Ch 8  ·  Q–
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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%
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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
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Introduction To Trigonometry | Mathematics Class -10
Introduction To Trigonometry | Mathematics Class -10 — Complete Notes & Solutions · academia-aeternum.com
🎓 Class -10 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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