Chapter 8: Introduction to Trigonometry - Board Exam Notes

Exam Weightage & Blueprint

Total: 8-10 Marks

This chapter is part of the Trigonometry Unit (12 Marks Total). Focus on ratios, specific angles, and proving identities.

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 $\infty$.

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.

Self-Assessment Mock Test (10 Marks)

Q1 (1M): Evaluate $2\tan^2 45^{\circ} + \cos^2 30^{\circ} - \sin^2 60^{\circ}$.

Q2 (2M): If $\sin A = 3/4$, calculate $\cos A$ and $\tan A$.

Q3 (3M): Prove that $\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$.

Q4 (4M): Evaluate $\frac{5\cos^2 60^{\circ} + 4\sec^2 30^{\circ} - \tan^2 45^{\circ}}{\sin^2 30^{\circ} + \cos^2 30^{\circ}}$.