}$

= $\frac{\sin^2\theta + 1 + 2\cos\theta + \cos^2\theta}{\sin\theta(1 + \cos\theta)}$

= $\frac{(\sin^2\theta + \cos^2\theta) + 1 + 2\cos\theta}{\sin\theta(1 + \cos\theta)}$

= $\frac{1 + 1 + 2\cos\theta}{\sin\theta(1 + \cos\theta)}$

= $\frac{2(1 + \cos\theta)}{\sin\theta(1 + \cos\theta)}$

= $\frac{2}{\sin\theta} = 2\csc\theta$ = RHS ✓

6. Heights and Distances

Angle of Elevation: Angle made by line of sight with horizontal when looking UP at an object.

Angle of Depression: Angle made by line of sight with horizontal when looking DOWN at an object.

[Diagram: Tower with observer on ground showing angle of elevation; Observer on building looking down showing angle of depression]

Key Properties:

Angle of elevation from A to B = Angle of depression from B to A
Always draw horizontal lines at eye level

Problem Types

Type	Given	Formula
Find Height	Distance, Angle	$\tan\theta = \frac{\text{Height}}{\text{Distance}}$ → Height = Distance × tan θ
Find Distance	Height, Angle	Distance = Height × cot θ
Two Angles	Moving towards	Height from $\tan\theta_1 - \tan\theta_2$ relation

Example 3: From a point 30m from the base of a tower, the angle of elevation of the top is 60°. Find the height.

Solution:

$\tan 60° = \frac{h}{30}$

$\sqrt{3} = \frac{h}{30}$

$h = 30\sqrt{3} \approx$ 51.96 m

Example 4: From the top of a 50m building, the angles of depression to the top and bottom of a tower are 45° and 60°. Find the height of the tower.

Solution:

Let tower height = h, horizontal distance = d

From 60° (to bottom of tower): $\tan 60° = \frac{50}{d}$ → $d = \frac{50}{\sqrt{3}}$

From 45° (to top of tower): $\tan 45° = \frac{50-h}{d}$

$1 = \frac{50-h}{50/\sqrt{3}}$

$\frac{50}{\sqrt{3}} = 50 - h$

$h = 50 - \frac{50}{\sqrt{3}} = 50\left(1 - \frac{1}{\sqrt{3}}\right) = 50 \times \frac{\sqrt{3}-1}{\sqrt{3}}$

$h \approx$ 21.13 m

7. Quick Reference Formulas

Category	Formulas
Pythagorean	$\sin^2\theta + \cos^2\theta = 1$, $\sec^2\theta - \tan^2\theta = 1$, $\csc^2\theta - \cot^2\theta = 1$
Reciprocal	$\sin\theta\cdot\csc\theta = 1$, $\cos\theta\cdot\sec\theta = 1$, $\tan\theta\cdot\cot\theta = 1$
Quotient	$\tan\theta = \frac{\sin\theta}{\cos\theta}$, $\cot\theta = \frac{\cos\theta}{\sin\theta}$
Complementary	$\sin(90°-\theta) = \cos\theta$, $\tan(90°-\theta) = \cot\theta$, $\sec(90°-\theta) = \csc\theta$

Exam Practice (PYQ Trends)

PYQ: 2023

BOARD Prove: $\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A} = 1 + \sec A \cdot \csc A$

PYQ: 2022

BOARD From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a 20m high building are 45° and 60° respectively. Find the height of the tower.

Additional

HOTS If $\sin\theta + \cos\theta = \sqrt{2}\cos\theta$, show that $\cos\theta - \sin\theta = \sqrt{2}\sin\theta$