Chapter 8: Introduction to Trigonometry

Overview

This page provides comprehensive Chapter 8: Introduction to Trigonometry – Board Exam Notes aligned with the latest CBSE 2025–26 syllabus. Covers trigonometric ratios, the trigonometric table (0°–90°), and applications of the fundamental identity $\sin^2 A + \cos^2 A = 1$.

T-Ratios • Trig Table • Identities • Proving Questions

Exam Weightage & Blueprint

Total: 8-10 Marks

This chapter falls under Unit V: Trigonometry (12 marks total). As per the latest syllabus, focus is on: Trigonometric ratios of an acute angle, values of the trigonometric ratios of 30°, 45° and 60°, and proof and applications of the identity $\sin^2 A + \cos^2 A = 1$.

Question Type	Marks	Frequency	Focus Topic
MCQ	1	High	Values of specific angles, Basic Identities
Short Answer	2 or 3	Medium	Evaluation problems, Finding ratios
Long Answer	4 or 5	High	Proving Trigonometric Identities

⏰ Last 24-Hour Checklist

Trigonometric Ratios: PBP/HHB (Pandit Badri Prasad...)
Table of Values: 0-, 30-, 45-, 60-, 90-.
Pythagoras Theorem: $H^2 = P^2 + B^2$.

Identity 1: $\sin^2\theta + \cos^2\theta = 1$.
Identity 2: $1 + \tan^2\theta = \sec^2\theta$.
Identity 3: $1 + \cot^2\theta = \text{cosec}^2\theta$.

📐 Trigonometric Ratios

Trigonometric Ratios: Ratios of sides of a right-angled triangle with respect to its acute angles.

Mnemonics

sin A

$\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

cos A

$\frac{\text{Base}}{\text{Hypotenuse}}$

tan A

$\frac{\text{Perpendicular}}{\text{Base}}$

Reciprocal Relations: $\text{cosec } A = \frac{1}{\sin A}, \quad \sec A = \frac{1}{\cos A}, \quad \cot A = \frac{1}{\tan A}$

📐 Trigonometric Table

Memorize this table. It's used in 30-40% of questions.

$\angle A$	$0^{\circ}$	$30^{\circ}$	$45^{\circ}$	$60^{\circ}$	$90^{\circ}$
sin A	0	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	1
cos A	1	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	0
tan A	0	$\frac{1}{\sqrt{3}}$	1	$\sqrt{3}$	ND

⚠️ Common Mistake: $\tan 90^{\circ}$ is Not Defined (ND). Don't write it as 0 or $\infty$.

$\tan(90^\circ - A) = \cot A$
$\cot(90^\circ - A) = \tan A$

$\sec(90^\circ - A) = \text{cosec } A$
$\text{cosec}(90^\circ - A) = \sec A$

Trigonometric Identities

An equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angle.

Identity 1

$$ \sin^2 A + \cos^2 A = 1 $$

Derived forms:

$$ \sin^2 A = 1 - \cos^2 A $$ $$ \cos^2 A = 1 - \sin^2 A $$

Identity 2 & 3

$$ 1 + \tan^2 A = \sec^2 A $$ $$ 1 + \cot^2 A = \text{cosec}^2 A $$

Useful for proving questions.

Solved Examples (Board Marking Scheme)

Q1. Given $\tan A = \frac{4}{3}$, find other trigonometric ratios of $\angle A$. (3 Marks)

Step 1: Draw Triangle 0.5 Mark

Let $\triangle ABC$ be right-angled at B. $\tan A = \frac{BC}{AB} = \frac{4k}{3k}$.

By Pythagoras Theorem: $AC^2 = (3k)^2 + (4k)^2 = 9k^2 + 16k^2 = 25k^2$.

$AC = 5k$.

Step 2: Calculate Ratios 2.5 Marks

$\sin A = \frac{BC}{AC} = \frac{4k}{5k} = \frac{4}{5}$.

$\cos A = \frac{AB}{AC} = \frac{3k}{5k} = \frac{3}{5}$.

$\cot A = \frac{3}{4}, \sec A = \frac{5}{3}, \text{cosec } A = \frac{5}{4}$.

Q2. Prove that $\frac{\sin \theta - 2\sin^3 \theta}{2\cos^3 \theta - \cos \theta} = \tan \theta$. (3 Marks)

Step 1: Simplify LHS 1 Mark

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

Step 2: Use Identity 1 Mark

Substitute $\sin^2 \theta = 1 - \cos^2 \theta$ in numerator.

$= \frac{\sin \theta[1 - 2(1 - \cos^2 \theta)]}{\cos \theta(2\cos^2 \theta - 1)}$

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

Step 3: Conclusion 1 Mark

$= \frac{\sin \theta}{\cos \theta} = \tan \theta = RHS$. Hence Proved.

Previous Year Questions (PYQs)

2023 (1 Mark): Value of $\sin 60^{\circ} \cos 30^{\circ} + \sin 30^{\circ} \cos 60^{\circ}$ is?
Ans: $(\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2}) + (\frac{1}{2})(\frac{1}{2}) = \frac{3}{4} + \frac{1}{4} = 1$.

2020 (3 Marks): If $\tan (A+B) = \sqrt{3}$ and $\tan (A-B) = \frac{1}{\sqrt{3}}$, find A and B.
Ans: $A+B = 60^{\circ}, A-B = 30^{\circ}$. Solving gives $A=45^{\circ}, B=15^{\circ}$.

2019 (4 Marks): Prove $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A$.
Ans: Take LCM $\rightarrow \frac{\cos^2 A + (1+\sin A)^2}{\cos A(1+\sin A)}$. Expand $(1+\sin A)^2$, simplify using $\sin^2 A + \cos^2 A = 1$.

Exam Strategy & Mistake Bank

⚠️ Mistake Bank

Notation Error: Writing $\sin A$ as $\sin \times A$. This is incorrect. $\sin$ has no meaning without angle.

Squaring: $(\sin A + \cos A)^2 \neq \sin^2 A + \cos^2 A$. It is $1 + 2\sin A \cos A$.

Reciprocals: Confusing $\sin^{-1} A$ with $(\sin A)^{-1} = \text{cosec } A$.

💡 Scoring Tips

Convert to Sin/Cos: In proving identities, if stuck, convert all terms ($\tan, \sec, \cot$) into $\sin$ and $\cos$.

Rationalize: For terms like $\frac{1}{1 - \sin A}$, multiply numerator and denominator by $(1 + \sin A)$.

Values: Memorize values of $\tan 30^{\circ}, \tan 45^{\circ}, \tan 60^{\circ}$ thoroughly.

📝 More Solved Board Questions

Q3. If $\sin(A-B) = 1/2, \cos(A+B) = 1/2$, $0^\circ < A+B \leq 90^\circ$, $A > B$, find A and B. 3 Marks

Sol. We know $\sin 30^\circ = 1/2 \Rightarrow A-B = 30^\circ$ ...(1)

And $\cos 60^\circ = 1/2 \Rightarrow A+B = 60^\circ$ ...(2)

Adding (1) and (2): $2A = 90^\circ \Rightarrow A = 45^\circ$.

Substituting in (2): $45^\circ + B = 60^\circ \Rightarrow B = 15^\circ$.

Answer: A = 45°, B = 15°

Q4. Prove that $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \text{cosec } \theta$. 5 Marks

Sol. Convert to $\sin$ and $\cos$:

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

$= \frac{\sin^3 - \cos^3}{\sin\cos(\sin-\cos)} = \frac{(\sin-\cos)(\sin^2+\cos^2+\sin\cos)}{\sin\cos(\sin-\cos)}$

$= \frac{1 + \sin\cos}{\sin\cos} = \frac{1}{\sin\cos} + 1 = \sec\theta\text{cosec }\theta + 1$.

LHS = RHS. Proved.

🎯 Board Pattern (2018–2025): The proving questions from Exercise 8.3 (Identity Questions) are almost guaranteed in Section D (5 Marks) or Section C (3 Marks). Practice all 10 parts of Question 4 in the NCERT textbook multiple times.

📋 Board Revision Checklist

✅ Ratios: $\sin = P/H, \cos = B/H, \tan = P/B$
✅ Reciprocals: $\text{cosec} = 1/\sin, \sec = 1/\cos, \cot = 1/\tan$
✅ Important Values: $\sin 30^\circ = 1/2, \tan 45^\circ = 1, \sin 60^\circ = \sqrt{3}/2$
✅ Identity 1: $\sin^2\theta + \cos^2\theta = 1$ (Crucial!)
✅ Identity 2: $1 + \tan^2\theta = \sec^2\theta$
✅ Identity 3: $1 + \cot^2\theta = \text{cosec}^2\theta$
✅ Strategy: If stuck in proof, convert everything to $\sin$ and $\cos$
✅ Deleted: $\sin(90-A) = \cos A$ and related complementary angle problems

💡 Exam Tip:
Always write the angle $\theta$ or $A$. Writing just "sin" or "cos" is mathematically incorrect and may lead to mark deductions.

Concept Mastery Quiz 🎯

Test your readiness for the board exam.

1. The value of $\sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ$ is:

2. If $9\sec^2 A - 9\tan^2 A = \dots$

3. The value of $(\sin^2 30^\circ + \cos^2 30^\circ)$ is:

4. Which of the following is equal to $\sec A$?

5. In a right triangle ABC, right-angled at B, if $\tan A = 1$, then $2\sin A\cos A = \dots$