Chap 8 Introduction of Trigonometry Page 227
CHAPTER 8
I ntroduction of Trigonometry
Ans : [Board 2020 SQP Standard]
ONE MARK QUESTION
We have sin θ + cos θ = 2 cos θ
Dividing both sides by cos θ , we get
Multiple Choice Question sin θ + cos θ = 2 cos θ
cos θ cos θ cos θ
1. Given that sin α = 3 and cos β = 0 , then the value tan θ + 1 = 2
2
of β − α is
tan θ = 2 −1
(a) 0c (b) 90c
Thus (a) is correct option.
(c) 60c (d) 30c
Ans : 4. If cos A = 4 , then the value of tan A is
[Board 2020 SQP Standard] 5
(a) 3 (b) 3
3 5 4
We have sin α =
2
(c) 4 (d) 5
3 3
sin α = sin 60c & α = 60c ...(1) Ans :
and cos β = 0
We have cos A = 4
cos β = cos 90c & β = 90c ...(2) 5
Now, β − α = 90c − 60c = 30c We know that, cos A = Base =4
Hypotenuse 5
Thus (d) is correct option.
Perpendicular = 52 − 42 = 25 − 16 = 3
2. If TABC is right angled at C , then the value of
Perpendicular
sec (A + B) is Now, tan A = =3
(a) 0 (b) 1 Base 4
2 Thus (b) is correct option.
(c) (d) not defined
3
5. If sin A = 1 , then the value of cot A is
Ans : [Board 2020 SQP Standard] 2
We have +C = 90c (a) 3 (b) 1
3
Since, +A + +B + +C = 180c 3
(c) (d) 1
2
+A + +B = 180c − +C Ans :
= 180c − 90c = 90c
We have sin A = 1
2
Now, sec (A + B) = sec 90c not defined
Perpendicular
Thus (d) is correct option. sin A = =1
Hypotenuse 2
3. If sin θ + cos θ = 2 cos θ , (θ ! 90c) then the value of
tan θ is Now, Base = 22 − 12 = 3
(a) 2 - 1 (b) 2 +1 Base
So, cot A = = 3 = 3
(c) 2 (d) - 2 Perpendicular 1
,Page 228 Introduction of Trigonometry Chap 8
Hence, the required value of cot A is 3. 10α = 90c & α = 9c
Thus (a) is correct option. tan 5α = tan ^5 # 9ch
6. If sin θ = a , then cos θ is equal to
b
= tan 45c = 1
6tan 45c = 1@
b Thus (c) is correct option.
(a) (b) b
b2 - a2 a
9. If TABC is right angled at C , then the value of
(c) b2 - a2 (d) a cos ^A + B h is
b b2 - a2
Ans : (a) 0 (b) 1
Perpendicular (c) 1 (d) 3
We have sin θ = a = 2 2
b Hypotenuse
Ans :
Base = b2 − a2 We know that in TABC ,
2 2
So, cos θ = Base = b −a
Hypotenuse b
Thus (c) is correct option.
7. If cos ^α + β h = 0 , then sin ^α - β h can be reduced to +A + +B + +C = 180c
(a) cos β (b) cos 2β But right angled at C i.e., +C = 90c, thus
(c) sin α (d) sin 2α +A + +B + 90c = 180c
Ans :
A + B = 90c
Given, cos ^α + β h = 0 = cos 90c 6cos 90c = 0@ cos ^A + B h = cos 90c = 0
α + β = 90c Thus (a) is correct option.
α = 90c − β 10. If sin α = 1 and cos β = 1 , then the value of ^α + β h
2 2
Now, sin ^α - β h = sin ^90c − b − b h is
(a) 0c (b) 30c
= sin ^90c − 2β h
(c) 60c (d) 90c
= cos 2β
Ans :
Thus (b) is correct option.
8. If cos 9a = sin a and 9α < 90c, then the value of Given, sin α = 1 = sin 30c & α = 30c
2
tan 5α is 1
and cos β = = cos 60c & β = 60c
2
(a) 1 (b) 3 α + β = 30c + 60c = 90c
3
Thus (d) is correct option.
(c) 1 (d) 0
Ans : 11. If 4 tan θ = 3 , then c 4 sin q − cos q m is equal to
4 sin q + cos q
We have cos 9α = sin α where 9α < 90c
(a) 2 (b) 1
sin ^90c - 9αh = sin α 3 3
90c - 9α = α (c) 1 (d) 3
2 4
Ans :
, Chap 8 Introduction of Trigonometry Page 229
14. If x = p sec θ and y = q tan θ , then
Given, 4 tan θ = 3
(a) x2 − y2 = p2 q2 (b) x2 q2 − y2 p2 = pq
tan θ = 3 ...(i) (c) x2 q2 − y2 p2 = 21 2 (d) x2 q2 − y2 p2 = p2 q2
4
pq
sin q
4 sin q − cos q = 4 cos q − 1 = 4 tan q − 1 Ans :
4 sin q + cos q sin q
4 cos q +1
4 tan q + 1
We know, sec2 θ − tan2 θ = 1
4` j − 13
= 3−1 = 2 = 1 sec θ = x
y
4
= 3 tan θ =
4` 4 j + 1 3+1 4 2 Substituting and in above
p q
equation we have
Thus (c) is correct option.
x 2 y 2
12. If sin q − cos q = 0 , then the value of ^sin 4 q + cos 4 qh is c p m - cq m = 1
(a) 1 (b) 3
4 x2 q2 - y2 p2 = p2 q2
(c) 1 (d) 1 Thus (d) is correct option.
2 4
Ans :
15. If b tan θ = a , the value of a sin θ − b cos θ is
a sin θ + b cos θ
Given, sin q - cos q = 0 (a) 2 a − b (b) a2 + b2
a + b2 a +b
sin θ = cos θ 2 2 2 2
(c) a2 + b2 (d) a2 − b2
sin θ = sin ^90c − θh a −b a +b
Ans :
θ = 90c − θ & θ = 45c
We have tan θ = a
Now, sin 4 q + cos 4 q = sin 4 45c + cos 4 45c b
sin θ
a sin θ − b cos θ = a cos θ − b
= d 1 n +d 1 n = 1 + 1 = 1
4 4
= a tan θ − b
2 2 4 4 2 a sin θ + b cos θ a cos
sin θ
θ +b
a tan θ + b
2 2
= 2a − b
Thus (c) is correct option. a + b2
13. In the adjoining figure, the length of BC is Thus (d) is correct option.
16. (cos 4 A - sin 4 A) is equal to
(a) 1 - 2 cos2 A (b) 2 sin2 A - 1
(c) sin2 A - cos2 A (d) 2 cos2 A - 1
Ans :
cos 4 A - sin 4 A = (cos2 A) 2 − (sin2 A) 2
= (cos2 A − sin2 A) (cos2 A + sin2 A)
(a) 2 3 cm (b) 3 3 cm = (cos2 A − sin2 A) (1)
(c) 4 3 cm (d) 3 cm = cos2 A − (1 − cos2 A)
Ans :
= 2 cos2 A − 1
In TABC , sin 30c = BC Thus (d) is correct option.
AC
17. If sec 5A = cosec ^A + 30ch , where 5A is an acute
1 = BC angle, then the value of A is
2 6
(a) 15c (b) 5c
BC = 3 cm (c) 20c (d) 10c
Thus (d) is correct option. Ans :
CHAPTER 8
I ntroduction of Trigonometry
Ans : [Board 2020 SQP Standard]
ONE MARK QUESTION
We have sin θ + cos θ = 2 cos θ
Dividing both sides by cos θ , we get
Multiple Choice Question sin θ + cos θ = 2 cos θ
cos θ cos θ cos θ
1. Given that sin α = 3 and cos β = 0 , then the value tan θ + 1 = 2
2
of β − α is
tan θ = 2 −1
(a) 0c (b) 90c
Thus (a) is correct option.
(c) 60c (d) 30c
Ans : 4. If cos A = 4 , then the value of tan A is
[Board 2020 SQP Standard] 5
(a) 3 (b) 3
3 5 4
We have sin α =
2
(c) 4 (d) 5
3 3
sin α = sin 60c & α = 60c ...(1) Ans :
and cos β = 0
We have cos A = 4
cos β = cos 90c & β = 90c ...(2) 5
Now, β − α = 90c − 60c = 30c We know that, cos A = Base =4
Hypotenuse 5
Thus (d) is correct option.
Perpendicular = 52 − 42 = 25 − 16 = 3
2. If TABC is right angled at C , then the value of
Perpendicular
sec (A + B) is Now, tan A = =3
(a) 0 (b) 1 Base 4
2 Thus (b) is correct option.
(c) (d) not defined
3
5. If sin A = 1 , then the value of cot A is
Ans : [Board 2020 SQP Standard] 2
We have +C = 90c (a) 3 (b) 1
3
Since, +A + +B + +C = 180c 3
(c) (d) 1
2
+A + +B = 180c − +C Ans :
= 180c − 90c = 90c
We have sin A = 1
2
Now, sec (A + B) = sec 90c not defined
Perpendicular
Thus (d) is correct option. sin A = =1
Hypotenuse 2
3. If sin θ + cos θ = 2 cos θ , (θ ! 90c) then the value of
tan θ is Now, Base = 22 − 12 = 3
(a) 2 - 1 (b) 2 +1 Base
So, cot A = = 3 = 3
(c) 2 (d) - 2 Perpendicular 1
,Page 228 Introduction of Trigonometry Chap 8
Hence, the required value of cot A is 3. 10α = 90c & α = 9c
Thus (a) is correct option. tan 5α = tan ^5 # 9ch
6. If sin θ = a , then cos θ is equal to
b
= tan 45c = 1
6tan 45c = 1@
b Thus (c) is correct option.
(a) (b) b
b2 - a2 a
9. If TABC is right angled at C , then the value of
(c) b2 - a2 (d) a cos ^A + B h is
b b2 - a2
Ans : (a) 0 (b) 1
Perpendicular (c) 1 (d) 3
We have sin θ = a = 2 2
b Hypotenuse
Ans :
Base = b2 − a2 We know that in TABC ,
2 2
So, cos θ = Base = b −a
Hypotenuse b
Thus (c) is correct option.
7. If cos ^α + β h = 0 , then sin ^α - β h can be reduced to +A + +B + +C = 180c
(a) cos β (b) cos 2β But right angled at C i.e., +C = 90c, thus
(c) sin α (d) sin 2α +A + +B + 90c = 180c
Ans :
A + B = 90c
Given, cos ^α + β h = 0 = cos 90c 6cos 90c = 0@ cos ^A + B h = cos 90c = 0
α + β = 90c Thus (a) is correct option.
α = 90c − β 10. If sin α = 1 and cos β = 1 , then the value of ^α + β h
2 2
Now, sin ^α - β h = sin ^90c − b − b h is
(a) 0c (b) 30c
= sin ^90c − 2β h
(c) 60c (d) 90c
= cos 2β
Ans :
Thus (b) is correct option.
8. If cos 9a = sin a and 9α < 90c, then the value of Given, sin α = 1 = sin 30c & α = 30c
2
tan 5α is 1
and cos β = = cos 60c & β = 60c
2
(a) 1 (b) 3 α + β = 30c + 60c = 90c
3
Thus (d) is correct option.
(c) 1 (d) 0
Ans : 11. If 4 tan θ = 3 , then c 4 sin q − cos q m is equal to
4 sin q + cos q
We have cos 9α = sin α where 9α < 90c
(a) 2 (b) 1
sin ^90c - 9αh = sin α 3 3
90c - 9α = α (c) 1 (d) 3
2 4
Ans :
, Chap 8 Introduction of Trigonometry Page 229
14. If x = p sec θ and y = q tan θ , then
Given, 4 tan θ = 3
(a) x2 − y2 = p2 q2 (b) x2 q2 − y2 p2 = pq
tan θ = 3 ...(i) (c) x2 q2 − y2 p2 = 21 2 (d) x2 q2 − y2 p2 = p2 q2
4
pq
sin q
4 sin q − cos q = 4 cos q − 1 = 4 tan q − 1 Ans :
4 sin q + cos q sin q
4 cos q +1
4 tan q + 1
We know, sec2 θ − tan2 θ = 1
4` j − 13
= 3−1 = 2 = 1 sec θ = x
y
4
= 3 tan θ =
4` 4 j + 1 3+1 4 2 Substituting and in above
p q
equation we have
Thus (c) is correct option.
x 2 y 2
12. If sin q − cos q = 0 , then the value of ^sin 4 q + cos 4 qh is c p m - cq m = 1
(a) 1 (b) 3
4 x2 q2 - y2 p2 = p2 q2
(c) 1 (d) 1 Thus (d) is correct option.
2 4
Ans :
15. If b tan θ = a , the value of a sin θ − b cos θ is
a sin θ + b cos θ
Given, sin q - cos q = 0 (a) 2 a − b (b) a2 + b2
a + b2 a +b
sin θ = cos θ 2 2 2 2
(c) a2 + b2 (d) a2 − b2
sin θ = sin ^90c − θh a −b a +b
Ans :
θ = 90c − θ & θ = 45c
We have tan θ = a
Now, sin 4 q + cos 4 q = sin 4 45c + cos 4 45c b
sin θ
a sin θ − b cos θ = a cos θ − b
= d 1 n +d 1 n = 1 + 1 = 1
4 4
= a tan θ − b
2 2 4 4 2 a sin θ + b cos θ a cos
sin θ
θ +b
a tan θ + b
2 2
= 2a − b
Thus (c) is correct option. a + b2
13. In the adjoining figure, the length of BC is Thus (d) is correct option.
16. (cos 4 A - sin 4 A) is equal to
(a) 1 - 2 cos2 A (b) 2 sin2 A - 1
(c) sin2 A - cos2 A (d) 2 cos2 A - 1
Ans :
cos 4 A - sin 4 A = (cos2 A) 2 − (sin2 A) 2
= (cos2 A − sin2 A) (cos2 A + sin2 A)
(a) 2 3 cm (b) 3 3 cm = (cos2 A − sin2 A) (1)
(c) 4 3 cm (d) 3 cm = cos2 A − (1 − cos2 A)
Ans :
= 2 cos2 A − 1
In TABC , sin 30c = BC Thus (d) is correct option.
AC
17. If sec 5A = cosec ^A + 30ch , where 5A is an acute
1 = BC angle, then the value of A is
2 6
(a) 15c (b) 5c
BC = 3 cm (c) 20c (d) 10c
Thus (d) is correct option. Ans :
