Chapter 8: Introduction to Trigonometry
Overview
Revise trigonometric ratios (\(\sin, \cos, \tan, \csc, \sec, \cot\)), right-triangle definitions, values at specific angles, and proofs of fundamental identities like \(\sin^2\theta + \cos^2\theta = 1\) for CBSE Board exams.
T-Ratios • Trig Table • Identities • Proving Questions
Exam Weightage & Blueprint
Total: 8-10 MarksThis chapter falls under Unit V: Trigonometry (12 marks total) in the 2026–27 syllabus. Focus areas include ratios of acute angles, the 30°, 45° and 60° values, and proof and applications of $\sin^2 A + \cos^2 A = 1$.
| 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 |
CBSE 2026–27 Syllabus Status: Introduction to Trigonometry
Unit V: Trigonometry • 7–8 MarksThis chapter is strictly structured according to the CBSE Rationalised Mathematics Syllabus. Focus your board preparation only on prescribed topics and avoid studying deleted sections:
In 2026–27 Board Syllabus:
Deleted / Not in Board Exam:
⏰ 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: $1 + \cot^2θ = cosec^2θ$.
📐 Trigonometric Ratios
Right-angled triangle showing Perpendicular (P), Base (B), and Hypotenuse (H)
Mnemonics
sin A
$\frac{Perpendicular}{Hypotenuse}$
cos A
$\frac{Base}{Hypotenuse}$
tan A
$\frac{Perpendicular}{Base}$
📐 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 |
Trigonometric Identities & Quotient Relations
An equation involving trigonometric ratios of an angle is called a trigonometric identity if it is true for all permissible values of the angle.
Identity 1
$$ \sin^2 A + \cos^2 A = 1 $$
$\sin^2 A = 1 - \cos^2 A$
$\cos^2 A = 1 - \sin^2 A$
Identity 2
$$ 1 + \tan^2 A = \sec^2 A $$
$\sec^2 A - \tan^2 A = 1$
$(\sec A - \tan A)(\sec A + \tan A) = 1$
Identity 3
$$ 1 + \cot^2 A = cosec^2 A $$
$cosec^2 A - \cot^2 A = 1$
$(cosec A - \cot A)(cosec A + \cot A) = 1$
If $\sec A + \tan A = p$, then $\sec A - \tan A = \frac{1}{p}$. Adding both gives $2\sec A = p + \frac{1}{p} \implies \sec A = \frac{p^2 + 1}{2p}$.
Similarly, if $cosec A + \cot A = q$, then $cosec A - \cot A = \frac{1}{q}$.
- Range & Limits of T-Ratios ($0^\circ \le \theta \le 90^\circ$):
• $0 \le \sin\theta \le 1$ and $0 \le \cos\theta \le 1$ (Hypotenuse is always the longest side).
• $\sec\theta \ge 1$ and $\csc\theta \ge 1$.
• As $\theta$ increases from $0^\circ \to 90^\circ$: $\sin\theta$ and $\tan\theta$ strictly increase, while $\cos\theta$ strictly decreases. - High-Frequency Algebraic Identity: $$\sin^4\theta - \cos^4\theta = (\sin^2\theta - \cos^2\theta)(\sin^2\theta + \cos^2\theta) = \sin^2\theta - \cos^2\theta = 2\sin^2\theta - 1 = 1 - 2\cos^2\theta$$
- Independence of Triangle Size: The values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same.
Solved Examples (Board Marking Scheme)
Q1. Given $\tan A = \frac{4}{3}$, find other trigonometric ratios of $\angle A$. (3 Marks)
Let $\triangle ABC$ be right-angled at B. $\tan A = \frac{BC}{AB} = \frac{4k}{3k}$.
By Pythagoras Theorem: $AC^2 = (3k)^2 + (4k)^2 = 9k^2 + 16k^2 = 25k^2$.
$AC = 5k$.
$\sin A = \frac{BC}{AC} = \frac{4k}{5k} = \frac{4}{5}$.
$\cos A = \frac{AB}{AC} = \frac{3k}{5k} = \frac{3}{5}$.
$\cot A = \frac{3}{4}, \sec A = \frac{5}{3}, cosec A = \frac{5}{4}$.
Q2. Prove that $$\frac{\sin θ - 2\sin^3 θ}{2\cos^3 θ - \cos θ}$ = \tan θ$. (3 Marks)
$LHS = \frac{\sin \theta(1 - 2\sin^2 \theta)}{\cos \theta(2\cos^2 \theta - 1)}$
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 θ(1 - 2 + 2\cos^2 θ)}{\cos θ(2\cos^2 θ - 1)} = \frac{\sin θ(2\cos^2 θ - 1)}{\cos θ(2\cos^2 θ - 1)}$
$= \frac{\sin \theta}{\cos \theta} = \tan \theta = RHS$. Hence Proved.
Previous Year Questions (PYQs)
Ans: $($\frac{$\sqrt{3}$}{2}$)($\frac{$\sqrt{3}$}{2}$) + ($\frac{1}{2}$)($\frac{1}{2}$) = $\frac{3}{4}$ + $\frac{1}{4}$ = 1$.
Ans: $A+B = 60^\circ, A-B = 30^\circ$. Solving gives $A=45°, B=15°$.
Ans: Take LCM $→ \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
💡 Scoring Tips
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 = \frac{3}{4}$, calculate $\cos A$ and $\tan A$.
Q3 (3M): If $\sec\theta + \tan\theta = p$, express $\sin\theta$ in terms of $p$.
Q4 (4M): Prove that: $(\sin A + cosec A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$.
📝 More Solved Board Questions
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°
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\thetacosec \theta + 1$.
LHS = RHS. Proved.
Sol. Rationalising the denominator inside the square root:
$$LHS = \sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 - \sin A)(1 + \sin A)}} = \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}}$$
Using $1 - \sin^2 A = \cos^2 A$:
$$= \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} = \frac{1 + \sin A}{\cos A} = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A = RHS \quad [Hence Proved]$$
Case Study: Barrier-Free Hospital Access Ramp Design
Scenario: A modern medical center is installing a wheelchair ramp at its primary entrance to comply with disability accessibility guidelines. The entrance floor is elevated 1.5 meters above ground level. Architectural safety guidelines stipulate that the angle of inclination of the ramp must not exceed 30° for safety.
Q1: If the contractor installs the ramp at exactly an inclination of 30°, write the trigonometric ratio connecting ramp length and vertical height. 1 Mark
Q2: Calculate the total length of the ramp required. 1 Mark
Q3: Find the horizontal distance along the ground from the base of the ramp to the building wall. (OR) If space constraints force the ground span to be 3√3 meters, find the resulting ramp inclination angle and check whether it complies with the 30° safety limit. 2 Marks
📋 Board Revision Checklist
- ✅ Ratios: $\sin = P/H, \cos = B/H, \tan = P/B$
- ✅ Reciprocals: $cosec = 1/\sin, \sec = 1/\cos, \cot = 1/\tan$
- ✅ 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: $1 + \cot^2\theta = cosec^2\theta$
- ✅ Strategy: If stuck in proof, convert everything to $\sin$ and $\cos$
- ✅ Deleted: $\sin(90-A) = \cos A$ and related complementary angle problems
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 $2\sin A\cos A = \dots$
Frequently Asked Questions: Class 10 Maths Chapter 8
What is the marks weightage of Chapter 8 in Class 10 Maths?
Introduction to Trigonometry contributes around 6 to 8 marks (part of Unit 5: Trigonometry, total 12 marks), featuring 1-mark ratio MCQs and 3-mark or 4-mark identity proofs.
Which trigonometric identities are in the syllabus?
The fundamental identity sin^2(theta) + cos^2(theta) = 1 is central to the syllabus along with 1 + tan^2(theta) = sec^2(theta) and 1 + cot^2(theta) = cosec^2(theta). Ratios of complementary angles have been removed.
What are the trigonometric values for standard angles?
Standard angle values must be memorized for 0, 30, 45, 60, and 90 degrees, especially sin 30 = 1/2, sin 45 = 1/sqrt(2), cos 60 = 1/2, and tan 45 = 1.
What is the best strategy to prove trigonometric identities?
Start with the more complicated side (usually LHS), convert all terms into sine and cosine (tan = sin/cos, sec = 1/cos), find common denominators, and apply standard Pythagorean identities.