175-NCERT-10-8-Introduction to Trigonometry - Ex- 8.2

NCERT

10th Mathematics

Exercise 8.2

Topic: 8 Introduction to Trigonometry

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Click here for ⇨ NCERT-10-8-Introduction to Trigonometry - Ex- 8.1

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 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 have

LHS ≠ 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° = 1/√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° = 1/√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° = 1/√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.

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Click here for ⇨ NCERT-10-8-introduction-to-trigonometry - Ex- 8.3

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