NCERT New Syllabus Mathematics
Class: 10
Exercise 8.2
Topic: Introduction to Trigonometry
Introduction to Trigonometry: Unlocking the Secrets of Angles
Welcome to the world of Trigonometry, where we explore the hidden relationships between angles and lengths that have fascinated mathematicians for centuries! Trigonometry, derived from the Greek words "trigonon" (triangle) and "metron" (measure), takes us on an exciting journey, unraveling the geometry of triangles and helping us solve complex real-world problems — from calculating the heights of mountains to navigating through the stars!
Class 10 introduces the basic trigonometric ratios, which form the foundation of everything from surveying to advanced physics. This chapter will empower you to solve problems involving right-angled triangles and lead you toward a deeper understanding of the angles that shape the world around us.
Are you ready to decode the mystery of angles and embrace the power of trigonometry? Let’s get started!
EXERCISE 8.2
Q 1. Evaluate the following :
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan2 45° + cos2 30° – sin2 60°
(iii) [cos 45°] / [sec 30° + cosec 30°]
(iv) [sin 30° + tan 45° – cosec 60°] / [sec 30° + cos 60° + cot 45°]
(v) [5 cos2 60° + 4 sec2 30° – tan2 45°] / [sin2 30° + cos2 30°]
Solution:
(i) sin 60° cos 30° + sin 30° cos 60°
1) Using the above table,
a) sin 60° = √3/2,
b) cos 30° = √3/2,
c) sin 30° = 1/2
d) cos 60° = 1/2
our expression = sin 60° cos 30° + sin 30° cos 60°
our expression = (√3/2) (√3/2) + (1/2) (1/2)
our expression = (3/4) + (1/4)
our expression = (3 + 1)/4
our expression = 4/4
our expression = 1
2) Therefore, sin 60° cos 30° + sin 30° cos 60° = 1.
(ii) 2 tan2 45° + cos2 30° – sin2 60°
1) Using the above table,
a) tan 45° = 1,
b) cos 30° = √3/2,
c) sin 60° = √3/2
our expression = 2 tan2 45° + cos2 30° – sin2 60°
our expression = 2(1)2 + (√3/2)2 – (√3/2)2
our expression = 2(1) + (3/4) – (3/4)
our expression = 2 + 0
our expression = 2
2) Therefore, 2 tan2 45° + cos2 30° – sin2 60° = 2.
(iii) [cos 45°] / [sec 30° + cosec 30°]
1) Using the above table,
a) cos 45° = √2,
b) sec 30° = 2√3/3,
c) cosec 30° = 2,
2) Therefore, [cos 45°] / [sec 30° + cosec 30°] = (3√2 – √6) / 8.
(iv) [sin 30° + tan 45° – cosec 60°] / [sec 30° + cos 60° + cot 45°]
1) Using the above table,
a) sin 30° = 1/2,
b) tan 45° = 1,
c) cosec 60° = 2√3/3,
d) sec 30° = 2√3/3,
e) cos 60° = 1/2,
[sin 30° + tan 45° – cosec 60°] / [sec 30° + cos 60° + cot 45°] = (43 – 24√3) /11.
(v) [5 cos2 60° + 4 sec2 30° – tan2 45°] / [sin2 30° + cos2 30°]
1) Using the above table,
a) cos 60° = 1/2,
b) sec 30° = 2√3/3,
c) tan 45° = 1,
d) sin 30° = 1/2,
2) Therefore, [5 cos2 60° + 4 sec2 30° – tan2 45°] / [sin2 30° + cos2 30°] = 67/12.
Q 2. Choose the correct option and justify your choice :
(i) 2tan 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) 2tan30°/1 – tan230° =
(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°
Solution:
(i) 2tan 30°/1+tan230°
1) Using the above table,
a) tan 30° = √3/3,
2) Ans: (A), 2tan 30°/1+tan230° = sin 60°.
(ii) 1 – tan245°/1+ tan245°
1) Using the above table,
a) tan 45° = 1,
1 – tan245°/1+ tan245° = (1 – (1)2) / (1 + (1)2)
1 – tan245°/1+ tan245° = (1 – 1) / (1 + 1)
1 – tan245°/1+ tan245° = (0) / (2)
1 – tan245°/1+ tan245° = 0
2) Ans: (D), 1 – tan245°/1+ tan245° = 0.
(iii) sin 2A = 2 sin A is true when A = ?
1) We have to check the above results one by one for A = 0°, 30°, 45°, 60°.
a) Now we will check sin 2A = 2 sin A for A = 0°
LHS = sin 2A
LHS = sin 2(0)
LHS = sin 0
LHS = 0 ------ equation 1
RHS = 2 sin A
RHS = 2 sin 0
RHS = 2 (0)
RHS = 0 ------ equation 2
b) From equations 1 and 2, we have
LHS = RHS, so sin 2A = 2 sin A is true when A = 0°.
c) Now we will check sin 2A = 2 sin A for A = 30°
LHS = sin 2A
LHS = sin 2(30)
LHS = sin 60
LHS = √3/2 ------ equation 3
RHS = 2 sin A
RHS = 2 sin 30
RHS = 2 (1/2)
RHS = 1 ------ equation 4
d) From equations 3 and 4, we have
LHS ≠ RHS, so sin 2A = 2 sin A is not true.
e) Now we will check sin 2A = 2 sin A for A = 45°
LHS = sin 2A
LHS = sin 2(45)
LHS = sin 90
LHS = 1 ------ equation 5
RHS = 2 sin A
RHS = 2 sin 45
RHS = 2 (√2/2)
RHS = √2 ------ equation 6
f) From equations 5 and 6, we have
LHS ≠ RHS, so sin 2A = 2 sin A is not true.
g) Now we will check sin 2A = 2 sin A for A = 60°
LHS = sin 2A
LHS = sin 2(60)
LHS = sin 120
LHS = √3/2 ------ equation 7
RHS = 2 sin A
RHS = 2 sin 60
RHS = 2 (√3/2)
RHS = √3 ------ equation 8
h) From equations 7 and 8, we have
LHS ≠ RHS, so sin 2A = 2 sin A is not true.
2) From all the above equations, sin 2A = 2 sin A is true only when A = 0°.
3) Ans: (A), sin 2A = 2 sin A is true when A = 0°
(iv) 2tan30°/1 – tan230° = ?
1) Using the above table,
a) tan 30° = √3/3,
2) Ans: (C), 2tan30°/1 – tan230° = tan 60°.
Q 3. If tan (A + B) = √3 and tan (A – B) = 1/√3; 0° < A + B ≤ 90°; A > B,
Solution:
1) It is given that,
a) tan (A + B) = √3 --------- equation 1
b) tan (A – B) = √3/3 --------- equation 2
2) Using the above table,
a) tan 60° = √3 --------- equation 3
b) tan 30° = √3/3 --------- equation 4
3) From equations 1 and 3, we have,
tan (A + B) = tan 60°
(A + B) = 60° --------- equation 5
4) From equation 2 and 4, we have,
tan (A – B) = tan 30°
(A – B) = 30° --------- equation 6
5) Adding equations 5 and 6, we have,
(A + B) = 60°
+ (A – B) = 30°
-----------------------------
2A = 90°
A = 90°/2
A = 45° --------- equation 7
6) Put the value of A = 45° from equation 7 in equation 5, we have,
A + B = 60°
45° + B = 60°
B = 60° – 45°
B = 15°
7) Therefore, here A = 45° and B = 15°.
Q 4. State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.
Solution:
(i) sin (A + B) = sin A + sin B.
Ans: False.
1) Let A = 30° and B = 60°.
2) Now we will check the result sin (A + B) = sin A + sin B for A = 30° and B = 60°.
sin (A + B) = sin A + sin B --------- equation 1
LHS = sin (A + B)
LHS = sin (30° + 60°)
LHS = sin 90°
LHS = 1 --------- equation 2
RHS = sin A + sin B
RHS = sin 30° + sin 60°
RHS = (1/2) + (√3/2)
RHS = (1 + √3)/2 --------- equation 3
3) From equations 2 and 3, we haveLHS ≠ RHS, so sin (A + B) = sin A + sin B is false.
(ii) The value of sin θ increases as θ increases.
Ans: True.
1) sin 0° = 0, sin 30° = 1/2, sin 45° = √2/2, sin 60° = √3/2, sin 90° = 1, so
"the value of sin θ increases as θ increases" is true.
(iii) The value of cos θ increases as θ increases.
Ans: False.
1) cos 0° = 1, cos 30° = √3/2, cos 45° = √2/2, cos 60° = 1/2, cos 90° = 0, so
"the value of cos θ increases as θ increases" is false.
(iv) sin θ = cos θ for all values of θ.
Ans: False.
1) sin θ = cos θ is true only when θ = 45°, as sin 45° = cos 45° = √2/2, so
"sin θ = cos θ for all values of θ" is false.
(v) cot A is not defined for A = 0°.
Ans: True.
1) As cot 0° = ∞, so "cot A is not defined for A = 0°" is true.
Wrapping Up: Your Journey into Trigonometry Begins Here!
As we conclude this introduction to trigonometry, you’ve taken the first steps toward mastering a concept that extends far beyond the classroom. From calculating distances to understanding the intricacies of wave patterns and even space exploration, trigonometry is your gateway to a world of endless possibilities. Keep exploring, keep practicing, and soon you'll find yourself confidently solving real-world problems with the power of trigonometric ratios.
Stay curious and remember — this is just the beginning! The more you engage with trigonometry, the clearer its magic becomes.
Let’s keep learning together and unravel the mysteries of triangles, one angle at a time!
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