Ch 8  ·  Q–
0%
Class 10 Mathematics Exercise 8.1 NCERT Solutions Olympiad Board Exam

Chapter 8 — Introduction to Trigonometry

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

📋11 questions
⏱Ideal time: 45-50 min
📍Now at: Q1
Begin Q1 → ↑ Jump to progress
Q1
NUMERIC3 marks

In \(\triangle ABC\), right-angled at B, AB = 24 cm, BC = 7 cm. Determine :
(i) sin A, cos A
(ii) sin C, cos C

\(\tiny\begin{aligned} P&\Rightarrow \text{Perpendicular}\\ B&\Rightarrow \text{Base}\\ H&\Rightarrow \text{Hypotenuse} \end{aligned}\)
B C A 7 cm 24 cm 25 cm
Right Triangle ABC
Concept Used

In a right-angled triangle:

\[ \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}}, \quad \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} \]

Also, by Pythagoras theorem:

\[ (\text{Hypotenuse})^2 = (\text{Base})^2 + (\text{Perpendicular})^2 \]

Solution Roadmap
  • Step 1: Identify given sides
  • Step 2: Find hypotenuse using Pythagoras theorem
  • Step 3: Use trigonometric ratios for angle A
  • Step 4: Use trigonometric ratios for angle C
Solution

Given:
AB = 24 cm, BC = 7 cm

Step 1: Find Hypotenuse AC

$$\begin{aligned} AC^{2} &= AB^{2} + BC^{2} \\ &= 24^{2} + 7^{2} \\ &= 576 + 49 \\ &= 625 \end{aligned}$$ $$\begin{aligned} AC &= \sqrt{625} \\ &= 25 \text{ cm} \end{aligned}$$

Step 2: Find sin A and cos A

For angle A:
Perpendicular = BC = 7
Base = AB = 24
Hypotenuse = AC = 25

$$\begin{aligned} \sin A &= \frac{BC}{AC} \\ &= \frac{7}{25} \end{aligned}$$ $$\begin{aligned} \cos A &= \frac{AB}{AC} \\ &= \frac{24}{25} \end{aligned}$$

Step 3: Find sin C and cos C

For angle C:
Perpendicular = AB = 24
Base = BC = 7
Hypotenuse = AC = 25

$$\begin{aligned} \sin C &= \frac{AB}{AC} \\ &= \frac{24}{25} \end{aligned}$$ $$\begin{aligned} \cos C &= \frac{BC}{AC} \\ &= \frac{7}{25} \end{aligned}$$

Final Answer:

\(\sin A = \frac{7}{25}, \quad \cos A = \frac{24}{25}\)
\(\sin C = \frac{24}{25}, \quad \cos C = \frac{7}{25}\)

Exam Significance
  • Very common 2–3 mark question in board exams
  • Tests understanding of side identification
  • Forms base for higher trigonometric identities
  • Important for competitive exams like JEE foundation and NDA
↑ Top
1 / 11  ·  9%
Q2 →
Q2
NUMERIC3 marks

In Fig. 8.13, find \( \tan P - \cot R \).

Q R P 12 13 5
Triangle PQR (Right-angled at Q)
Concept Used

In a right-angled triangle:

\[ \begin{aligned} \tan \theta = \frac{\text{Perpendicular}}{\text{Base}}, \\\\ \cot \theta = \frac{\text{Base}}{\text{Perpendicular}} \end{aligned} \]

Also,

\[ (\text{Hypotenuse})^2 = (\text{Base})^2 + (\text{Perpendicular})^2 \]

Solution Roadmap
  • Step 1: Identify given sides
  • Step 2: Find missing side using Pythagoras theorem
  • Step 3: Compute tan P
  • Step 4: Compute cot R
  • Step 5: Subtract values
Solution

Given:
PQ = 12, PR = 13

Step 1: Find QR using Pythagoras theorem

$$\begin{aligned} QR^{2} &= PR^{2} - PQ^{2} \\ &= 13^{2} - 12^{2} \\ &= 169 - 144 \\ &= 25 \end{aligned}$$ $$\begin{aligned} QR &= \sqrt{25} \\ &= 5 \end{aligned}$$

Step 2: Find \( \tan P \)

For angle P:
Perpendicular = QR = 5
Base = PQ = 12

$$\begin{aligned} \tan P &= \frac{QR}{PQ} \\ &= \frac{5}{12} \end{aligned}$$

Step 3: Find \( \cot R \)

For angle R:
Perpendicular = PQ = 12
Base = QR = 5

$$\begin{aligned} \cot R &= \frac{\text{Base}}{\text{Perpendicular}} \\ &= \frac{QR}{PQ} \\ &= \frac{5}{12} \end{aligned}$$

Step 4: Compute required expression

$$\begin{aligned} \tan P - \cot R &= \frac{5}{12} - \frac{5}{12} \\ &= 0 \end{aligned}$$

Final Answer: \(0\)

Exam Significance
  • Tests understanding of complementary angles in right triangle
  • Common conceptual MCQ in competitive exams
  • Important identity insight: \( \tan \theta = \cot (90^\circ - \theta) \)
  • Frequently appears as simplification-based 2 mark question
← Q1
2 / 11  ·  18%
Q3 →
Q3
NUMERIC3 marks

If \( \sin A = \frac{3}{4} \), calculate \( \cos A \) and \( \tan A \).

B C A 3k √7 k 4k
Right Triangle Representation
Concept Used

Trigonometric ratios represent ratios of sides in a right triangle.

\[ \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \]

When a ratio is given, we assume sides proportional using a constant \(k\).

Solution Roadmap
  • Step 1: Express sides using ratio and constant \(k\)
  • Step 2: Find base using Pythagoras theorem
  • Step 3: Calculate cos A
  • Step 4: Calculate tan A
  • Step 5: Rationalize answer (important for exams)
Solution

Step 1: Interpret given ratio

$$\begin{aligned} \sin A &= \frac{3}{4} \\ \sin A &= \frac{P}{H} \end{aligned}$$

Comparing:

$$\begin{aligned} P &= 3k \\ H &= 4k \end{aligned}$$

Step 2: Find Base using Pythagoras theorem

$$\begin{aligned} B^{2} &= H^{2} - P^{2} \\ &= (4k)^{2} - (3k)^{2} \\ &= 16k^{2} - 9k^{2} \\ &= 7k^{2} \end{aligned}$$ $$\begin{aligned} B &= \sqrt{7k^{2}} \\ &= k\sqrt{7} \end{aligned}$$

Step 3: Calculate \( \cos A \)

$$\begin{aligned} \cos A &= \frac{B}{H} \\ &= \frac{k\sqrt{7}}{4k} \\ &= \frac{\sqrt{7}}{4} \end{aligned}$$

Step 4: Calculate \( \tan A \)

$$\begin{aligned} \tan A &= \frac{P}{B} \\ &= \frac{3k}{k\sqrt{7}} \\ &= \frac{3}{\sqrt{7}} \end{aligned}$$

Step 5: Rationalize \( \tan A \)

$$\begin{aligned} \tan A &= \frac{3}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} \\ &= \frac{3\sqrt{7}}{7} \end{aligned}$$

Final Answer:

\(\cos A = \frac{\sqrt{7}}{4}\)
\(\tan A = \frac{3\sqrt{7}}{7}\)

Exam Significance
  • Very common board question based on ratio method
  • Tests conceptual clarity of side relationships
  • Rationalization is frequently required in answers
  • Foundation for trigonometric identities and heights & distances
← Q2
3 / 11  ·  27%
Q4 →
Q4
NUMERIC3 marks

Given \(15 \cot A = 8\), find \( \sin A \) and \( \sec A \).

B C A 15k 8k 17k
Right Triangle Representation
Concept Used

\[ \begin{aligned} \cot A &= \frac{\text{Base}}{\text{Perpendicular}}, \\\\ \sin A &= \frac{\text{Perpendicular}}{\text{Hypotenuse}}, \\\\ \sec A &= \frac{\text{Hypotenuse}}{\text{Base}} \end{aligned} \]

Ratio method is used by assuming sides proportional to a constant \(k\).

Solution Roadmap
  • Step 1: Simplify given equation
  • Step 2: Express sides using ratio method
  • Step 3: Find hypotenuse using Pythagoras theorem
  • Step 4: Calculate required ratios
Solution

Step 1: Simplify given equation

$$\begin{aligned} 15 \cot A &= 8 \\ \cot A &= \frac{8}{15} \end{aligned}$$

Step 2: Interpret ratio

$$\begin{aligned} \cot A &= \frac{B}{P} = \frac{8}{15} \end{aligned}$$ $$\begin{aligned} B &= 8k \\ P &= 15k \end{aligned}$$

Step 3: Find Hypotenuse

$$\begin{aligned} H^{2} &= B^{2} + P^{2} \\ &= (8k)^{2} + (15k)^{2} \\ &= 64k^{2} + 225k^{2} \\ &= 289k^{2} \end{aligned}$$ $$\begin{aligned} H &= \sqrt{289k^{2}} \\ &= 17k \end{aligned}$$

Step 4: Find \( \sin A \)

$$\begin{aligned} \sin A &= \frac{P}{H} \\ &= \frac{15k}{17k} \\ &= \frac{15}{17} \end{aligned}$$

Step 5: Find \( \sec A \)

$$\begin{aligned} \sec A &= \frac{H}{B} \\ &= \frac{17k}{8k} \\ &= \frac{17}{8} \end{aligned}$$

Final Answer:

\(\sin A = \frac{15}{17}\)
\(\sec A = \frac{17}{8}\)

Exam Significance
  • Classic board question based on cot ratio conversion
  • Tests ability to convert between trigonometric ratios
  • Very important for identity-based simplifications
  • Common in NDA, SSC and foundation-level competitive exams
← Q3
4 / 11  ·  36%
Q5 →
Q5
NUMERIC3 marks

Question text...

B C A 5k 12k 13k
Right Triangle Representation

Q5. Given \( \sec \theta = \frac{13}{12} \), calculate all other trigonometric ratios.

Concept Used

\[ \sec \theta = \frac{\text{Hypotenuse}}{\text{Base}} \]

Use ratio method by assuming sides proportional to a constant \(k\), then apply Pythagoras theorem.

Solution Roadmap
  • Step 1: Express sides using ratio
  • Step 2: Find perpendicular using Pythagoras theorem
  • Step 3: Compute all trigonometric ratios
Solution

Step 1: Interpret given ratio

$$\begin{aligned} \sec \theta &= \frac{13}{12} \\ \sec \theta &= \frac{H}{B} \end{aligned}$$ $$\begin{aligned} H &= 13k \\ B &= 12k \end{aligned}$$

Step 2: Find Perpendicular

$$\begin{aligned} P^{2} &= H^{2} - B^{2} \\ &= (13k)^{2} - (12k)^{2} \\ &= 169k^{2} - 144k^{2} \\ &= 25k^{2} \end{aligned}$$ $$\begin{aligned} P &= \sqrt{25k^{2}} \\ &= 5k \end{aligned}$$

Step 3: All Trigonometric Ratios

$$\begin{aligned} \sin \theta &= \frac{P}{H} = \frac{5k}{13k} = \frac{5}{13} \\\\ \cos \theta &= \frac{B}{H} = \frac{12k}{13k} = \frac{12}{13} \\\\ \tan \theta &= \frac{P}{B} = \frac{5k}{12k} = \frac{5}{12} \\\\ \cot \theta &= \frac{B}{P} = \frac{12k}{5k} = \frac{12}{5} \\\\ \text{cosec }\theta &= \frac{H}{P} = \frac{13k}{5k} = \frac{13}{5} \end{aligned}$$

Final Answer:

\(\sin \theta = \frac{5}{13}\), \(\cos \theta = \frac{12}{13}\), \(\tan \theta = \frac{5}{12}\),
\(\cot \theta = \frac{12}{5}\), \(\text{cosec }\theta = \frac{13}{5}\)

Exam Significance
  • Very important 3–4 mark question in board exams
  • Tests full understanding of all six trigonometric ratios
  • Frequently used in identity-based simplifications
  • Highly relevant for NDA, SSC, and foundation-level competitive exams
← Q4
5 / 11  ·  45%
Q6 →
Q6
NUMERIC3 marks

If \( \angle A \) and \( \angle B \) are acute angles such that \( \cos A = \cos B \), then show that \( \angle A = \angle B \).

B C A
Equal Sides ⇒ Equal Angles
Concept Used

In a right triangle:

\[ \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} \]

Also, for acute angles:
Cosine function is one-to-one (injective), meaning equal cosine values imply equal angles.

Solution Roadmap
  • Step 1: Express cos A and cos B in triangle form
  • Step 2: Equate ratios
  • Step 3: Deduce equality of sides
  • Step 4: Apply triangle property
Solution

Step 1: Given condition

$$\begin{aligned} \cos A = \cos B \end{aligned}$$

Step 2: Express using trigonometric ratio

$$\begin{aligned} \cos A &= \frac{\text{Base}_1}{\text{Hypotenuse}} \\\\ \cos B &= \frac{\text{Base}_2}{\text{Hypotenuse}} \end{aligned}$$

Step 3: Equate ratios

$$\begin{aligned} \frac{\text{Base}_1}{\text{Hypotenuse}} = \frac{\text{Base}_2}{\text{Hypotenuse}} \end{aligned}$$ $$\begin{aligned} \Rightarrow \text{Base}_1 = \text{Base}_2 \end{aligned}$$

Step 4: Conclude using triangle property

In a triangle, angles opposite equal sides are equal.

$$\begin{aligned} \therefore \angle A = \angle B \end{aligned}$$

Key Insight:

Since A and B are acute angles and \( \cos A = \cos B \), cosine function gives a unique value. Hence, directly:
\( A = B \)

Exam Significance
  • Important reasoning-based question in board exams
  • Tests understanding of trigonometric function properties
  • Frequently used in identity proofs
  • Concept used in higher mathematics and calculus monotonicity
← Q5
6 / 11  ·  55%
Q7 →
Q7
NUMERIC3 marks

If \( \cot \theta = \frac{7}{8} \), evaluate:

(i) \( \dfrac{ (1 + \sin \theta)(1 - \sin \theta) }{(1 + \cos \theta)(1 - \cos \theta)} \)

(ii) \( \cot^2 \theta \)

B C A 8k 7k √113 k
Right Triangle Representation
Concept Used

Identity:

\[ (1 + x)(1 - x) = 1 - x^2 \]

Also,

\[ \sin^2 \theta + \cos^2 \theta = 1 \]

Solution Roadmap
  • Step 1: Use ratio method to find sin θ and cos θ
  • Step 2: Apply identity to simplify expression
  • Step 3: Compute required values
Solution

Step 1: Interpret given ratio

$$\begin{aligned} \cot \theta &= \frac{7}{8} = \frac{B}{P} \end{aligned}$$ $$\begin{aligned} B &= 7k,\quad P = 8k \end{aligned}$$

Step 2: Find Hypotenuse

$$\begin{aligned} H^{2} &= (7k)^2 + (8k)^2 \\ &= 49k^2 + 64k^2 \\ &= 113k^2 \end{aligned}$$ $$\begin{aligned} H = k\sqrt{113} \end{aligned}$$

Step 3: Find sin θ and cos θ

$$\begin{aligned} \sin \theta &= \frac{P}{H} = \frac{8k}{k\sqrt{113}} = \frac{8}{\sqrt{113}} \\\\ \cos \theta &= \frac{B}{H} = \frac{7k}{k\sqrt{113}} = \frac{7}{\sqrt{113}} \end{aligned}$$

(i) Evaluate expression

Key Identity Shortcut:

$$\begin{aligned} \frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)} &= \frac{1 - \sin^2\theta}{1 - \cos^2\theta} \end{aligned}$$ $$\begin{aligned} &= \frac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta \end{aligned}$$ $$\begin{aligned} \cot^2\theta = \left(\frac{7}{8}\right)^2 = \frac{49}{64} \end{aligned}$$

(ii) Direct calculation

$$\begin{aligned} \cot^2\theta = \left(\frac{7}{8}\right)^2 = \frac{49}{64} \end{aligned}$$

Final Answer:

(i) \( \frac{49}{64} \)
(ii) \( \frac{49}{64} \)

Exam Significance
  • Identity-based simplification is a high-frequency board question
  • Shortcut method saves significant exam time
  • Common in MCQs and competitive exams
  • Builds foundation for advanced trigonometric identities
← Q6
7 / 11  ·  64%
Q8 →
Q8
NUMERIC3 marks

If \( 3 \cot A = 4 \), check whether \( \dfrac{1-\tan^2 A}{1+\tan^2 A} = \cos^2 A - \sin^2 A \) or not.

B C A 3k 4k 5k
Right Triangle Representation
Concept Used

Identity:

\[ \begin{aligned} \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} &= \cos 2\theta \quad \text{and} \\\\ \cos^2 \theta - \sin^2 \theta &= \cos 2\theta \end{aligned} \]

Hence both expressions are theoretically equal.

Solution Roadmap
  • Step 1: Find basic trigonometric ratios
  • Step 2: Evaluate RHS
  • Step 3: Evaluate LHS
  • Step 4: Compare both sides
Solution

Step 1: Interpret given condition

$$\begin{aligned} 3 \cot A &= 4 \\ \cot A &= \frac{4}{3} = \frac{B}{P} \end{aligned}$$ $$\begin{aligned} B = 4k,\quad P = 3k \end{aligned}$$

Step 2: Find Hypotenuse

$$\begin{aligned} H^2 &= (4k)^2 + (3k)^2 \\ &= 16k^2 + 9k^2 \\ &= 25k^2 \end{aligned}$$ $$\begin{aligned} H = 5k \end{aligned}$$

Step 3: Evaluate RHS

$$\begin{aligned} \cos A &= \frac{4}{5}, \quad \sin A = \frac{3}{5} \end{aligned}$$ $$\begin{aligned} \cos^2 A - \sin^2 A &= \frac{16}{25} - \frac{9}{25} \\ &= \frac{7}{25} \end{aligned}$$

Step 4: Evaluate LHS

$$\begin{aligned} \tan A &= \frac{1}{\cot A} = \frac{3}{4} \end{aligned}$$ $$\begin{aligned} \tan^2 A = \frac{9}{16} \end{aligned}$$ $$\begin{aligned} \frac{1 - \tan^2 A}{1 + \tan^2 A} &= \frac{1 - \frac{9}{16}}{1 + \frac{9}{16}} \\ &= \frac{\frac{7}{16}}{\frac{25}{16}} \\ &= \frac{7}{25} \end{aligned}$$

Step 5: Comparison

$$\begin{aligned} \text{LHS} = \text{RHS} = \frac{7}{25} \end{aligned}$$

Conclusion: \( \dfrac{1-\tan^2 A}{1+\tan^2 A} = \cos^2 A - \sin^2 A \) is verified.

Exam Significance
  • Classic identity verification question (3–4 marks)
  • Tests LHS vs RHS structured approach
  • Important for simplifying trigonometric expressions
  • Common in CBSE boards and competitive exams
← Q7
8 / 11  ·  73%
Q9 →
Q9
NUMERIC3 marks

In triangle ABC, right-angled at B, if \( \tan A = \frac{1}{\sqrt{3}} \), find:
(i) \( \sin A \cos C + \cos A \sin C \)
(ii) \( \cos A \cos C - \sin A \sin C \)

B C A k √3 k 2k
Right Triangle ABC
Concept Used

Since triangle is right-angled at B:

\[ A + C = 90^\circ \]

Identities:

\[ \sin A \cos C + \cos A \sin C = \sin(A + C) \]

\[ \cos A \cos C - \sin A \sin C = \cos(A + C) \]

Solution Roadmap
  • Step 1: Find basic trigonometric ratios
  • Step 2: Use identity shortcut
  • Step 3: Verify using direct substitution
Solution

Step 1: Interpret given ratio

$$\begin{aligned} \tan A &= \frac{1}{\sqrt{3}} \\&= \frac{P}{B} \end{aligned}$$ $$\begin{aligned} P &= k,\\ B &= \sqrt{3}k \end{aligned}$$

Step 2: Find Hypotenuse

$$\begin{aligned} H^2 &= k^2 + 3k^2 \\&= 4k^2 \end{aligned}$$ $$\begin{aligned} H = 2k \end{aligned}$$

Step 3: Trigonometric ratios

$$\begin{aligned} \sin A &= \frac{1}{2}, \\ \cos A &= \frac{\sqrt{3}}{2} \end{aligned}$$ $$\begin{aligned} \sin C &= \frac{\sqrt{3}}{2}, \\ \cos C &= \frac{1}{2} \end{aligned}$$

(i) Using identity

$$\begin{aligned} \sin A \cos C + \cos A \sin C = \sin(A + C) \end{aligned}$$ $$\begin{aligned} = \sin 90^\circ = 1 \end{aligned}$$

(ii) Using identity

$$\begin{aligned} \cos A \cos C - \sin A \sin C = \cos(A + C) \end{aligned}$$ $$\begin{aligned} = \cos 90^\circ = 0 \end{aligned}$$

Final Answer:

(i) \(1\)
(ii) \(0\)

Exam Significance
  • Classic identity + complementary angle question
  • Shortcut avoids lengthy calculations
  • Very common in CBSE boards (3–4 marks)
  • Highly useful for JEE foundation & NDA exams
← Q8
9 / 11  ·  82%
Q10 →
Q10
NUMERIC3 marks

In \( \triangle PQR \), right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine \( \sin P \), \( \cos P \) and \( \tan P \).

Q R P 5 12 13
Triangle PQR
Concept Used

Pythagoras theorem:

\[ (\text{Hypotenuse})^2 = (\text{Base})^2 + (\text{Perpendicular})^2 \]

Trigonometric ratios:

\[ \begin{aligned} \sin \theta &= \frac{P}{H},\\ \cos \theta &= \frac{B}{H},\\ \tan \theta &= \frac{P}{B} \end{aligned} \]

Solution Roadmap
  • Step 1: Assume unknown side
  • Step 2: Form equation using given condition
  • Step 3: Apply Pythagoras theorem
  • Step 4: Compute trigonometric ratios
Solution

Step 1: Let QR = x

Then, PR = 25 − x

Step 2: Apply Pythagoras theorem

$$\begin{aligned} (25 - x)^2 &= PQ^2 + QR^2 \\ &= 5^2 + x^2 \end{aligned}$$ $$\begin{aligned} 625 - 50x + x^2 &= 25 + x^2 \end{aligned}$$ $$\begin{aligned} 625 - 50x &= 25 \\ -50x &= 25 - 625 \\ -50x &= -600 \\ x &= 12 \end{aligned}$$

Therefore,
QR = 12 cm
PR = 13 cm

Step 3: Find trigonometric ratios for angle P

Perpendicular = QR = 12
Base = PQ = 5
Hypotenuse = PR = 13

$$\begin{aligned} \sin P &= \frac{QR}{PR} \\&= \frac{12}{13} \end{aligned}$$ $$\begin{aligned} \cos P &= \frac{PQ}{PR} \\&= \frac{5}{13} \end{aligned}$$ $$\begin{aligned} \tan P &= \frac{QR}{PQ} \\&= \frac{12}{5} \end{aligned}$$

Final Answer:

\(\sin P = \frac{12}{13}\),
\(\cos P = \frac{5}{13}\),
\(\tan P = \frac{12}{5}\)

Exam Significance
  • Multi-concept question (algebra + geometry + trigonometry)
  • Tests equation formation skill (very important)
  • Common 3–4 mark board question
  • Useful for competitive exams (problem-solving type)
← Q9
10 / 11  ·  91%
Q11 →
Q11
NUMERIC3 marks

State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) \( \sec A = \frac{12}{5} \) for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) \( \sin \theta = \frac{4}{3} \) for some angle \( \theta \).

Solution

(i) False
\[ \tan A = \frac{\text{Perpendicular}}{\text{Base}} \] Since perpendicular and base can take different values:
If \(P > B\), then \( \tan A > 1 \)
If \(P = B\), then \( \tan A = 1 \)
If \(P < B\), then \( \tan A < 1 \)
Hence, \( \tan A \) is not always less than 1.

(ii) True
\[ \sec A = \frac{\text{Hypotenuse}}{\text{Base}} \] Since hypotenuse is always greater than base:
\( \sec A \geq 1 \)
A triangle with sides in ratio 5 : 12 : 13 gives: \[ \sec A = \frac{13}{5} \approx 2.6 \] Hence values greater than 1 such as \( \frac{12}{5} \) are possible.

(iii) False
“cos A” denotes cosine of angle A: \[ \cos A = \frac{\text{Base}}{\text{Hypotenuse}} \] but, cosecant is written as: \[ \text{cosec }A = \frac{\text{Hypotenuse}}{\text{Perpendicular}} \] Therefore, cos A is not cosecant.

(iv) False
“cot A” represents a trigonometric ratio: \[ \cot A = \frac{\text{Base}}{\text{Perpendicular}} \] It is a single function, not a multiplication of “cot” and “A”.

(v) False
\[ \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \] Since hypotenuse is always the largest side: \[ 0 \leq \sin \theta \leq 1 \] Therefore, \( \frac{4}{3} > 1 \) is not possible.

Key Concept: Range of trigonometric ratios:
\( 0 \leq \sin \theta \leq 1,\quad 0 \leq \cos \theta \leq 1,\quad \sec \theta \geq 1,\quad \tan \theta \in (0,\infty) \)

Exam Significance
  • Concept-based MCQs frequently asked in boards
  • Tests understanding of range of trigonometric ratios
  • Important for avoiding conceptual mistakes
  • Highly useful for competitive exams (SSC, NDA, etc.)
← Q10
11 / 11  ·  100%
↑ Back to top
🎓

Chapter Complete!

All 11 solutions for Introduction to Trigonometry covered.

↑ Review from the top
NCERT · Class X · Chapter 8 · Exercise 8.1

Introduction to Trigonometry

Complete AI Learning Engine — Concepts, Formulas, Solver & Interactive Modules

Trigonometry is the study of relationships between the sides and angles of a right-angled triangle. These ratios, called trigonometric ratios, form the language of waves, engineering, astronomy, and beyond.

📐 What is a Trigonometric Ratio?

Consider a right-angled triangle with the right angle at vertex C. For an acute angle θ at vertex A, the three sides are labelled with respect to angle θ:

Opposite
BC (side opposite θ)
The side facing the angle θ
Adjacent
AB (side next to θ)
The non-hypotenuse side touching θ
Hypotenuse
AC (longest side)
Side opposite the right angle; always longest
📊 Values of Trigonometric Ratios

Standard angle values for NCERT Exercise 8.1 — memorise these; they appear in almost every problem.

Angle (θ) sin θ cos θ tan θ cosec θ sec θ cot θ
0° 0 1 0 — 1 —
30° 1/2 √3/2 1/√3 2 2/√3 √3
45° 1/√2 1/√2 1 √2 √2 1
60° √3/2 1/2 √3 2/√3 2 1/√3
90° 1 0 — 1 — 0

— denotes undefined (division by zero)

🔗 Reciprocal Relationships
cosec θ = 1 / sin θ
Cosecant is reciprocal of sine
sec θ = 1 / cos θ
Secant is reciprocal of cosine
cot θ = 1 / tan θ
Cotangent is reciprocal of tangent
tan θ = sin θ / cos θ
Tangent expressed via sin and cos
cot θ = cos θ / sin θ
Cotangent expressed via cos and sin
⚠️ Defined Only for 0° < θ ≤ 90°

Trigonometric ratios are defined for acute angles (0° to 90°) in right-angled triangles in this chapter. The values of sin and cos are always between 0 and 1 inclusive. tan and cot can take any non-negative value. The trigonometric ratios of an angle are independent of the size of the right triangle — they depend only on the angle.

All essential formulas from NCERT Class X Chapter 8 — definitions, identities, reciprocals, and Pythagorean identities with derivation hints.

📏 Six Basic Trigonometric Ratios
Sine
sin θ = Opposite / Hypotenuse
= BC / AC
Cosine
cos θ = Adjacent / Hypotenuse
= AB / AC
Tangent
tan θ = Opposite / Adjacent
= BC / AB = sin θ / cos θ
Cosecant
cosec θ = Hypotenuse / Opposite
= AC / BC = 1 / sin θ
Secant
sec θ = Hypotenuse / Adjacent
= AC / AB = 1 / cos θ
Cotangent
cot θ = Adjacent / Opposite
= AB / BC = 1 / tan θ
🔺 Pythagorean Identities
Identity I — Core
sin²θ + cos²θ = 1
Derived from Pythagoras theorem: divide a²+b²=c² by c²
Identity II
1 + tan²θ = sec²θ
Derived from Identity I: divide throughout by cos²θ
Identity III
1 + cot²θ = cosec²θ
Derived from Identity I: divide throughout by sin²θ
🔄 Complementary Angle Relationships
sin(90°−θ) = cos θ
Sine and cosine are complementary
cos(90°−θ) = sin θ
tan(90°−θ) = cot θ
cot(90°−θ) = tan θ
sec(90°−θ) = cosec θ
cosec(90°−θ) = sec θ
📐 Common Pythagorean Triples

Recognise these triples instantly to avoid square-root calculations in exams.

3 — 4 — 5
3²+4²=5² → sin A=3/5, cos A=4/5
5 — 12 — 13
5²+12²=13²
8 — 15 — 17
8²+15²=17²
7 — 24 — 25
7²+24²=25²
Multiples work too
e.g. 6-8-10, 9-12-15, etc.
📈 Range of Trigonometric Ratios (0° to 90°)
sin θ
0 ≤ sin θ ≤ 1
Increases as θ increases
cos θ
0 ≤ cos θ ≤ 1
Decreases as θ increases
tan θ
0 ≤ tan θ < ∞
Increases from 0 to ∞
cosec θ
cosec θ ≥ 1
Decreases from ∞ to 1
sec θ
sec θ ≥ 1
Increases from 1 to ∞
cot θ
0 < cot θ < ∞
Decreases from ∞ to 0

Select a problem type, enter your values, and get an instant detailed step-by-step solution — no external API required.

🔍 Step-by-Step AI Solver
📚 Worked Examples Library

Click any example to see its complete solution instantly.

W1
In △ABC, right-angled at B, if tan A = 1/√3, find sin A, cos A, sin C, cos C.
▼
1
Set up sides using tan A.
tan A = opposite/adjacent = BC/AB = 1/√3
Let BC = 1k, AB = √3k. By Pythagoras: AC² = BC² + AB² = 1 + 3 = 4 → AC = 2kBC = 1, AB = √3, AC = 2 (for k=1)
2
Find ratios for angle A.sin A = BC/AC = 1/2    cos A = AB/AC = √3/2
3
Ratios for angle C (complement of A since B = 90°).
Angle C is at vertex C. For angle C: opposite = AB = √3, adjacent = BC = 1, hyp = AC = 2sin C = AB/AC = √3/2    cos C = BC/AC = 1/2
✦ sin A = 1/2, cos A = √3/2, sin C = √3/2, cos C = 1/2
W2
Evaluate: (sin 30° + tan 45° − cosec 60°) ÷ (sec 30° + cos 60° + cot 45°)
▼
1
Substitute standard values.sin 30° = 1/2   tan 45° = 1   cosec 60° = 2/√3
sec 30° = 2/√3   cos 60° = 1/2   cot 45° = 1
2
Compute numerator.1/2 + 1 − 2/√3 = 3/2 − 2/√3Rationalise: 2/√3 = 2√3/3= 3/2 − 2√3/3 = (9 − 4√3)/6
3
Compute denominator.2/√3 + 1/2 + 1 = 2√3/3 + 3/2 = (4√3 + 9)/6
4
Divide.Result = (9 − 4√3) / (9 + 4√3)Rationalise denominator:× (9 − 4√3)/(9 − 4√3) = (9−4√3)² / (81−48) = (81−72√3+48)/33 = (129−72√3)/33
✦ (129 − 72√3) / 33
W3
If sin A = 3/5, find cos A and tan A using Pythagorean identity.
▼
1
Apply identity sin²A + cos²A = 1.(3/5)² + cos²A = 1 → 9/25 + cos²A = 1
2
Solve for cos A.cos²A = 1 − 9/25 = 16/25 → cos A = 4/5
3
Find tan A.tan A = sin A / cos A = (3/5) / (4/5) = 3/4
✦ cos A = 4/5, tan A = 3/4

Original concept-building questions organised by topic — not repeated from the textbook. Full step-by-step solutions included. Difficulty is tagged for exam preparation.

⬡ Concept A — Identifying & Setting Up Ratios
1
In △PQR, right-angled at Q, PQ = 7 cm and QR = 24 cm. Find sin P, cos P, and tan R.
Easy
▼
1
Find hypotenuse PR.
Using Pythagoras: PR² = PQ² + QR² = 49 + 576 = 625PR = 25 cm
2
Ratios for angle P (opposite = QR = 24, adjacent = PQ = 7, hyp = PR = 25):sin P = 24/25    cos P = 7/25
3
Ratios for angle R (opposite = PQ = 7, adjacent = QR = 24, hyp = PR = 25):tan R = PQ/QR = 7/24
✦ sin P = 24/25, cos P = 7/25, tan R = 7/24
2
In △ABC, ∠B = 90°. If AB = 5k and BC = 12k for some positive constant k, find all six trigonometric ratios for angle A without finding AC explicitly first.
Easy
▼
1
Find AC using Pythagoras.AC² = AB² + BC² = 25k² + 144k² = 169k² → AC = 13k
2
Six ratios for angle A (opposite = BC = 12k, adjacent = AB = 5k, hyp = AC = 13k):sin A = 12/13   cos A = 5/13   tan A = 12/5
cosec A = 13/12   sec A = 13/5   cot A = 5/12
3
Key insight: The factor k cancels in all ratios — ratios are independent of triangle size.
✦ sin A=12/13, cos A=5/13, tan A=12/5, cosec A=13/12, sec A=13/5, cot A=5/12
⬡ Concept B — Given One Ratio, Find Others
3
Given cot θ = 7/8, find (1 + sin θ)(1 − sin θ) / (1 + cos θ)(1 − cos θ).
Medium
▼
1
Simplify the expression first using difference of squares:(1 + sin θ)(1 − sin θ) = 1 − sin²θ = cos²θ
(1 + cos θ)(1 − cos θ) = 1 − cos²θ = sin²θ
2
Expression reduces to:cos²θ / sin²θ = cot²θ
3
Substitute cot θ = 7/8:cot²θ = (7/8)² = 49/64
✦ The expression = 49/64
4
If 15 cot A = 8, find sin A and sec A. Then verify sin²A + cos²A = 1.
Medium
▼
1
Find cot A:cot A = 8/15 → adjacent/opposite = 8/15
Let AB = 8k, BC = 15k → AC = √(64+225)k = √289 k = 17k
2
Find sin A and sec A:sin A = BC/AC = 15/17    cos A = AB/AC = 8/17
sec A = 1/cos A = 17/8
3
Verify identity:sin²A + cos²A = (15/17)² + (8/17)² = 225/289 + 64/289 = 289/289 = 1 ✓
✦ sin A = 15/17, sec A = 17/8
⬡ Concept C — Standard Angle Evaluations
5
Evaluate: 2 tan²45° + cos²30° − sin²60°
Easy
▼
1
Substitute values:tan 45° = 1, cos 30° = √3/2, sin 60° = √3/2
2
Evaluate:2(1)² + (√3/2)² − (√3/2)² = 2 + 3/4 − 3/4 = 2
3
Observation: cos²30° and sin²60° cancel since cos 30° = sin 60°.
✦ Value = 2
6
Show that: tan 48° · tan 23° · tan 42° · tan 67° = 1
Medium
▼
1
Use complementary angle identity: tan(90°−θ) = cot θ
2
Identify pairs:tan 48° = tan(90°−42°) = cot 42°
tan 23° = tan(90°−67°) = cot 67°
3
Substitute and simplify:cot 42° · cot 67° · tan 42° · tan 67°
= (cot 42° · tan 42°) · (cot 67° · tan 67°)
= 1 · 1 = 1 ✓
✦ LHS = 1 = RHS (Proved)
⬡ Concept D — Pythagorean Identity Applications
7
Prove: (cosec A − sin A)(sec A − cos A) = 1/(tan A + cot A)
Hard
▼
1
Simplify LHS.cosec A − sin A = 1/sin A − sin A = (1 − sin²A)/sin A = cos²A/sin A
sec A − cos A = 1/cos A − cos A = (1 − cos²A)/cos A = sin²A/cos A
2
Multiply the two simplified factors:LHS = (cos²A/sin A) × (sin²A/cos A) = sin A · cos A
3
Simplify RHS.tan A + cot A = sin A/cos A + cos A/sin A = (sin²A + cos²A)/(sin A · cos A) = 1/(sin A · cos A)So RHS = 1/(tan A + cot A) = sin A · cos A
4
LHS = RHS = sin A · cos A ✓
✦ Proved: LHS = RHS = sin A · cos A
8
If cos A + sin A = √2 cos A, show that cos A − sin A = √2 sin A.
Hard
▼
1
Start from the given condition.cos A + sin A = √2 cos A
⟹ sin A = √2 cos A − cos A = cos A(√2 − 1)
2
Rationalise by multiplying by (√2+1)/(√2+1):cos A = sin A · (√2+1)/[(√2−1)(√2+1)] = sin A(√2+1)/(2−1) = sin A(√2+1)
3
Compute cos A − sin A:cos A − sin A = sin A(√2+1) − sin A = sin A · √2 = √2 sin A ✓
✦ Proved: cos A − sin A = √2 sin A
⬡ Concept E — Complementary Angles
9
Without using tables or calculator, evaluate: cos 38° / sin 52° + sin 15° cos 75° + cos 15° sin 75°
Medium
▼
1
First term:sin 52° = sin(90°−38°) = cos 38°
∴ cos 38°/sin 52° = cos 38°/cos 38° = 1
2
Second and third terms form sin(A+B) pattern:sin 15° cos 75° + cos 15° sin 75° = sin(15°+75°) = sin 90° = 1
3
Total:1 + 1 = 2
✦ Value = 2
10
If tan 2A = cot(A − 18°), where 2A is an acute angle, find the value of A.
Medium
▼
1
Use tan θ = cot(90°−θ):tan 2A = cot(90°−2A)
2
Equate arguments:90° − 2A = A − 18°
108° = 3A → A = 36°
3
Verify: 2A = 72° (acute ✓)   A − 18° = 18° (acute ✓)
✦ A = 36°

Exam-tested strategies, memory tricks, and the most common errors students make — know these before your exam.

💡 Tips & Tricks for Trigonometry
  • 🔢
    SOH-CAH-TOA Memory Aid:
    Sin = Opposite/Hypotenuse  |  Cos = Adjacent/Hypotenuse  |  Tan = Opposite/Adjacent.
    Remember: "Some Old Hippie Caught Another Hippie Tripping On Acid"
  • 📈
    Sin Table Trick (0° to 90°):
    sin 0°, 30°, 45°, 60°, 90° = √0/2, √1/2, √2/2, √3/2, √4/2 = 0, 1/2, 1/√2, √3/2, 1 — note the numerators under the radical are 0, 1, 2, 3, 4.
  • 🔄
    Cos is reverse of Sin: cos table for 0°→90° is the same as the sin table read backwards (90°→0°). So cos 0°=1, cos 30°=√3/2, cos 45°=1/√2, cos 60°=1/2, cos 90°=0.
  • ⚡
    Rationalise Denominators Early: When you see 1/√3, 2/√3 etc., convert immediately: 1/√3 = √3/3. This prevents messy calculations later.
  • 🔺
    Pythagorean Triple Recognition: Instantly recognise 3-4-5, 5-12-13, 8-15-17 triples. If AB = 5, BC = 12, you know AC = 13 without calculating. Saves 30+ seconds per question!
  • 🧮
    Identity-first Approach: For complex expressions, try applying sin²θ + cos²θ = 1 (or its derived forms) before substituting values. This often simplifies work dramatically.
  • 🪞
    Complementary Pairs to Memorise: sin ↔ cos, tan ↔ cot, sec ↔ cosec — these are always complementary pairs. For any angle θ, f(θ) = co-f(90°−θ).
  • 🔗
    Convert everything to sin/cos: For proving identities, convert all ratios to sin and cos first. Most identities simplify beautifully once in terms of just two functions.
⚠️ Common Mistakes to Avoid
  • ✗
    Confusing sin⁻¹ with 1/sin:
    sin⁻¹A means "angle whose sine is A" (inverse sine), NOT 1/sin A (which is cosec A).
    ❌ Wrong

    sin⁻¹A = 1/sin A = cosec A

    ✓ Correct

    sin⁻¹A = arcsin(A)  |  1/sin A = cosec A (these are different!)

  • ✗
    Wrong side labelling depending on angle:
    The "opposite" and "adjacent" sides change depending on which angle you are considering!
    ❌ Wrong

    Using BC as "opposite" for both angles A and C in △ABC

    ✓ Correct

    BC is opposite to A but adjacent to C — always re-label for each angle

  • ✗
    sin²A means (sin A)², not sin(A²):
    ❌ Wrong

    sin²30° = sin(900°) = sin(180°) = 0

    ✓ Correct

    sin²30° = (sin 30°)² = (1/2)² = 1/4

  • ✗
    Forgetting that tan 90° is undefined:
    tan 90° = sin 90°/cos 90° = 1/0 → undefined. Similarly cosec 0° and sec 90° are undefined. Never use them in calculations.
  • ✗
    sin(A + B) ≠ sin A + sin B:
    ❌ Wrong

    sin 90° = sin 30° + sin 60° = 1/2 + √3/2 ≠ 1

    ✓ Correct

    sin(A+B) = sin A cos B + cos A sin B — use the compound angle formula

  • ✗
    Not squaring the ratio when setting up sides:
    If sin A = 3/5, some students write opposite = 3, hyp = 5 and then guess adjacent = 2. Always use Pythagoras: adjacent = √(25−9) = 4.

Six interactive modules to solidify your understanding through active learning — quizzes, matching, fill-in-the-blanks, and a live triangle visualiser.

🎯
MCQ Quiz
10 questions on ratios, identities & standard values
Question 1 of 10
Score: 0 / 0
🔗
Match the Ratios
Click to match each expression with its simplified value

Click a card on the left, then its match on the right.

✏️
Fill in the Blanks
Complete the trigonometric identities and values
📐
Triangle Visualiser
Drag sliders to see how ratios change live
A B C opp adj hyp θ
Angle θ30°
⚖️
True or False
Test your conceptual understanding quickly
⚡
Rapid Fire Values
Type the value before time runs out!
Score: 0
NCERT Mathematics Class X · Chapter 8 Exercise 8.1 · Introduction to Trigonometry  ·  All concepts © NCERT; engine design is original.
📚
ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025
Sharing this chapter
NCERT Class 10 Trigonometry Ex 8.1 Solutions
NCERT Class 10 Trigonometry Ex 8.1 Solutions — Complete Notes & Solutions · academia-aeternum.com
Chapter 8, Introduction to Trigonometry, marks a fundamental transition in Class X Mathematics, enabling learners to interpret and relate the sides and angles of right-angled triangles through precise trigonometric ratios. The textbook exercises curated by NCERT are structured to guide students progressively—from identifying basic ratios such as sine, cosine, and tangent to applying identities, evaluating standard angles, deriving relationships, and solving angle-dependent problems with…
🎓 Class 10 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
Share on
WhatsApp
Telegram
X / Twitter
Facebook
Instagram
Threads
Pinterest
LinkedIn
Quora
Discord
Reddit
Email
academia-aeternum.com/class-10/mathematics/introduction-to-trigonometry/exercises/exercise-8.1/ Copy link
💡
Exam tip: Sharing chapter notes with your study group creates a reinforcement loop. Teaching a concept is the fastest path to mastering it.

Recent posts

    Get in Touch

    Let's Connect

    Questions, feedback, or suggestions?
    We'd love to hear from you.