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ICSE • Class 9 • Mathematics • Ch 21
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Trigonometry

In ICSE Class 9 Mathematics, "Trigonometry" (derived from Greek *trigonon* = triangle and *metron* = measure) opens the profound mathematical bridge connecting angular rotation with linear distance. The chapter formally defines the six fundamental trigonometric ratios of an acute angle $\theta$ ($0^\circ \le \theta \le 90^\circ$) in a right-angled triangle where the legs are designated as Perpendicular ($P$, side opposite to $\theta$), Base ($B$, side adjacent to $\theta$), and Hypotenuse ($H$, side opposite the $90^\circ$ right angle): $\sin\theta = \frac{P}{H}$, $\cos\theta = \frac{B}{H}$, $\tan\theta = \frac{P}{B}$, and their reciprocals $\csc\theta = \frac{H}{P} = \frac{1}{\sin\theta}$, $\sec\theta = \frac{H}{B} = \frac{1}{\cos\theta}$, $\cot\theta = \frac{B}{P} = \frac{1}{\tan\theta}$. The curriculum derives the exact values of trigonometric ratios for standard benchmark angles ($0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$) using equilateral and isosceles right-angled geometric constructions. Students master the three fundamental Pythagorean identities: (1) $\sin^2\theta + \cos^2\theta = 1$; (2) $1 + \tan^2\theta = \sec^2\theta \iff \sec^2\theta - \tan^2\theta = 1$; and (3) $1 + \cot^2\theta = \csc^2\theta \iff \csc^2\theta - \cot^2\theta = 1$. The chapter emphasizes algebraic manipulations of trigonometric proofs, evaluating expressions without tables, and using complementary angle relationships.

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Overview Core Concepts Formulas & Identities Summary Self Check Take MCQ Test

How Eratosthenes Measured the Circumference of Planet Earth with a Wooden Stick and Sunlight Shadows

In 240 BCE in ancient Alexandria, the Greek polymath Eratosthenes accomplished one of the most astonishing scientific feats of antiquity: he calculated the circumference of planet Earth to within $1\%$ accuracy, without ever leaving Egypt! On the summer solstice at noon in the southern city of Syene (Aswan), the sun shone directly down to the bottom of a deep water well, casting zero shadow ($0^\circ$ zenith angle). Meanwhile, 800 kilometers north in Alexandria, Eratosthenes placed a vertical wooden stick in the ground and measured the shadow it cast. Using the right-angled triangle formed by the stick and its shadow, he computed $\tan\theta = \frac{\text{Shadow}}{\text{Stick}}$, revealing an angle of $7.2^\circ$. Recognizing that $7.2^\circ$ is exactly $\frac{1}{50}$ of a full $360^\circ$ circle, he multiplied the $800\text{ km}$ distance by $50$, arriving at $40,000\text{ kilometers}$—the exact circumference of the Earth! That is the astonishing power of Trigonometry: it enables human beings to measure the height of Mount Everest or the distance to the Moon without ever climbing them! Let us master the ratios of the triangle!

Why This Chapter Matters

Trigonometry is the universal mathematical engine of satellite GPS navigation, astronomy, sound wave synthesis, seismic earthquake modeling, 3D video game rendering, and alternating electrical currents.

Before You Begin (Prerequisites)

  • Pythagorean theorem ($P^2 + B^2 = H^2$) from Chapter 12.
  • Radical arithmetic and fraction simplification from Chapter 1.

What You Will Learn (Core Objectives)

  • Define the six trigonometric ratios with respect to an acute angle in a right triangle.
  • Derive and memorize the exact values of trigonometric ratios for $0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$.
  • State and prove the three fundamental Pythagorean trigonometric identities.
  • Prove complex trigonometric identities by transforming LHS to RHS using algebraic manipulations.
  • Evaluate numerical trigonometric expressions without using statistical or four-figure tables.

Chapter Roadmap & Progression

1 1. The Six Trigonometric Ratios
2 2. Values of Standard Benchmark Ang...
3 3. Fundamental Pythagorean Trigonom...
4 4. Worked ICSE Problem Archetypes

Complete Concept Guide (100% Curriculum Coverage)

1. The Six Trigonometric Ratios

Fundamental Definitions

In a right-angled triangle $\triangle ABC$ with $\angle B = 90^\circ$, for acute angle $\angle C = \theta$:

  • Hypotenuse ($H$) = $AC$ (side opposite $90^\circ$)
  • Perpendicular ($P$) = $AB$ (side opposite angle $\theta$)
  • Base ($B$) = $BC$ (side adjacent to angle $\theta$)
Primary RatioDefinition ($P, B, H$)Reciprocal RatioReciprocal Definition
$\sin\theta$ (Sine)$\frac{P}{H} = \frac{\text{Opposite}}{\text{Hypotenuse}}$$\csc\theta$ (Cosecant)$\frac{H}{P} = \frac{1}{\sin\theta}$
$\cos\theta$ (Cosine)$\frac{B}{H} = \frac{\text{Adjacent}}{\text{Hypotenuse}}$$\sec\theta$ (Secant)$\frac{H}{B} = \frac{1}{\cos\theta}$
$\tan\theta$ (Tangent)$\frac{P}{B} = \frac{\sin\theta}{\cos\theta}$$\cot\theta$ (Cotangent)$\frac{B}{P} = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}$

2. Values of Standard Benchmark Angles

Standard Angle Table
Ratio / $\theta$$0^\circ$$30^\circ$$45^\circ$$60^\circ$$90^\circ$
$\sin\theta$$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$$\sqrt{3}$$\text{Not Defined}$
$\csc\theta$$\text{Not Defined}$$2$$\sqrt{2}$$\frac{2}{\sqrt{3}}$$1$
$\sec\theta$$1$$\frac{2}{\sqrt{3}}$$\sqrt{2}$$2$$\text{Not Defined}$
$\cot\theta$$\text{Not Defined}$$\sqrt{3}$$1$$\frac{1}{\sqrt{3}}$$0$

3. Fundamental Pythagorean Trigonometric Identities

Pythagorean Identities
Identity 1: $\sin^2\theta + \cos^2\theta = 1$

Proof: In right $\triangle ABC$: $P^2 + B^2 = H^2$. Divide by $H^2$:

$$\frac{P^2}{H^2} + \frac{B^2}{H^2} = \frac{H^2}{H^2} \implies \left(\frac{P}{H}\right)^2 + \left(\frac{B}{H}\right)^2 = 1 \implies \mathbf{\sin^2\theta + \cos^2\theta = 1}$$

Corollaries: $\sin^2\theta = 1 - \cos^2\theta$ and $\cos^2\theta = 1 - \sin^2\theta$.

Identity 2: $1 + \tan^2\theta = \sec^2\theta$

Divide $P^2 + B^2 = H^2$ by $B^2$: $\frac{P^2}{B^2} + 1 = \frac{H^2}{B^2} \implies \tan^2\theta + 1 = \sec^2\theta$.

Corollary: $\mathbf{\sec^2\theta - \tan^2\theta = 1 \iff (\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1}$.

Identity 3: $1 + \cot^2\theta = \csc^2\theta$

Divide $P^2 + B^2 = H^2$ by $P^2$: $1 + \frac{B^2}{P^2} = \frac{H^2}{P^2} \implies 1 + \cot^2\theta = \csc^2\theta$.

Corollary: $\mathbf{\csc^2\theta - \cot^2\theta = 1}$.

4. Worked ICSE Problem Archetypes

Exemplary Solutions
Problem 1: If $\sin\theta = \frac{3}{5}$ and $\theta$ is an acute angle, find the value of $\frac{\tan\theta + \sec\theta}{\cot\theta + \csc\theta}$.

Solution:

Since $\sin\theta = \frac{P}{H} = \frac{3}{5}$, let $P = 3k$ and $H = 5k$.

By Pythagoras Theorem: $B = \sqrt{H^2 - P^2} = \sqrt{(5k)^2 - (3k)^2} = \sqrt{16k^2} = 4k$.

Now find all required ratios:

$$\tan\theta = \frac{P}{B} = \frac{3}{4}, \quad \sec\theta = \frac{H}{B} = \frac{5}{4}, \quad \cot\theta = \frac{4}{3}, \quad \csc\theta = \frac{5}{3}$$

Compute Numerator and Denominator:

$$\text{Numerator} = \tan\theta + \sec\theta = \frac{3}{4} + \frac{5}{4} = \frac{8}{4} = 2$$ $$\text{Denominator} = \cot\theta + \csc\theta = \frac{4}{3} + \frac{5}{3} = \frac{9}{3} = 3$$ $$\mathbf{\text{Value} = \frac{2}{3}}$$
Problem 2: Prove the identity: $\frac{\cos\theta}{1 - \sin\theta} = \sec\theta + \tan\theta$.

Proof:

Start with the Left-Hand Side (LHS) and multiply numerator and denominator by $(1 + \sin\theta)$:

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

Using the Pythagorean identity $1 - \sin^2\theta = \cos^2\theta$:

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

Split the fraction:

$$= \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta} = \mathbf{\sec\theta + \tan\theta} = \text{RHS} \quad \blacksquare$$

Key Formulas, Identities & Theorems

Pythagorean Identity 1
$$\sin^2\theta + \cos^2\theta = 1$$
Valid for all acute angles.
Pythagorean Identity 2
$$\sec^2\theta - \tan^2\theta = 1$$
Reciprocal relation: sec θ - tan θ = 1/(sec θ + tan θ).
Pythagorean Identity 3
$$\csc^2\theta - \cot^2\theta = 1$$
Relates cosecant and cotangent.
Quotient Identities
$$\tan\theta = \frac{\sin\theta}{\cos\theta}, \quad \cot\theta = \frac{\cos\theta}{\sin\theta}$$
Connects tangent and cotangent to sine and cosine.

Mathematics: Right Triangle Trigonometric Ratios & Unit Circle Overview

Trigonometry: Right Triangle Ratios & Pythagorean Identities Right Triangle Ratios (sin, cos, tan) A B (90°) C θ Perpendicular P Base B (Adjacent) Hypotenuse H sinθ = P/H  |  cosθ = B/H  |  tanθ = P/B Mnemonic: "Some People Have, Curly Brown Hair..." The Three Pythagorean Identities 1. sin²θ + cos²θ = 1 ⇒ sin²θ = 1 - cos²θ  |  cos²θ = 1 - sin²θ 2. 1 + tan²θ = sec²θ ⇒ sec²θ - tan²θ = 1  |  (secθ - tanθ)(secθ + tanθ) = 1 3. 1 + cot²θ = csc²θ ⇒ csc²θ - cot²θ = 1  |  (cscθ - cotθ)(cscθ + cotθ) = 1 Reciprocal & Quotient Rules: tanθ = sinθ/cosθ

Chapter Summary & 10 Key Takeaways

Takeaway 1
Trigonometry studies the quantitative relationships between side lengths and angles in triangles.
Takeaway 2
Primary ratios: sin θ = P/H, cos θ = B/H, tan θ = P/B.
Takeaway 3
Reciprocal ratios: csc θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ.
Takeaway 4
Quotient identities: tan θ = sin θ / cos θ and cot θ = cos θ / sin θ.
Takeaway 5
Standard values: sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2, sin 90° = 1.
Takeaway 6
Standard values: cos 30° = √3/2, cos 45° = 1/√2, cos 60° = 1/2, cos 90° = 0.
Takeaway 7
Standard values: tan 30° = 1/√3, tan 45° = 1, tan 60° = √3.
Takeaway 8
Pythagorean identity 1: sin^2 θ + cos^2 θ = 1.
Takeaway 9
Pythagorean identity 2: sec^2 θ - tan^2 θ = 1.
Takeaway 10
Pythagorean identity 3: csc^2 θ - cot^2 θ = 1.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
If $\tan\theta = \frac{4}{3}$, find the value of $\frac{1 - \sin\theta}{1 + \sin\theta}$.
Reveal Answer & Explanation
Answer: • Given $\tan\theta = \frac{P}{B} = \frac{4}{3}$.
• Hypotenuse $H = \sqrt{P^2 + B^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
• Therefore, $\sin\theta = \frac{P}{H} = \frac{4}{5}$.
• Evaluate the expression:
$$\frac{1 - \sin\theta}{1 + \sin\theta} = \frac{1 - \frac{4}{5}}{1 + \frac{4}{5}} = \frac{\frac{1}{5}}{\frac{9}{5}} = \mathbf{\frac{1}{9}}$$
Find H = 5. sin θ = 4/5. (1 - 4/5)/(1 + 4/5) = (1/5)/(9/5) = 1/9.
2
Evaluate without using trigonometric tables: $\frac{\sin 30^\circ + \tan 45^\circ - \csc 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$.
Reveal Answer & Explanation
Answer:

• Substitute exact standard angle values:
- $\sin 30^\circ = \frac{1}{2}, \quad \tan 45^\circ = 1, \quad \csc 60^\circ = \frac{2}{\sqrt{3}}$
- $\sec 30^\circ = \frac{2}{\sqrt{3}}, \quad \cos 60^\circ = \frac{1}{2}, \quad \cot 45^\circ = 1$
• Numerator:

$$\frac{1}{2} + 1 - \frac{2}{\sqrt{3}} = \frac{3}{2} - \frac{2}{\sqrt{3}} = \frac{3\sqrt{3} - 4}{2\sqrt{3}}$$


• Denominator:

$$\frac{2}{\sqrt{3}} + \frac{1}{2} + 1 = \frac{2}{\sqrt{3}} + \frac{3}{2} = \frac{4 + 3\sqrt{3}}{2\sqrt{3}} = \frac{3\sqrt{3} + 4}{2\sqrt{3}}$$


• Ratio:

$$\frac{3\sqrt{3} - 4}{3\sqrt{3} + 4}$$


• Rationalise the denominator by multiplying by $(3\sqrt{3} - 4)$:

$$\frac{(3\sqrt{3} - 4)^2}{(3\sqrt{3})^2 - 4^2} = \frac{27 + 16 - 24\sqrt{3}}{27 - 16} = \mathbf{\frac{43 - 24\sqrt{3}}{11}}$$


Substitute standard values. Numerator = (3√3 - 4)/(2√3), Denominator = (3√3 + 4)/(2√3). Rationalise.
3
Prove that: $(\sin\theta + \csc\theta)^2 + (\cos\theta + \sec\theta)^2 = 7 + \tan^2\theta + \cot^2\theta$.
Reveal Answer & Explanation
Answer: • Expand both binomial squares:
$$\text{LHS} = (\sin^2\theta + \csc^2\theta + 2\sin\theta\csc\theta) + (\cos^2\theta + \sec^2\theta + 2\cos\theta\sec\theta)$$
• Use reciprocal products $\sin\theta\csc\theta = 1$ and $\cos\theta\sec\theta = 1$:
$$= \sin^2\theta + \csc^2\theta + 2(1) + \cos^2\theta + \sec^2\theta + 2(1)$$
• Group $\sin^2\theta + \cos^2\theta = 1$ and add the constants ($2 + 2 = 4$):
$$= (\sin^2\theta + \cos^2\theta) + 4 + \sec^2\theta + \csc^2\theta$$
$$= 1 + 4 + \sec^2\theta + \csc^2\theta = 5 + \sec^2\theta + \csc^2\theta$$
• Substitute $\sec^2\theta = 1 + \tan^2\theta$ and $\csc^2\theta = 1 + \cot^2\theta$:
$$= 5 + (1 + \tan^2\theta) + (1 + \cot^2\theta) = \mathbf{7 + \tan^2\theta + \cot^2\theta} = \text{RHS} \quad \blacksquare$$
Expand squares. sin^2 + cos^2 = 1, cross terms give 2 + 2 = 4. Replace sec^2 with 1 + tan^2 and csc^2 with 1 + cot^2.
4
If $\sec\theta + \tan\theta = p$, prove that $\sin\theta = \frac{p^2 - 1}{p^2 + 1}$.
Reveal Answer & Explanation
Answer: • We know the identity: $\sec^2\theta - \tan^2\theta = 1$.
$$(\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1$$
• Substitute $\sec\theta + \tan\theta = p$:
$$\sec\theta - \tan\theta = \frac{1}{p} \quad \text{--- (2)}$$
• Adding (1) and (2):
$$2\sec\theta = p + \frac{1}{p} = \frac{p^2 + 1}{p} \implies \sec\theta = \frac{p^2 + 1}{2p} \implies \cos\theta = \frac{2p}{p^2 + 1}$$
• Subtracting (2) from (1):
$$2\tan\theta = p - \frac{1}{p} = \frac{p^2 - 1}{p} \implies \tan\theta = \frac{p^2 - 1}{2p}$$
• Now compute $\sin\theta = \tan\theta \times \cos\theta$:
$$\sin\theta = \left(\frac{p^2 - 1}{2p}\right) \times \left(\frac{2p}{p^2 + 1}\right) = \mathbf{\frac{p^2 - 1}{p^2 + 1}} \quad \blacksquare$$
Use sec θ - tan θ = 1/p. Add and subtract to find sec θ and tan θ, then sin θ = tan θ * cos θ.
5
Prove that: $\sqrt{\frac{1 + \cos\theta}{1 - \cos\theta}} = \csc\theta + \cot\theta$.
Reveal Answer & Explanation
Answer: • Rationalise the expression under the square root by multiplying by $(1 + \cos\theta)$:
$$\text{LHS} = \sqrt{\frac{(1 + \cos\theta)(1 + \cos\theta)}{(1 - \cos\theta)(1 + \cos\theta)}} = \sqrt{\frac{(1 + \cos\theta)^2}{1 - \cos^2\theta}}$$
• Use $1 - \cos^2\theta = \sin^2\theta$:
$$= \sqrt{\frac{(1 + \cos\theta)^2}{\sin^2\theta}} = \frac{1 + \cos\theta}{\sin\theta}$$
• Split the fraction:
$$= \frac{1}{\sin\theta} + \frac{\cos\theta}{\sin\theta} = \mathbf{\csc\theta + \cot\theta} = \text{RHS} \quad \blacksquare$$
Multiply numerator and denominator under the root by (1 + cos θ) to get √((1 + cos θ)^2 / sin^2 θ).
6
Find the value of $\theta$ ($0^\circ \le \theta \le 90^\circ$) if $2\cos 3\theta = 1$.
Reveal Answer & Explanation
Answer: • Isolate $\cos 3\theta$:
$$\cos 3\theta = \frac{1}{2}$$
• We know that $\cos 60^\circ = \frac{1}{2}$:
$$3\theta = 60^\circ \implies \theta = \frac{60^\circ}{3} = \mathbf{20^\circ}$$
cos 3θ = 1/2 = cos 60°. 3θ = 60° -> θ = 20°.
7
Can $\sin\theta$ or $\cos\theta$ ever be greater than $1$ for any real angle? Explain.
Reveal Answer & Explanation
Answer:

• No. In any right-angled triangle, the hypotenuse $H$ is strictly the longest side ($H > P$ and $H > B$).
• Since $\sin\theta = \frac{P}{H}$ and $\cos\theta = \frac{B}{H}$, the numerator is always less than or equal to the denominator.
• Therefore, for all real angles, $\sin\theta \le 1$ and $\cos\theta \le 1$. (Their values strictly lie in the closed interval $[-1, 1]$).


No, because the hypotenuse is always the longest side, so P/H and B/H can never exceed 1.
8
Prove that: $(1 + \cot\theta - \csc\theta)(1 + \tan\theta + \sec\theta) = 2$.
Reveal Answer & Explanation
Answer: • Convert all ratios to $\sin\theta$ and $\cos\theta$:
$$1 + \frac{\cos\theta}{\sin\theta} - \frac{1}{\sin\theta} = \frac{\sin\theta + \cos\theta - 1}{\sin\theta}$$
$$1 + \frac{\sin\theta}{\cos\theta} + \frac{1}{\cos\theta} = \frac{\cos\theta + \sin\theta + 1}{\cos\theta}$$
• Multiply both fractions:
$$\text{Numerator} = [(\sin\theta + \cos\theta) - 1][(\sin\theta + \cos\theta) + 1] = (\sin\theta + \cos\theta)^2 - 1^2$$
$$= (\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta) - 1 = 1 + 2\sin\theta\cos\theta - 1 = 2\sin\theta\cos\theta$$
• Divide by denominator $(\sin\theta\cos\theta)$:
$$\frac{2\sin\theta\cos\theta}{\sin\theta\cos\theta} = \mathbf{2} = \text{RHS} \quad \blacksquare$$
Convert to sin and cos. Numerator is ((sin+cos) - 1)((sin+cos) + 1) = (sin+cos)^2 - 1 = 2sin*cos.
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