NCERT
10th Mathematics
Exercise 8.2
Topic: 8 Introduction to Trigonometry
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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) So, 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) So, 2 tan2 45° + cos2 30° – sin2 60° = 2.
(iii) [cos 45°] / [sec 30° + cosec 30°]
1) Using the above table,
a) cos 45° = 1/√2,
b) sec 30° = 2/√3,
c) cosec 30° = 2,
2) So, [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,
d) sec 30° = 2/√3,
e) cos 60° = 1/2,
f) cot 45° = 1,
(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,
c) tan 45° = 1,
d) sin 30° = 1/2,
e) cos 30° = √3/2,
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° = 1/√3,
(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
Ans: (D) 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 (1/√2)
RHS = 2/√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°
(iv) 2tan30°/1-tan230° = ?
1) Using the above table,
a) tan 30° = 1/√3,
Q 3. If tan (A + B) = √3 and tan (A – B) = 1/√3; 0° < A + B ≤ 90°; A > B, find A and
B.
Solution:
1) It is given that,
a) tan (A + B) = √3 --------- equation 1
b) tan (A – B) = 1/√3 --------- equation 2
2) Using the above table,
a) tan 60° = √3 --------- equation 3
b) tan 30° = 1/√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) So, 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 ฮธ increases as ฮธ increases.
(iii) The value of cos ฮธ increases as ฮธ increases.
(iv) sin ฮธ = cos ฮธ for all values of ฮธ.
(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 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 ฮธ increases as ฮธ increases.
Ans: True.
1) sin 0° = 0, sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2, sin 90° = 1, so
"the value of sin ฮธ increases as ฮธ increases" is true.
(iii) The value of cos ฮธ increases as ฮธ increases.
Ans: False.
1) cos 0° = 1, cos 30° = √3/2, cos 45° = 1/√2, cos 60° = 1/2, cos 90° = 0, so
"the value of cos ฮธ increases as ฮธ increases" is false.
(iv) sin ฮธ = cos ฮธ for all values of ฮธ.
Ans: False.
1) sin ฮธ = cos ฮธ is true only when ฮธ = 45°, as sin 45° = cos 45° = 1/√2, so
"sin ฮธ = cos ฮธ for all values of ฮธ" 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.
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