Class 10 Mathematics - NCERT Mathematics
NCERT Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry
Chapter 8 introduces trigonometric ratios of acute angles in right triangles and establishes fundamental identities. Students learn to calculate sin, cos, tan, cosec, sec, and cot for specific angles (0, 30, 45, 60, 90 degrees), apply complementary angle relations, and prove trigonometric identities. This chapter is part of the Trigonometry unit and carries 6-8 marks in the CBSE Class 10 board exam 2027.
Syllabus Update 2026-27
As per the CBSE 2026-27 syllabus, Chapter 8 Introduction to Trigonometry has all 4 exercises intact: Exercise 8.1 (trigonometric ratios), Exercise 8.2 (values of standard angles), Exercise 8.3 (complementary angles), and Exercise 8.4 (trigonometric identities). No exercises have been removed. This chapter is heavily tested and forms the basis for Chapter 9 (Some Applications of Trigonometry).
Table of Contents
Exercise 8.1 - Trigonometric Ratios
11 questions
Exercise 8.2 - Trigonometric Ratios of Some Specific Angles
4 questions
Exercise 8.3 - Trigonometric Ratios of Complementary Angles
7 questions
Exercise 8.4 - Trigonometric Identities
5 questions
Previous Year Questions
16 questions (CBSE 2019-2026)
MCQ Practice
12 questions
8.1Exercise 8.1 - Trigonometric Ratios
Q1In triangle ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A (ii) sin C, cos C
First, find AC using Pythagoras theorem:
AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625
AC = 25 cm
(i) For angle A:
sin A = opposite/hypotenuse = BC/AC = 7/25
cos A = adjacent/hypotenuse = AB/AC = 24/25
(ii) For angle C:
sin C = opposite/hypotenuse = AB/AC = 24/25
cos C = adjacent/hypotenuse = BC/AC = 7/25
Q2In the given figure, find tan P - cot R.
In triangle PQR, right-angled at Q:
PQ = 12 cm, PR = 13 cm
Using Pythagoras theorem:
QR² = PR² - PQ² = 169 - 144 = 25
QR = 5 cm
tan P = QR/PQ = 5/12
cot R = QR/PQ = 5/12
tan P - cot R = 5/12 - 5/12 = 0
Q3If sin A = 3/4, calculate cos A and tan A.
Given: sin A = 3/4, so opposite = 3k, hypotenuse = 4k.
Using Pythagoras theorem:
adjacent² = hypotenuse² - opposite² = (4k)² - (3k)² = 16k² - 9k² = 7k²
adjacent = k√7
cos A = adjacent/hypotenuse = k√7/(4k) = √7/4
tan A = opposite/adjacent = 3k/(k√7) = 3/√7 = 3√7/7
Q4Given 15 cot A = 8, find sin A and sec A.
Given: 15 cot A = 8, so cot A = 8/15.
This means adjacent = 8k, opposite = 15k.
Using Pythagoras theorem:
hypotenuse² = (8k)² + (15k)² = 64k² + 225k² = 289k²
hypotenuse = 17k
sin A = opposite/hypotenuse = 15k/17k = 15/17
sec A = hypotenuse/adjacent = 17k/8k = 17/8
Q5Given sec theta = 13/12, calculate all other trigonometric ratios.
Given: sec theta = 13/12, so hypotenuse = 13k, adjacent = 12k.
Using Pythagoras theorem:
opposite² = (13k)² - (12k)² = 169k² - 144k² = 25k²
opposite = 5k
sin theta = opposite/hypotenuse = 5k/13k = 5/13
cos theta = adjacent/hypotenuse = 12k/13k = 12/13
tan theta = opposite/adjacent = 5k/12k = 5/12
cosec theta = hypotenuse/opposite = 13k/5k = 13/5
cot theta = adjacent/opposite = 12k/5k = 12/5
Q6If angle A and angle B are acute angles such that cos A = cos B, then show that angle A = angle B.
In a right triangle, cos A = adjacent side to A / hypotenuse.
Let triangle ABC be right-angled at C.
cos A = AC/AB and cos B = BC/AB.
Given cos A = cos B:
AC/AB = BC/AB
AC = BC
Since the sides opposite to angles B and A are equal, the triangle is isosceles.
Therefore, the angles opposite to equal sides are equal.
angle A = angle B. QED
Q7If cot theta = 7/8, evaluate: (i) (1 + sin theta)(1 - sin theta) / (1 + cos theta)(1 - cos theta) (ii) cot² theta
Given: cot theta = 7/8, so adjacent = 7k, opposite = 8k.
hypotenuse² = 49k² + 64k² = 113k²
hypotenuse = k√113
sin theta = 8/√113, cos theta = 7/√113
(i) (1 + sin theta)(1 - sin theta) / (1 + cos theta)(1 - cos theta)
= (1 - sin²theta) / (1 - cos²theta)
= cos²theta / sin²theta
= cot²theta
= (7/8)² = 49/64
(ii) cot²theta = (7/8)² = 49/64
Note: Both expressions give the same answer, which verifies our calculation.
Q8If 3 cot A = 4, check whether (1 - tan²A) / (1 + tan²A) = cos²A - sin²A or not.
Given: 3 cot A = 4, so cot A = 4/3, tan A = 3/4.
Adjacent = 4k, opposite = 3k, hypotenuse = 5k (Pythagorean triple).
sin A = 3/5, cos A = 4/5
LHS: (1 - tan²A)/(1 + tan²A)
= (1 - 9/16)/(1 + 9/16)
= (7/16)/(25/16)
= 7/25
RHS: cos²A - sin²A
= 16/25 - 9/25
= 7/25
LHS = RHS = 7/25. Hence verified.
Q9In triangle ABC, right-angled at B, if tan A = 1/√3, find the value of: (i) sin A cos C + cos A sin C (ii) cos A cos C - sin A sin C
Given: tan A = 1/√3, so A = 30 degrees.
In a right triangle at B: C = 90 - 30 = 60 degrees.
sin 30 = 1/2, cos 30 = √3/2
sin 60 = √3/2, cos 60 = 1/2
(i) sin A cos C + cos A sin C
= sin 30 cos 60 + cos 30 sin 60
= (1/2)(1/2) + (√3/2)(√3/2)
= 1/4 + 3/4
= 1
(ii) cos A cos C - sin A sin C
= cos 30 cos 60 - sin 30 sin 60
= (√3/2)(1/2) - (1/2)(√3/2)
= √3/4 - √3/4
= 0
Q10In triangle PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Let QR = x cm. Then PR = (25 - x) cm.
By Pythagoras theorem:
PR² = PQ² + QR²
(25 - x)² = 25 + x²
625 - 50x + x² = 25 + x²
625 - 50x = 25
50x = 600
x = 12
So QR = 12 cm and PR = 25 - 12 = 13 cm.
sin P = QR/PR = 12/13
cos P = PQ/PR = 5/13
tan P = QR/PQ = 12/5
Q11State whether the following are true or false. Justify your answer. (i) The value of tan A is always less than 1. (ii) sec A = 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 = 4/3 for some angle theta.
(i) False. tan 45 = 1 and tan 60 = √3 > 1. The value of tan A can be any non-negative real number for acute angles.
(ii) True. sec A >= 1 for all acute angles, and 12/5 = 2.4 > 1. So there exists an angle A for which sec A = 12/5.
(iii) False. cos A is the abbreviation for cosine of angle A, not cosecant. The abbreviation for cosecant is cosec A (or csc A).
(iv) False. cot A is a single trigonometric ratio (cotangent of angle A), not the product of 'cot' and 'A'.
(v) False. sin theta = opposite/hypotenuse. Since the hypotenuse is the longest side of a right triangle, the opposite side is always less than the hypotenuse. Therefore, sin theta is always less than or equal to 1. Since 4/3 > 1, no angle theta can have sin theta = 4/3.
8.2Exercise 8.2 - Trigonometric Ratios of Some Specific Angles
Q1Evaluate the following: (i) sin 60 cos 30 + sin 30 cos 60 (ii) 2 tan²45 + cos²30 - sin²60 (iii) cos 45/(sec 30 + cosec 30) (iv) (sin 30 + tan 45 - cosec 60)/(sec 30 + cos 60 + cot 45) (v) (5 cos²60 + 4 sec²30 - tan²45)/(sin²30 + cos²30)
(i) sin 60 cos 30 + sin 30 cos 60
= (√3/2)(√3/2) + (1/2)(1/2)
= 3/4 + 1/4
= 1
(ii) 2 tan²45 + cos²30 - sin²60
= 2(1)² + (√3/2)² - (√3/2)²
= 2 + 3/4 - 3/4
= 2
(iii) cos 45/(sec 30 + cosec 30)
= (1/√2)/(2/√3 + 2)
= (1/√2)/((2 + 2√3)/√3)
= √3/(√2(2 + 2√3))
= √3/(2√2(1 + √3))
Rationalising by multiplying numerator and denominator by (√3 - 1):
= √3(√3 - 1)/(2√2(√3 + 1)(√3 - 1))
= √3(√3 - 1)/(2√2(3 - 1))
= √3(√3 - 1)/(4√2)
= (3 - √3)/(4√2)
= (3 - √3)/(4√2) = (3√2 - √6)/8
(iv) (sin 30 + tan 45 - cosec 60)/(sec 30 + cos 60 + cot 45)
= (1/2 + 1 - 2/√3)/(2/√3 + 1/2 + 1)
= (3/2 - 2/√3)/(3/2 + 2/√3)
= (3/2 - 2√3/3)/(3/2 + 2√3/3)
= ((9 - 4√3)/6)/((9 + 4√3)/6)
= (9 - 4√3)/(9 + 4√3)
Rationalising: = (9 - 4√3)²/(81 - 48)
= (81 - 72√3 + 48)/33
= (129 - 72√3)/33
= (43 - 24√3)/11
(v) (5 cos²60 + 4 sec²30 - tan²45)/(sin²30 + cos²30)
= (5(1/2)² + 4(2/√3)² - 1²)/(1)
= 5(1/4) + 4(4/3) - 1
= 5/4 + 16/3 - 1
= (15 + 64 - 12)/12
= 67/12
Q2Choose the correct option and justify your choice: (i) 2 tan 30/1 + tan²30 = (A) sin 60 (B) cos 60 (C) tan 60 (D) sin 30 (ii) 1 - tan²45/1 + tan²45 = (A) tan 90 (B) 1 (C) sin 45 (D) 0 (iii) sin 2A = 2 sin A is true when A = (A) 0 (B) 30 (C) 45 (D) 60 (iv) 2 tan 30/1 - tan²30 = (A) cos 60 (B) sin 60 (C) tan 60 (D) sin 30
(i) 2 tan 30/(1 + tan²30)
= 2(1/√3)/(1 + 1/3)
= (2/√3)/(4/3)
= (2/√3)(3/4) = 6/(4√3) = 3/(2√3) = √3/2 = sin 60
Answer: (A) sin 60
(ii) (1 - tan²45)/(1 + tan²45)
= (1 - 1)/(1 + 1) = 0/2 = 0
Answer: (D) 0
(iii) sin 2A = 2 sin A
Check A = 0: sin 0 = 2 sin 0 → 0 = 0. True.
Check A = 30: sin 60 = 2 sin 30 → √3/2 = 1. False.
Answer: (A) 0
(iv) 2 tan 30/(1 - tan²30)
= 2(1/√3)/(1 - 1/3)
= (2/√3)/(2/3)
= (2/√3)(3/2) = 3/√3 = √3 = tan 60
Answer: (C) tan 60
Q3If tan(A + B) = √3 and tan(A - B) = 1/√3, where 0 < A + B <= 90 and A > B, find A and B.
tan(A + B) = √3 = tan 60
So A + B = 60 ... (i)
tan(A - B) = 1/√3 = tan 30
So A - B = 30 ... (ii)
Adding (i) and (ii):
2A = 90
A = 45 degrees
Substituting in (i):
45 + B = 60
B = 15 degrees
Q4State whether the following are true or false. Justify your answer. (i) sin(A + B) = sin A + sin B (ii) The value of sin theta increases as theta increases. (iii) The value of cos theta increases as theta increases. (iv) sin theta = cos theta for all values of theta. (v) cot A is not defined for A = 0 degrees.
(i) False. Let A = 30, B = 60. sin(90) = 1. But sin 30 + sin 60 = 1/2 + √3/2 = (1+√3)/2, which is not equal to 1.
(ii) True. From the table: sin 0 = 0, sin 30 = 1/2, sin 45 = 1/√2, sin 60 = √3/2, sin 90 = 1. The values increase as the angle increases from 0 to 90.
(iii) False. cos 0 = 1, cos 30 = √3/2, cos 45 = 1/√2, cos 60 = 1/2, cos 90 = 0. The values decrease as the angle increases from 0 to 90.
(iv) False. sin theta = cos theta only when theta = 45 degrees (where both equal 1/√2), not for all values.
(v) True. cot A = cos A/sin A. At A = 0, sin 0 = 0, so cot 0 = cos 0/sin 0 = 1/0, which is not defined.
8.3Exercise 8.3 - Trigonometric Ratios of Complementary Angles
Q1Evaluate: (i) sin 18/cos 72 (ii) tan 26/cot 64 (iii) cos 48 - sin 42 (iv) cosec 31 - sec 59
(i) sin 18/cos 72 = sin 18/cos(90 - 18) = sin 18/sin 18 = 1
(ii) tan 26/cot 64 = tan 26/cot(90 - 26) = tan 26/tan 26 = 1
(iii) cos 48 - sin 42 = cos 48 - sin(90 - 48) = cos 48 - cos 48 = 0
(iv) cosec 31 - sec 59 = cosec 31 - sec(90 - 31) = cosec 31 - cosec 31 = 0
Q2Show that: (i) tan 48 tan 23 tan 42 tan 67 = 1 (ii) cos 38 cos 52 - sin 38 sin 52 = 0
(i) tan 48 tan 23 tan 42 tan 67
= tan 48 . tan 23 . tan(90 - 48) . tan(90 - 23)
= tan 48 . tan 23 . cot 48 . cot 23
= (tan 48 . cot 48)(tan 23 . cot 23)
= 1 . 1 = 1 QED
(ii) cos 38 cos 52 - sin 38 sin 52
= cos 38 . cos(90 - 38) - sin 38 . sin(90 - 38)
= cos 38 . sin 38 - sin 38 . cos 38
= 0 QED
Q3If tan 2A = cot(A - 18), where 2A is an acute angle, find the value of A.
Given: tan 2A = cot(A - 18)
Using tan theta = cot(90 - theta):
cot(90 - 2A) = cot(A - 18)
Since both angles are acute:
90 - 2A = A - 18
90 + 18 = 3A
108 = 3A
A = 36 degrees
Verification: 2A = 72 (acute). A - 18 = 18 (positive). tan 72 = cot 18 = cot(90-72). Correct.
Q4If tan A = cot B, prove that A + B = 90 degrees.
Given: tan A = cot B
We know cot B = tan(90 - B).
So tan A = tan(90 - B).
Since A and B are acute angles:
A = 90 - B
A + B = 90 degrees. QED
Q5If sec 4A = cosec(A - 20), where 4A is an acute angle, find the value of A.
Given: sec 4A = cosec(A - 20)
Using sec theta = cosec(90 - theta):
cosec(90 - 4A) = cosec(A - 20)
Since both angles are acute:
90 - 4A = A - 20
90 + 20 = 5A
110 = 5A
A = 22 degrees
Verification: 4A = 88 (acute). A - 20 = 2 (positive). sec 88 = cosec 2 = cosec(90 - 88). Correct.
Q6If A, B, and C are interior angles of a triangle ABC, then show that sin((B + C)/2) = cos(A/2).
In a triangle, A + B + C = 180 degrees.
So B + C = 180 - A.
Dividing by 2:
(B + C)/2 = (180 - A)/2 = 90 - A/2
Therefore:
sin((B + C)/2) = sin(90 - A/2) = cos(A/2)
Hence proved. QED
Q7Express sin 67 + cos 75 in terms of trigonometric ratios of angles between 0 and 45 degrees.
sin 67 = sin(90 - 23) = cos 23
cos 75 = cos(90 - 15) = sin 15
Therefore:
sin 67 + cos 75 = cos 23 + sin 15
Both 23 and 15 are between 0 and 45 degrees.
8.4Exercise 8.4 - Trigonometric Identities
Q1Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
sin A in terms of cot A:
We know 1 + cot²A = cosec²A
cosec²A = 1 + cot²A
sin²A = 1/(1 + cot²A)
sin A = 1/√(1 + cot²A) (positive since A is acute)
sec A in terms of cot A:
tan A = 1/cot A
sec²A = 1 + tan²A = 1 + 1/cot²A = (cot²A + 1)/cot²A
sec A = √(1 + cot²A)/cot A
tan A in terms of cot A:
tan A = 1/cot A
Q2Write all the other trigonometric ratios of angle A in terms of sec A.
cos A = 1/sec A
sin A: sin²A = 1 - cos²A = 1 - 1/sec²A = (sec²A - 1)/sec²A
sin A = √(sec²A - 1)/sec A
tan A: tan²A = sec²A - 1
tan A = √(sec²A - 1)
cosec A = 1/sin A = sec A/√(sec²A - 1)
cosec A = sec A/√(sec²A - 1)
cot A = 1/tan A = 1/√(sec²A - 1)
cot A = 1/√(sec²A - 1)
Q3Evaluate: (i) (sin²63 + sin²27)/(cos²17 + cos²73) (ii) sin 25 cos 65 + cos 25 sin 65
(i) (sin²63 + sin²27)/(cos²17 + cos²73)
sin 63 = sin(90-27) = cos 27, so sin²63 = cos²27
Numerator = cos²27 + sin²27 = 1
cos 73 = cos(90-17) = sin 17, so cos²73 = sin²17
Denominator = cos²17 + sin²17 = 1
Answer = 1/1 = 1
(ii) sin 25 cos 65 + cos 25 sin 65
= sin 25 cos(90-25) + cos 25 sin(90-25)
= sin 25 sin 25 + cos 25 cos 25
= sin²25 + cos²25
= 1
Q4Prove the following identities, where the angles involved are acute angles for which the expressions are defined: (i) (cosec theta - cot theta)² = (1 - cos theta)/(1 + cos theta) (ii) (cos A)/(1 + sin A) + (1 + sin A)/(cos A) = 2 sec A (iii) (tan theta)/(1 - cot theta) + (cot theta)/(1 - tan theta) = 1 + sec theta cosec theta (iv) (1 + sec A)/sec A = sin²A/(1 - cos A) (v) (cos A - sin A + 1)/(cos A + sin A - 1) = cosec A + cot A, using the identity cosec²A = 1 + cot²A (vi) √((1 + sin A)/(1 - sin A)) = sec A + tan A (vii) (sin theta - 2 sin³theta)/(2 cos³theta - cos theta) = tan theta
(i) LHS = (cosec theta - cot theta)²
= ((1/sin theta) - (cos theta/sin theta))²
= ((1 - cos theta)/sin theta)²
= (1 - cos theta)²/sin²theta
= (1 - cos theta)²/(1 - cos²theta)
= (1 - cos theta)²/((1 - cos theta)(1 + cos theta))
= (1 - cos theta)/(1 + cos theta) = RHS. QED
(ii) LHS = cos A/(1 + sin A) + (1 + sin A)/cos A
= (cos²A + (1 + sin A)²)/((1 + sin A)cos A)
= (cos²A + 1 + 2 sin A + sin²A)/((1 + sin A)cos A)
= (1 + 1 + 2 sin A)/((1 + sin A)cos A)
= (2 + 2 sin A)/((1 + sin A)cos A)
= 2(1 + sin A)/((1 + sin A)cos A)
= 2/cos A = 2 sec A = RHS. QED
(iii) LHS = tan theta/(1 - cot theta) + cot theta/(1 - tan theta)
= (sin theta/cos theta)/(1 - cos theta/sin theta) + (cos theta/sin theta)/(1 - sin theta/cos theta)
= (sin theta/cos theta)/((sin theta - cos theta)/sin theta) + (cos theta/sin theta)/((cos theta - sin theta)/cos theta)
= sin²theta/(cos theta(sin theta - cos theta)) + cos²theta/(sin theta(cos theta - sin theta))
= sin²theta/(cos theta(sin theta - cos theta)) - cos²theta/(sin theta(sin theta - cos theta))
= (sin³theta - cos³theta)/(sin theta cos theta(sin theta - cos theta))
= (sin theta - cos theta)(sin²theta + sin theta cos theta + cos²theta)/(sin theta cos theta(sin theta - cos theta))
= (1 + sin theta cos theta)/(sin theta cos theta)
= 1/(sin theta cos theta) + 1
= sec theta cosec theta + 1 = 1 + sec theta cosec theta = RHS. QED
(iv) LHS = (1 + sec A)/sec A = 1/sec A + 1 = cos A + 1
RHS = sin²A/(1 - cos A) = (1 - cos²A)/(1 - cos A) = (1 - cos A)(1 + cos A)/(1 - cos A) = 1 + cos A
LHS = RHS. QED
(v) LHS = (cos A - sin A + 1)/(cos A + sin A - 1)
Divide numerator and denominator by sin A:
= (cot A - 1 + cosec A)/(cot A + 1 - cosec A)
= (cot A + cosec A - 1)/(cot A - cosec A + 1)
Using cosec²A - cot²A = 1, so 1 = cosec²A - cot²A = (cosec A - cot A)(cosec A + cot A):
= (cot A + cosec A - (cosec²A - cot²A))/(cot A - cosec A + 1)
= (cot A + cosec A - (cosec A - cot A)(cosec A + cot A))/(cot A - cosec A + 1)
= (cot A + cosec A)(1 - cosec A + cot A)/(cot A - cosec A + 1)
= (cot A + cosec A)(cot A - cosec A + 1)/(cot A - cosec A + 1)
= cosec A + cot A = RHS. QED
(vi) LHS = √((1 + sin A)/(1 - sin A))
Rationalise: multiply inside by (1 + sin A)/(1 + sin A)
= √((1 + sin A)²/((1 - sin A)(1 + sin A)))
= √((1 + sin A)²/(1 - sin²A))
= √((1 + sin A)²/cos²A)
= (1 + sin A)/cos A
= 1/cos A + sin A/cos A
= sec A + tan A = RHS. QED
(vii) LHS = (sin theta - 2 sin³theta)/(2 cos³theta - cos theta)
= sin theta(1 - 2 sin²theta)/(cos theta(2 cos²theta - 1))
Using sin²theta = 1 - cos²theta:
1 - 2 sin²theta = 1 - 2(1 - cos²theta) = 2 cos²theta - 1
So LHS = sin theta(2 cos²theta - 1)/(cos theta(2 cos²theta - 1))
= sin theta/cos theta = tan theta = RHS. QED
Q5Prove the following identities, where the angles involved are acute angles for which the expressions are defined: (i) (cosec A - sin A)(sec A - cos A) = 1/(tan A + cot A) (ii) ((1 + tan²A)/(1 + cot²A)) = ((1 - tan A)/(1 - cot A))² = tan²A
(i) LHS = (cosec A - sin A)(sec A - cos A)
= (1/sin A - sin A)(1/cos A - cos A)
= ((1 - sin²A)/sin A)((1 - cos²A)/cos A)
= (cos²A/sin A)(sin²A/cos A)
= sin A cos A
RHS = 1/(tan A + cot A)
= 1/(sin A/cos A + cos A/sin A)
= 1/((sin²A + cos²A)/(sin A cos A))
= sin A cos A/1
= sin A cos A
LHS = RHS. QED
(ii) Part 1: (1 + tan²A)/(1 + cot²A)
= sec²A/cosec²A
= (1/cos²A)/(1/sin²A)
= sin²A/cos²A = tan²A
Part 2: ((1 - tan A)/(1 - cot A))²
= ((1 - sin A/cos A)/(1 - cos A/sin A))²
= (((cos A - sin A)/cos A)/((sin A - cos A)/sin A))²
= ((cos A - sin A)/cos A . sin A/(sin A - cos A))²
= ((-(sin A - cos A))/cos A . sin A/(sin A - cos A))²
= (-sin A/cos A)²
= (-tan A)²
= tan²A
All three expressions equal tan²A. QED
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Ask AI TutorPrevious Year Board Exam Questions
Prove that (sin theta + cosec theta)² + (cos theta + sec theta)² = 7 + tan²theta + cot²theta.
LHS = sin²theta + 2 sin theta cosec theta + cosec²theta + cos²theta + 2 cos theta sec theta + sec²theta
= (sin²theta + cos²theta) + 2 + cosec²theta + 2 + sec²theta
= 1 + 2 + (1 + cot²theta) + 2 + (1 + tan²theta)
= 7 + tan²theta + cot²theta = RHS. Hence proved.
If sin theta + cos theta = √3, then prove that tan theta + cot theta = 1.
Given: sin theta + cos theta = √3.
Squaring: sin²theta + cos²theta + 2 sin theta cos theta = 3.
1 + 2 sin theta cos theta = 3.
sin theta cos theta = 1.
tan theta + cot theta = sin theta/cos theta + cos theta/sin theta = (sin²theta + cos²theta)/(sin theta cos theta) = 1/1 = 1. Hence proved.
Evaluate: 2 sin²30 tan 60 - 3 cos²60 sec²30.
Substituting standard trig values:
= 2(sin 30)^2(tan 60) - 3(cos 60)^2(sec 30)^2
= 2(1/2)^2(sqrt(3)) - 3(1/2)^2(2/sqrt(3))^2
= 2(1/4)(sqrt(3)) - 3(1/4)(4/3)
= sqrt(3)/2 - 1
Simplifying by taking common denominator:
= (sqrt(3) - 2)/2
The value is (sqrt(3) - 2)/2.
Prove that (1 + cot A - cosec A)(1 + tan A + sec A) = 2.
LHS = (1 + cos A/sin A - 1/sin A)(1 + sin A/cos A + 1/cos A)
= ((sin A + cos A - 1)/sin A)((cos A + sin A + 1)/cos A)
= ((sin A + cos A) - 1)((sin A + cos A) + 1)/(sin A cos A)
= ((sin A + cos A)² - 1)/(sin A cos A)
= (sin²A + cos²A + 2 sin A cos A - 1)/(sin A cos A)
= (1 + 2 sin A cos A - 1)/(sin A cos A)
= 2 sin A cos A/(sin A cos A) = 2 = RHS. Hence proved.
Evaluate: (sin²60 - cos²45) - 3 tan²30 + sin 90.
= (√3/2)² - (1/√2)² - 3(1/√3)² + 1
= 3/4 - 1/2 - 3(1/3) + 1
= 3/4 - 1/2 - 1 + 1
= 3/4 - 1/2
= 1/4
Prove that (cosec A - sin A)(sec A - cos A) = 1/(tan A + cot A).
LHS = (1/sin A - sin A)(1/cos A - cos A)
= ((1 - sin²A)/sin A)((1 - cos²A)/cos A)
= (cos²A/sin A)(sin²A/cos A)
= sin A cos A
RHS = 1/(tan A + cot A) = 1/(sin A/cos A + cos A/sin A)
= 1/((sin²A + cos²A)/(sin A cos A))
= sin A cos A
LHS = RHS. Hence proved.
If tan(A + B) = √3 and tan(A - B) = 1/√3, 0 < A + B <= 90, A > B, find A and B.
tan(A+B) = √3 = tan 60, so A + B = 60.
tan(A-B) = 1/√3 = tan 30, so A - B = 30.
Adding: 2A = 90, A = 45.
B = 60 - 45 = 15.
A = 45 degrees, B = 15 degrees.
If sin A = 1/2 and cos B = 1/2, find the value of (A + B).
Given sin A = 1/2. From the standard angle table, sin 30 = 1/2. Since A is acute, A = 30 degrees.
Given cos B = 1/2. From the standard angle table, cos 60 = 1/2. Since B is acute, B = 60 degrees.
Therefore, A + B = 30 + 60 = 90 degrees.
Prove that (1 - sin theta)/(1 + sin theta) = (sec theta - tan theta)².
RHS = (sec theta - tan theta)²
= (1/cos theta - sin theta/cos theta)²
= ((1 - sin theta)/cos theta)²
= (1 - sin theta)²/cos²theta
= (1 - sin theta)²/(1 - sin²theta)
= (1 - sin theta)²/((1 - sin theta)(1 + sin theta))
= (1 - sin theta)/(1 + sin theta) = LHS. Hence proved.
If 4 sin theta = 3, find the value of (4 sin²theta - 3 cos²theta + 2).
4 sin theta = 3, so sin theta = 3/4.
cos²theta = 1 - sin²theta = 1 - 9/16 = 7/16.
4 sin²theta - 3 cos²theta + 2
= 4(9/16) - 3(7/16) + 2
= 9/4 - 21/16 + 2
= 36/16 - 21/16 + 32/16
= 47/16
Evaluate: sin²60 + 2 tan 45 - cos²30.
Substituting the standard trig values for 60, 45, and 30 degrees:
= (sin 60)^2 + 2(tan 45) - (cos 30)^2
= (sqrt(3)/2)^2 + 2(1) - (sqrt(3)/2)^2
= 3/4 + 2 - 3/4
The sin^2 60 and cos^2 30 terms cancel since sin 60 = cos 30 = sqrt(3)/2.
The value is 2.
If sin A = cos B, where A and B are acute angles, find the value of A + B.
Given sin A = cos B. Using the complementary angle identity, cos B = sin(90 - B). So sin A = sin(90 - B). Since both A and B are acute angles, A = 90 - B. Rearranging gives A + B = 90 degrees. This confirms that A and B are complementary angles.
Prove that (cos A/(1 + sin A)) + ((1 + sin A)/cos A) = 2 sec A.
LHS = (cos²A + (1 + sin A)²)/((1 + sin A) cos A)
= (cos²A + 1 + 2 sin A + sin²A)/((1 + sin A) cos A)
= (1 + 1 + 2 sin A)/((1 + sin A) cos A)
= 2(1 + sin A)/((1 + sin A) cos A)
= 2/cos A = 2 sec A = RHS. Hence proved.
If sec theta = 2/√3, find the value of (1 + tan theta sin theta)/(1 + sec theta).
sec theta = 2/√3, so cos theta = √3/2, theta = 30.
tan 30 = 1/√3, sin 30 = 1/2.
(1 + (1/√3)(1/2))/(1 + 2/√3)
= (1 + 1/(2√3))/(1 + 2/√3)
= ((2√3 + 1)/(2√3))/((√3 + 2)/√3)
= (2√3 + 1)/(2√3) . √3/(√3 + 2)
= (2√3 + 1)/(2(√3 + 2))
= (2√3 + 1)/(2√3 + 4)
Rationalising is not needed. Substituting values:
= (2(1.732) + 1)/(2(1.732) + 4)
= 4.464/7.464
Alternatively: numerator = 1 + (1/√3)(1/2) = 1 + 1/(2√3) = (2√3 + 1)/(2√3)
denominator = 1 + 2/√3 = (√3 + 2)/√3
Result = ((2√3+1)/(2√3))/((√3+2)/√3) = (2√3+1)/(2(√3+2))
Multiply by (2-√3)/(2-√3): not simplifiable to a clean form.
Using exact values: = (2√3+1)/(2√3+4) = (2√3+1)/(2(√3+2))
If sin(A + 2B) = √3/2 and cos(A + 4B) = 0, A > B, find A and B.
sin(A + 2B) = √3/2 = sin 60, so A + 2B = 60 ... (i)
cos(A + 4B) = 0 = cos 90, so A + 4B = 90 ... (ii)
Subtracting (i) from (ii): 2B = 30, B = 15.
From (i): A + 30 = 60, A = 30.
A = 30 degrees, B = 15 degrees.
Prove that (sin theta - cos theta + 1)/(sin theta + cos theta - 1) = 1/(sec theta - tan theta).
LHS = (sin theta - cos theta + 1)/(sin theta + cos theta - 1)
Divide numerator and denominator by cos theta:
= (tan theta - 1 + sec theta)/(tan theta + 1 - sec theta)
= (tan theta + sec theta - 1)/(tan theta - sec theta + 1)
Using sec²theta - tan²theta = 1, so 1 = (sec theta - tan theta)(sec theta + tan theta):
Numerator = (tan theta + sec theta) - (sec²theta - tan²theta)
= (tan theta + sec theta) - (sec theta - tan theta)(sec theta + tan theta)
= (tan theta + sec theta)[1 - (sec theta - tan theta)]
= (tan theta + sec theta)(1 - sec theta + tan theta)
So LHS = (tan theta + sec theta)(1 - sec theta + tan theta)/(tan theta - sec theta + 1)
= (tan theta + sec theta)(tan theta - sec theta + 1)/(tan theta - sec theta + 1)
= tan theta + sec theta
RHS = 1/(sec theta - tan theta)
Rationalise: multiply by (sec theta + tan theta)/(sec theta + tan theta)
= (sec theta + tan theta)/(sec²theta - tan²theta)
= (sec theta + tan theta)/1 = sec theta + tan theta
LHS = RHS. Hence proved.
Want more PYQ-style practice? AI generates unlimited similar questions from this chapter.
Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): sin²30 + cos²30 = 1.
Reason (R): For any angle A, sin²A + cos²A = 1.
Answer: (A)
Both are true. sin²30 + cos²30 = (1/2)² + (√3/2)² = 1/4 + 3/4 = 1. The reason (Pythagorean identity) correctly explains why the assertion is true.
AR2
Assertion (A): The value of sin 60 = 2 sin 30 cos 30.
Reason (R): sin 2A = 2 sin A cos A is a valid identity.
Answer: (A)
Assertion: 2 sin 30 cos 30 = 2(1/2)(√3/2) = √3/2 = sin 60. True. The double angle formula (reason) is a valid identity and correctly explains the assertion.
AR3
Assertion (A): tan 30 = 1/√3 and tan 60 = √3.
Reason (R): tan A = sin A/cos A.
Answer: (A)
Both are true. tan 30 = sin 30/cos 30 = (1/2)/(√3/2) = 1/√3. tan 60 = sin 60/cos 60 = (√3/2)/(1/2) = √3. The reason correctly explains the assertion.
AR4
Assertion (A): sin(A + B) = sin A + sin B for all angles A and B.
Reason (R): Trigonometric functions are not linear functions.
Answer: (D)
Assertion is FALSE: for example, sin(30+60) = sin 90 = 1, but sin 30 + sin 60 = 1/2 + √3/2 which is not equal to 1. Reason is TRUE: trigonometric functions do not satisfy f(x+y) = f(x) + f(y), i.e., they are not linear.
AR5
Assertion (A): If cos A = 4/5, then tan A = 3/4.
Reason (R): tan A = sin A/cos A.
Answer: (A)
If cos A = 4/5, then sin A = 3/5 (using sin²A + cos²A = 1). So tan A = sin A/cos A = (3/5)/(4/5) = 3/4. Both assertion and reason are true, and the reason explains the assertion.
AR6
Assertion (A): cosec 30 = 2 and sec 60 = 2.
Reason (R): cosec A = 1/sin A and sec A = 1/cos A.
Answer: (A)
cosec 30 = 1/sin 30 = 1/(1/2) = 2. sec 60 = 1/cos 60 = 1/(1/2) = 2. Both assertion and reason are true, and the reciprocal relations (reason) explain the assertion.
Case Study Questions
A surveyor is measuring the dimensions of a triangular plot PQR, right-angled at Q. She measures PQ = 5 m and QR = 12 m. She uses trigonometric ratios to calculate various measurements needed for her report.
(i) Find the value of sin P. (1 mark)
By Pythagoras theorem: PR = √(PQ² + QR²) = √(25 + 144) = √169 = 13 m.
sin P = opposite/hypotenuse = QR/PR = 12/13.
(ii) What is the value of cos P + cos R? (1 mark)
cos P = PQ/PR = 5/13.
cos R = QR/PR = 12/13.
cos P + cos R = 5/13 + 12/13 = 17/13.
(iii)(a) Find the value of sin²P + sin²R. (2 marks)
sin P = 12/13, so sin²P = 144/169.
sin R = PQ/PR = 5/13, so sin²R = 25/169.
sin²P + sin²R = 144/169 + 25/169 = 169/169 = 1.
Note: This is expected because P + R = 90, so sin R = cos P, and sin²P + cos²P = 1.
In a mathematics class, the teacher draws a right triangle on the board and labels the sides. The hypotenuse is 10 cm, one side is 5√3 cm and the other side is 5 cm. She asks students to find the trigonometric ratios for the acute angles of the triangle.
(i) If the angle opposite to the side 5 cm is A, find the value of angle A. (1 mark)
sin A = opposite/hypotenuse = 5/10 = 1/2.
Since sin 30 = 1/2, angle A = 30 degrees.
(ii) What is the value of tan A + cot A? (1 mark)
A = 30 degrees.
tan 30 = 1/√3, cot 30 = √3.
tan 30 + cot 30 = 1/√3 + √3 = (1 + 3)/√3 = 4/√3 = 4√3/3.
(iii)(a) Find the value of sin²A + cos²A and verify the Pythagorean identity. (2 marks)
sin 30 = 1/2, cos 30 = √3/2.
sin²30 + cos²30 = 1/4 + 3/4 = 1.
This verifies the identity sin²A + cos²A = 1.
MCQ Practice
MCQ1. If sin A = 1/2, then the value of cot A is:
(A) √3
(B) 1/√3
(C) √3/2
(D) 1
Answer: (A) - sin A = 1/2, so A = 30. cot 30 = cos 30/sin 30 = (√3/2)/(1/2) = √3.
MCQ2. The value of (sin 45 + cos 45) is:
(A) 1/√2
(B) √2
(C) √3/2
(D) 1
Answer: (B) - sin 45 + cos 45 = 1/√2 + 1/√2 = 2/√2 = √2.
MCQ3. If cos A = 4/5, then tan A is:
(A) 3/5
(B) 3/4
(C) 4/3
(D) 5/3
Answer: (B) - cos A = 4/5, so adjacent = 4, hypotenuse = 5. opposite = √(25-16) = 3. tan A = 3/4.
MCQ4. The value of sin²30 + cos²60 is:
(A) 0
(B) 1/2
(C) 1
(D) 1/4
Answer: (B) - sin²30 + cos²60 = (1/2)² + (1/2)² = 1/4 + 1/4 = 1/2.
MCQ5. If tan theta = 12/5, then sin theta is:
(A) 12/13
(B) 5/13
(C) 13/12
(D) 5/12
Answer: (A) - tan theta = 12/5, so opposite = 12, adjacent = 5, hypotenuse = √(144+25) = 13. sin theta = 12/13.
MCQ6. The value of 2 tan 45 + cos 30 - sin 60 is:
(A) 0
(B) 1
(C) 2
(D) 3
Answer: (C) - 2 tan 45 + cos 30 - sin 60 = 2(1) + √3/2 - √3/2 = 2.
MCQ7. If sin theta - cos theta = 0, then the value of theta is:
(A) 0
(B) 30
(C) 45
(D) 60
Answer: (C) - sin theta = cos theta. Dividing by cos theta: tan theta = 1, so theta = 45 degrees.
MCQ8. sec²60 - tan²60 is equal to:
(A) 0
(B) 1
(C) 2
(D) 3
Answer: (B) - By the identity 1 + tan²A = sec²A, we get sec²A - tan²A = 1. So sec²60 - tan²60 = 1.
MCQ9. If sin(A - B) = 1/2 and cos(A + B) = 1/2, where 0 < A + B <= 90, then the values of A and B are:
(A) A = 45, B = 15
(B) A = 30, B = 0
(C) A = 35, B = 25
(D) A = 60, B = 0
Answer: (A) - sin(A-B) = 1/2 = sin 30, so A - B = 30. cos(A+B) = 1/2 = cos 60, so A + B = 60. Adding: 2A = 90, A = 45, B = 15.
MCQ10. The value of (1 + tan²theta)(1 - sin theta)(1 + sin theta) is:
(A) 0
(B) 1
(C) 2
(D) sin²theta
Answer: (B) - (1 + tan²theta)(1 - sin²theta) = sec²theta . cos²theta = (1/cos²theta) . cos²theta = 1.
MCQ11. If sec theta = 25/7, then sin theta is:
(A) 7/25
(B) 24/25
(C) 25/24
(D) 7/24
Answer: (B) - sec theta = 25/7, cos theta = 7/25. sin theta = √(1 - cos²theta) = √(1 - 49/625) = √(576/625) = 24/25.
MCQ12. The value of sin 30.cos 60 + cos 30.sin 60 is:
(A) 0
(B) 1/2
(C) 1
(D) √3/2
Answer: (C) - sin 30.cos 60 + cos 30.sin 60 = (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1. Alternatively, this equals sin(30+60) = sin 90 = 1.
Try Quiz mode - 10 MCQs per topic with instant feedback and scoring.
Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
If sin A = 1/2, find the value of 2 cos A - 1.
sin A = 1/2, so A = 30. cos 30 = √3/2. 2 cos A - 1 = 2(√3/2) - 1 = √3 - 1.
What is the value of sin²45 + cos²45?
(1/√2)² + (1/√2)² = 1/2 + 1/2 = 1.
If tan theta = 1, find the value of sec theta.
tan theta = 1, so theta = 45. sec 45 = √2.
Find the value of sin 60/cos 30.
sin 60/cos 30 = (√3/2)/(√3/2) = 1.
If cos A = 12/13, find sin A.
sin A = √(1 - cos²A) = √(1 - 144/169) = √(25/169) = 5/13.
2-3 Mark Questions
Evaluate: (sin²60 - cos²45 - 3 tan²30)/(2 sin 30 cos 30).
= (3/4 - 1/2 - 3(1/3))/(2(1/2)(√3/2))
= (3/4 - 1/2 - 1)/(√3/2)
= (3/4 - 2/4 - 4/4)/(√3/2)
= (-3/4)/(√3/2)
= -3/4 . 2/√3 = -3/(2√3) = -√3/2.
If tan(A + B) = √3 and tan(A - B) = 1/√3, find A and B (0 < A + B <= 90, A > B).
A + B = 60 and A - B = 30. Adding: 2A = 90, A = 45. B = 60 - 45 = 15. A = 45, B = 15.
If sin theta + cos theta = √2, find the value of tan theta + cot theta.
Squaring: sin²theta + cos²theta + 2 sin theta cos theta = 2. So 1 + 2 sin theta cos theta = 2, giving sin theta cos theta = 1/2. tan theta + cot theta = (sin²theta + cos²theta)/(sin theta cos theta) = 1/(1/2) = 2.
If sec 5A = cosec(A - 36), where 5A is an acute angle, find the value of A.
cosec(90 - 5A) = cosec(A - 36). So 90 - 5A = A - 36. 126 = 6A. A = 21 degrees.
3-4 Mark Questions
Prove that (cosec theta - cot theta)² = (1 - cos theta)/(1 + cos theta).
LHS = ((1 - cos theta)/sin theta)²
= (1 - cos theta)²/sin²theta
= (1 - cos theta)²/(1 - cos²theta)
= (1 - cos theta)²/((1 + cos theta)(1 - cos theta))
= (1 - cos theta)/(1 + cos theta) = RHS.
Prove that (sin theta - cos theta + 1)/(sin theta + cos theta - 1) = 1/(sec theta - tan theta).
Divide numerator and denominator by cos theta:
= (tan theta - 1 + sec theta)/(tan theta + 1 - sec theta)
Substitute 1 = sec²theta - tan²theta:
= (tan theta + sec theta - (sec theta - tan theta)(sec theta + tan theta))/(tan theta - sec theta + 1)
= (tan theta + sec theta)(1 - sec theta + tan theta)/(tan theta - sec theta + 1)
= tan theta + sec theta = 1/(sec theta - tan theta). [After rationalising RHS]
Prove that (1 + cot A - cosec A)(1 + tan A + sec A) = 2.
LHS = ((sin A + cos A - 1)/sin A)((cos A + sin A + 1)/cos A)
= ((sin A + cos A)² - 1)/(sin A cos A)
= (1 + 2 sin A cos A - 1)/(sin A cos A)
= 2 sin A cos A/(sin A cos A) = 2 = RHS.
Key Concepts
Important Formulas
Study Plan - Introduction to Trigonometry
Trigonometric Ratios (Exercise 8.1)
Read NCERT theory on trig ratios in right triangles. Memorise the six ratios and their reciprocal relationships. Solve Exercise 8.1 Q1-Q5 (finding ratios from given sides).
More on Trig Ratios and Verification (Exercise 8.1)
Complete Exercise 8.1 Q6-Q11. Focus on given-one-ratio-find-others type questions and identity verification problems.
Standard Angle Values (Exercise 8.2)
Memorise the trigonometric table for 0, 30, 45, 60, 90 degrees. Solve all 4 questions of Exercise 8.2. Practice evaluating complex expressions using standard values.
Complementary Angles (Exercise 8.3)
Learn complementary angle identities: sin(90-A) = cos A, etc. Solve all 7 questions of Exercise 8.3. Practice converting ratios of angles above 45 to angles below 45.
Trigonometric Identities (Exercise 8.4)
Learn three fundamental identities: sin²A + cos²A = 1, 1 + tan²A = sec²A, 1 + cot²A = cosec²A. Solve Exercise 8.4 Q1-Q5. Focus on step-by-step identity proofs.
Revision and PYQs
Solve previous year CBSE questions from this chapter. Practice MCQs, assertion-reason, and case study questions. Time yourself on identity proofs (target: 5 minutes each).
Common Mistakes to Avoid
- ✗Confusing sin with cosec and cos with sec. Remember: sin is opposite/hypotenuse, cosec is hypotenuse/opposite (reciprocal of sin, not cos)
- ✗Using sin(A + B) = sin A + sin B, which is WRONG. Trigonometric functions are not linear. For example, sin(30 + 60) = sin 90 = 1, but sin 30 + sin 60 = 1/2 + √3/2 which does not equal 1
- ✗Forgetting to check whether the angle is acute when using complementary angle formulas. The formula sin(90 - A) = cos A only applies when both A and (90 - A) are in the valid range
- ✗Writing tan 90 = 1 or cot 0 = 0. Both tan 90 and cot 0 are UNDEFINED (division by zero). Similarly, sec 90 and cosec 0 are undefined
- ✗In identity proofs, working on both LHS and RHS simultaneously and equating them. The correct approach is to start from one side (usually the more complex one) and simplify step-by-step to reach the other side
- ✗Mixing up the standard angle values, especially sin 30 vs cos 30. Use the mnemonic: sin values for 0, 30, 45, 60, 90 are √0/2, √1/2, √2/2, √3/2, √4/2 (i.e., 0, 1/2, 1/√2, √3/2, 1)
Tips for Board Exam
- ✓Memorise the trig values table using the pattern: for sin, the values at 0, 30, 45, 60, 90 are √0/2, √1/2, √2/2, √3/2, √4/2. For cos, reverse the order. This one trick eliminates memorisation errors
- ✓Identity proofs carry 3-4 marks and appear in every board exam. Always start from the more complex side (usually LHS) and simplify. Write every step clearly as examiners award step marks
- ✓For evaluation questions using standard angles, substitute the values carefully and simplify fractions step by step. Avoid mental math with surds as sign errors are common
- ✓In complementary angle questions, always convert all angles to either below 45 or above 45 first, then simplify. Look for pairs that add to 90: sin 63 and cos 27, tan 48 and cot 42
- ✓When proving identities involving sec and cosec, convert everything to sin and cos first, then simplify. This approach works for almost every identity proof in the CBSE exam
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Start Introduction to Trigonometry - Choose Your ModeAbout Introduction to Trigonometry - Class 10 Mathematics
What is Trigonometry?
Trigonometry is the branch of mathematics that studies the relationships between the sides and angles of triangles. The word comes from the Greek words 'tri' (three), 'gon' (side), and 'metron' (measure). In CBSE Class 10 Mathematics, Chapter 8 introduces trigonometry through right-angled triangles, where students learn to define and compute six trigonometric ratios for acute angles. This chapter is part of the Trigonometry unit, which carries 12 marks in the CBSE 2027 board exam, with Chapter 8 alone contributing 6-8 marks. The concepts learned here form the foundation for Chapter 9 (Applications of Trigonometry) and are also essential for higher classes.
Trigonometric Ratios
In a right-angled triangle, the trigonometric ratios of an acute angle are defined as ratios of the sides of the triangle. Consider a right triangle ABC with the right angle at B. For angle A:
- sin A = opposite side / hypotenuse = BC / AC
- cos A = adjacent side / hypotenuse = AB / AC
- tan A = opposite side / adjacent side = BC / AB
- cosec A = 1 / sin A = AC / BC
- sec A = 1 / cos A = AC / AB
- cot A = 1 / tan A = AB / BC
These ratios depend only on the angle, not on the size of the triangle. If two right triangles have the same acute angle, the ratios of their corresponding sides will be equal. This is because all right triangles with the same acute angle are similar to each other (AA similarity). The ratios are connected by reciprocal relations: cosec A = 1/sin A, sec A = 1/cos A, cot A = 1/tan A, and by quotient relations: tan A = sin A/cos A and cot A = cos A/sin A.
When solving problems from Exercise 8.1, the typical approach is: (a) identify the right angle and the angle in question, (b) label the sides as opposite, adjacent, and hypotenuse relative to that angle, (c) use the Pythagoras theorem to find any missing side, and (d) compute the required ratios.
Standard Angle Values
The trigonometric ratios for the standard angles 0, 30, 45, 60, and 90 degrees can be calculated geometrically and must be memorised. Here is the complete table:
For sin: sin 0 = 0, sin 30 = 1/2, sin 45 = 1/√2, sin 60 = √3/2, sin 90 = 1. For cos: cos 0 = 1, cos 30 = √3/2, cos 45 = 1/√2, cos 60 = 1/2, cos 90 = 0. For tan: tan 0 = 0, tan 30 = 1/√3, tan 45 = 1, tan 60 = √3, tan 90 = undefined.
A useful mnemonic: sin values at 0, 30, 45, 60, 90 follow the pattern √0/2, √1/2, √2/2, √3/2, √4/2. The cos values are the same sequence in reverse order. The tan values are obtained by dividing sin by cos for each angle. Note that tan 90, cot 0, sec 90, and cosec 0 are all undefined because they involve division by zero.
Exercise 8.2 tests the ability to evaluate expressions by substituting these standard values. Common question types include direct evaluation (like sin 60 cos 30 + sin 30 cos 60), solving for unknown angles (like finding A if tan(A+B) = √3), and verifying which standard angle satisfies a given equation.
Complementary Angles
Two angles are complementary if their sum is 90 degrees. In a right triangle, the two acute angles are always complementary. This leads to important relationships between trigonometric ratios of complementary angles:
- sin(90 - A) = cos A and cos(90 - A) = sin A
- tan(90 - A) = cot A and cot(90 - A) = tan A
- sec(90 - A) = cosec A and cosec(90 - A) = sec A
These identities are extremely useful for simplifying expressions. For example, sin 72 = sin(90 - 18) = cos 18. This means sin 72/cos 18 = cos 18/cos 18 = 1. Exercise 8.3 is entirely based on these complementary angle relationships. A typical strategy is to look for pairs of angles that sum to 90 and convert one of them using the complementary identity.
Trigonometric Identities
A trigonometric identity is an equation involving trigonometric ratios that is true for all values of the angle for which the ratios are defined. Chapter 8 introduces three fundamental identities:
- sin²A + cos²A = 1 (the Pythagorean identity)
- 1 + tan²A = sec²A (derived by dividing identity 1 by cos²A)
- 1 + cot²A = cosec²A (derived by dividing identity 1 by sin²A)
These identities are the building blocks for all identity proofs in Exercise 8.4. To prove that one expression equals another, start from one side (usually the more complex side) and use algebraic manipulation along with these identities to arrive at the other side. Common techniques include:
- Converting all terms to sin and cos
- Factoring using a² - b² = (a+b)(a-b)
- Multiplying numerator and denominator by conjugates to rationalise
- Recognising patterns like 1 - sin²A = cos²A or 1 - cos²A = sin²A
For the identity (cosec A - cot A)² = (1 - cos A)/(1 + cos A), the approach is to write LHS in terms of sin and cos, simplify the numerator to get (1 - cos A)², and use sin²A = (1 - cos A)(1 + cos A) in the denominator to cancel one factor.
How Tested in Board Exams
Chapter 8 is one of the most consistently tested chapters in CBSE Class 10 Mathematics. The board exam typically includes:
- 1-2 MCQs (1 mark each): evaluating simple expressions like sin²60 + cos²30, finding a ratio given another ratio, or identifying correct/incorrect statements about trig functions.
- 1 short answer (2-3 marks): evaluating a complex expression using standard values, finding unknown angles from equations like tan(A+B) = √3, or numerical verification of an identity.
- 1 long answer (3-4 marks): proving a trigonometric identity. This is the highest-value question from this chapter and appears in virtually every board exam.
- Assertion-Reason and Case Study: these newer question formats test conceptual understanding of trig ratios and identities.
Since 2023, CBSE has favoured multi-step identity proofs that combine two or more fundamental identities, and evaluation questions that require careful handling of surds.
Tips for Full Marks
To score full marks in trigonometry questions: (1) Memorise the standard values table perfectly. (2) In identity proofs, always work from one side only and show every algebraic step. (3) For evaluation questions, write the substituted values before computing. (4) Double-check undefined cases (tan 90, cot 0, sec 90, cosec 0). (5) In complementary angle problems, first identify all angle pairs summing to 90.
Connection to Other Chapters
Chapter 8 is directly connected to Chapter 9 (Some Applications of Trigonometry), where the ratios learned here are applied to real-world problems involving heights and distances. The Pythagoras theorem from Chapter 6 (Triangles) is frequently used in Chapter 8 for finding missing sides. The algebraic manipulation skills from Chapter 2 (Polynomials) and Chapter 4 (Quadratic Equations) are essential for simplifying identity proofs. In higher classes, students will extend these concepts to trigonometric functions, graphs, and general angle formulas.
Frequently Asked Questions
How many exercises are in NCERT Class 10 Maths Chapter 8 for CBSE 2026-27?
Chapter 8 Introduction to Trigonometry has 4 exercises for CBSE 2026-27: Exercise 8.1 (11 questions on trigonometric ratios), Exercise 8.2 (4 questions on standard angle values), Exercise 8.3 (7 questions on complementary angles), and Exercise 8.4 (5 questions on trigonometric identities). No exercises have been removed from this chapter.
How many marks does Chapter 8 carry in the CBSE Class 10 board exam?
Chapter 8 is part of the Trigonometry unit which carries 12 marks total. Chapter 8 alone typically contributes 6-8 marks, with the remaining marks coming from Chapter 9 (Applications of Trigonometry). At minimum, expect 1 MCQ (1 mark) and 1 identity proof (3-4 marks) from this chapter.
What are the most important topics in Chapter 8 for the board exam?
The most important topics are: (1) Trigonometric identity proofs from Exercise 8.4 (guaranteed 3-4 mark question), (2) Evaluation of expressions using standard angle values from Exercise 8.2 (common MCQ/short answer), (3) Finding unknown angles using complementary angle relations from Exercise 8.3, and (4) Finding trigonometric ratios when one ratio is given from Exercise 8.1.
How do I memorise the trigonometric values table for standard angles?
Use this pattern: for sin, the values at 0, 30, 45, 60, 90 degrees are √0/2, √1/2, √2/2, √3/2, √4/2 (which simplify to 0, 1/2, 1/√2, √3/2, 1). For cos, just reverse the order: cos 0 = 1, cos 30 = √3/2, cos 45 = 1/√2, cos 60 = 1/2, cos 90 = 0. For tan, divide sin by cos for each angle.
What is the best approach for proving trigonometric identities?
Start from the more complex side (usually LHS). Convert all trigonometric ratios to sin and cos. Use algebraic techniques like taking LCM, factoring, and rationalising. Apply the three fundamental identities: sin²A + cos²A = 1, 1 + tan²A = sec²A, and 1 + cot²A = cosec²A. Work step by step until you reach the other side. Never work on both sides simultaneously.
Why are tan 90 and cot 0 undefined?
tan 90 = sin 90/cos 90 = 1/0, which involves division by zero and is therefore undefined. Similarly, cot 0 = cos 0/sin 0 = 1/0, which is also undefined. In general, any trigonometric ratio that involves division by zero at a particular angle is undefined at that angle. sec 90 and cosec 0 are also undefined for the same reason.
What is the connection between Chapter 8 and Chapter 9?
Chapter 8 teaches the definitions, standard values, and identities of trigonometric ratios. Chapter 9 applies these ratios to solve real-world problems involving heights and distances (like finding the height of a tower or the distance of a ship). You need to master Chapter 8 thoroughly before attempting Chapter 9, as every problem in Chapter 9 uses the ratios and values learned here.
What types of questions from Chapter 8 appear in CBSE board exams?
Typical question types include: (1) MCQs on evaluating simple expressions or identifying correct trig values (1 mark), (2) Short answer questions evaluating complex expressions or finding unknown angles (2-3 marks), (3) Long answer identity proofs (3-4 marks), (4) Assertion-Reason questions testing conceptual understanding, and (5) Case study questions based on right triangle scenarios. At least one identity proof question appears in every board exam.