TRIGONOMETRY
Q.1. From the given figure, write down the values of :
(i) sin B
(ii) tan B
(iii) cos C
(iv) cot C
(v) (sin B cos C + cos B sin C)
(vi) (sec 2C − tan 2C)
Ans. In right-angled ∆ABC, BC = 17 units, AB = 15 units
Now, BC 2 = AB2 + AC 2
(Using Pythagoras Theorem)
⇒ (17)2 = (15)2 + AC2
⇒ 289 = 225 + AC2
⇒ AC2 = 289 − 225 = 64 = (8)2
∴ AC = 8 units
AC 8
AC 8
AC 8
(i) sin B =
=
(ii) tan B =
=
(iii) cos C =
=
BC 17
AB 15
BC 17
AC 8
(iv) cot C =
=
(v) sin B cos C + cos B sin C
AB 15
AC 8
AC 8
sin B =
= , cos C =
=
BC 17
BC 17
AB 15
AB 15
cos B =
= , sin C =
=
BC 17
BC 17
8 8 15 15
sin B cos C + cos B sin C = × + ×
17 17 17 17
64 225 64 + 225 289
=
+
=
=
=1
289 289
289
289
BC 17
AB 15
(vi) sec2 C − tan 2 C , sec C =
= , tan C =
=
AC 8
AC 8
2
2
289 225 289 − 225 64
 17   15 
2
2
−
=
=
=1
∴ sec C − tan C =   −   =
64
64
64
64
 8 8
Math Class IX
1
Question Bank
Q.2. From the following figure, find the values of :
(i) cos A
(ii) cosec A
(iii) tan 2 A − sec 2 A
(v) sec A
1
cot 2C − 2
sin C
Ans. In ∆ABD,
(iv) sin C
(vi)
AB2 = AD 2 + BD 2
(Using Pythagoras Theorem)
AB = AD 2 + BD 2 = (3)2 + (4) 2 = 9 + 16 = 25 = 5
In ∆BCD,
⇒
BC 2 = BD 2 + DC2
DC 2 = BC2 − BD2
⇒
(Using Pythagoras Theorem)
DC = BC 2 − BD2
= (12)2 − (4)2 = 144 − 16 = 128 = 64 × 2 = 8 2
AD 3
AB 5
(ii) cosec A =
(i) cos A =
=
=
AB 5
BD 4
(iii) tan 2 A − sec 2 A
2
2
 4   5  16 25 16 − 25 −9
∴ tan A − sec A =   −   = −
=
=
= −1
9 9
9
9
 3 3
BD 4 1
BC
12
3
=
=
(v) sec C =
(iv) sin C =
=
=
BC 12 3
DC 8 2 2 2
2
2
DC 8 2
1
=
= 2 2 and sin C =
[Using part (iv)]
BD
4
3
1
1
1
cot 2 C − 2 = (2 2)2 −
=
4
×
2
−
= 8 − 9 = −1
2
1
sin C
1
 
9
3
(vi) cot C =
∴
Math Class IX
2
Question Bank
Q.3. From the following figure find the values of :
(i) sin B
(ii) tan C
(iii) sec 2B − tan 2B (iv) sin 2C + cos2C
Ans. In right-angled ∆ABD,
⇒ AB2 = BD 2 + AD 2
⇒ AD 2 = AB2 − BD2
(Using Pythagoras Theorem)
⇒ AD = AB2 − BD 2 = (13)2 − (5)2 = 169 − 25 = 144 = 12
In right angled ∆ACD,
AC 2 = AD 2 + DC2
⇒ AC = AD 2 + DC 2 = (12)2 + (16)2 = 144 + 256 = 400 = 200
AD 12
=
(i) sin B =
AB 13
AD 12 3
(ii) tan C =
=
=
DC 16 4
AB 13
=
(iii) sec2 B − tan 2 B , sec B =
BD 5
AD 12
tan B =
=
BD 5
2
2
 13   12  169 144 169 − 144 25
2
2
−
=
=
=1
∴ sec B − tan B =   −   =
25 25
25
25
5  5
(iv) sin 2 C + cos2 C
AD 12 3
sin C =
=
=
AC 20 5
DC 16 4
cos C =
=
=
AC 20 5
2
∴
2
9 16 25
3 4
+
=
=1
sin C + cos C =   +   =
25 25 25
5 5
2
Math Class IX
2
3
Question Bank
1
, find the values of other trigonometrical ratios for θ.
2
BC
1
Ans. sin θ =
=
AC
2
∴ BC = 1 and AC = 2
In right ∆ABC, ∠B = 90°
Q.4. If sin θ =
(Using Pythagoras Theorem)
AC 2 = AB2 + BC 2
⇒ ( 2)2 = (AB)2 + (1)2 ⇒ 2 = AB2 + 1
⇒ AB2 = 2 − 1 = 1 = (1)2 ∴ AB = 1
1
1
AB
1
Now, cosθ =
, tan θ = = 1 , cot θ = = 1
=
1
1
AC
2
2
2
secθ =
= 2 , cosec θ =
= 2
1
1
8
Q.5. If tan θ = , find the values of other trigonometrical ratios for θ.
15
BC 8
tan θ =
=
∴ BC = 8 and AB = 15
Ans.
AB 15
In right ∆ABC, AC 2 = AB2 + BC 2
(Using Pythagoras Theorem)
∴
= (15) 2 + (8)2 = 225 + 64 = 289 = (17)2
BC 8
=
AC = 17 . Now sin θ =
AC 17
AB 15
AB 15
cosθ =
=
cot θ =
=
AC 17
BC 8
AC 17
AC 17
secθ =
=
cosec θ =
=
AB 15
BC 8
Q.6. If cosec θ = 10, find the values of other trigonometrical ratios for θ.
Ans. cosec θ =
AC
10
=
∴ AC = 10 and BC = 1
BC
1
In ∆ABC,
(Using Pythagoras Theorem)
AC 2 = AB2 + BC 2
2
2
2
⇒ ( 10) = AB + (1) ⇒ 10 = AB2 + 1
Math Class IX
4
Question Bank
⇒ AB2 = 10 − 1 = 9 = (3)2 ∴ AB = 3
BC
1
AB
3
Now, sin θ =
, cosθ =
=
=
AC
AC
10
10
BC 1
AB 3
tan θ =
= ,
cot θ =
=
AB 3
BC 1
AC
10
sec θ =
=
AB
3
3
Q.7. If sin θ = and θ is an acute angle, find the values of cos θ and tan θ.
5
3 BC
Ans. sin θ = =
∴ BC = 3 and AC = 5
5 AC
In ∆ABC, AC 2 = AB2 + BC 2
⇒ (5)2 = AB2 + (3)2 ⇒ 25 = AB2 + 9
⇒ AB2 = 25 − 9 = 16 = (4)2 ∴ AB = 4
AB 4
BC 3
cosθ =
= ,
tan θ =
=
AC 5
AB 4
5
Q.8. If tan θ =
and θ is acute, find the values of sin θ and cos θ.
12
BC 5
Ans. tan θ =
=
AB 12
∴ BC = 5, AB = 12
Now in ∆ABC,
∴
(Using Pythagoras Theorem)
AC 2 = AB2 + BC 2
2
2
= (12) + (5) = 144 + 25 = 169 = (13)2
AC = 13
BC 5
AB 12
sin θ =
= , cosθ =
=
AC 13
AC 13
Math Class IX
5
Question Bank
3
, find the value of (cosec θ + cot θ).
2
BC
3
Ans. sin θ =
=
∴ BC = 3 and AC = 2
AC
2
Now in ∆ABC,
Q.9. If sin θ =
(Using Pythagoras Theorem)
AC 2 = AB2 + BC 2
⇒ (2)2 = AB2 + ( 3)2 ⇒ 4 = AB2 + 3
⇒ AB2 = 4 − 3 = 1 = (1) 2 ∴ AB = 1
2
1
, and cot θ =
Now, cosec θ =
3
3
2
1
3
∴ cosec θ + cot θ =
+
=
= 3.
3
3
3
(5 sin θ − 2 cos θ)
.
tan θ
5
BC 5
=
Ans. 13sin θ = 5 ⇒ sin θ = . But sin θ =
13
AC 13
∴ BC = 5, AC = 13
Now in ∆ABC,
Q.10. If 13 sin θ = 5, find the value of
AC 2 = AB2 + BC 2
(Using Pythagoras Theorem)
⇒ (13)2 = AB2 + (5)2 ⇒ 169 = AB2 + 25
∴ AB = 12
⇒ AB2 = 169 − 25 = 144 ⇒ AB2 = (12)2
AB 12
BC 5
= , tan θ =
=
Now cosθ =
AC 13
AB 12
5
12 25 24
1
5× − 2×
−
5sin θ − 2cos θ
13
13 = 13 13 = 13 = 1 × 12 = 12
∴
=
5
5
5 13 5 65
tan θ
12
12
12
Math Class IX
6
Question Bank
 1 − cos 2θ  3
= .

2 
 2 − sin θ  5
1
Base
1
AB
Ans. cot θ =
. But cot θ =
=
=
Perp.
3
3 BC
∴ AB = 1 and BC = 3
Now in ∆ABC,
Q.11. If cot θ =
1
, show that
3
AC 2 = AB2 + BC 2 = (1)2 + ( 3)2
∴
(Using Pythagoras Theorem)
= 1 + 3 = 4 = (2)2
AC = 2
BC
3
AB 1
Now sin θ =
=
and cosθ =
=
AC
2
AC 2
2
∴
1
1 3
1−  
1−
2
1 − cos θ
2
  =
4 = 4 = 3×4 = 3
=
2
2
3 5 4 5 5
2 − sin θ
 3
2−
2−

4 4
 2 
5
sin A − cot A
1
. Evaluate : (i)
(ii) cot A +
13
2 tan A
cos A
Ans. In ∆ABC, ∠B = 90°
Base
AB
cos A =
=
Hypotenuse AC
5
Also, cos A =
13
Let, AB = 5 x units
AC = 13 x units
Q.12. Given : cos A =
BC = AC2 − AB2 = (13 x )2 − (5 x )2
= 169 x 2 − 25 x 2 = 144 x 2 = 12 x units
BC 12 x 12
AB 5 x
5
sin A =
=
= , cot A =
=
=
AC 13x 13
BC 12 x 12
BC 12 x 12
tan A =
=
=
AB 5 x
5
Math Class IX
7
Question Bank
12 5
−
395
sin A − cot A 13 12 144 − 65 79 5
=
×
=
=
=
12
156
156 24 3744
2 tan A
2×
24
5
5
5
5
(ii) cos A =
also cot A =
13
12
cot A +
1
5
1
5 13 25 + 156 181
=
.
= +
= + =
60
cos A 12 5 12 5
60
13
 2 sin θ − 3 cos θ 
13
, show that 
 = 3.
5
 4 sin θ − 9 cos θ) 
13 AC
∴ AC = 13, AB = 5
Ans. secθ = =
5 AB
Now in right ∆ABC,
Q.13. If sec θ =
AC 2 = AB2 + BC 2 ⇒ (13)2 = (5)2 + (BC)2
⇒ 169 = 25 + BC 2 ⇒ BC2 = 169 − 25 = 144 = (12)2
∴ BC = 12
BC 12
AB 5
Now, sin θ =
= , cosθ =
=
AC 13
AC 13
2sin θ − 3cosθ
Now LHS =
4sin θ − 9cosθ
12
5
24 15
9
2 × − 3×
−
13
13 = 13 13 = 13 = 9 × 13 = 3
=
12
5
48 45 3 13 3
4× − 9×
−
13
13 13 13 13
= RHS
Math Class IX
8
Question Bank
 3 sin θ + 2 cos θ 
Q.14. If 3 tan θ = 4, show that 
 = 3.
 3 sin θ − 2 cos θ 
4
Ans. 3tan θ = 4 ∴ tan θ =
3
sin θ
cos θ
3
+
2
3sin θ + 2 cosθ
cos θ = 3tan θ + 2
Now LHS =
= cosθ
3sin θ − 2cos θ 3 sin θ − 2 cosθ 3tan θ − 2
cos θ
cosθ
(Dividing numerator and denominator by cosθ )
Putting the value of 3tan θ = 4
4+2 6
=
= = 3 = RHS
4−2 2
Q.15. If cot θ =
 p sin θ − q cos θ   p 2 − q 2 
q
.
, show that 
= 2
2
p
 p sin θ + q cos θ   p + q 
q
p
p sin θ − q cosθ
LHS =
p sin θ + q cosθ
sin θ
cosθ
p
−q
sin θ
= sin θ
sin θ
cosθ
p
+q
sin θ
sin θ
Ans. cot θ =
(Dividing numerator and denominator by sin θ )
q
q2
p −q×
p−
p − q cot θ
p
p
=
=
=
q
p + q cot θ
q2
p + q×
p
+
p
p
=
p2 − q2
p
p2 − q2
× 2
=
= RHS
p
p + q2 p2 + q2
Math Class IX
9
Question Bank
1
tan A − tan B
1
and sin B =
.
, find the value of :
2
1 + tan A tan B
2
1
Base
Ans. (i) cos A = =
2 Hypotenuse
AC 1
=
⇒
AB 2
∴ AC = 1, AB = 2
Q.16. If cos A =
But AB2 = AC 2 + BC 2 (Using Pythagoras Theorem)
⇒ (2)2 = (1)2 + BC2 ⇒ 4 = 1 + BC 2
⇒ BC 2 = 4 − 1 = 3 ∴ BC = 3
BC
3
=
= 3
∴ tan A =
AC
1
1
Perpendicular AC
Again sin B =
=
=
∴ AC = 1, AB = 2
Hypotenuse
AB
2
But AB2 = AC 2 + BC 2
2
2
(Using Pythagoras Theorem)
2
⇒ ( 2) = (1) + BC ⇒ 2 = 1 + BC 2
⇒ BC 2 = 2 − 1 = 1 = (1) 2 ∴ BC = 1
AC 1
∴ tan B =
= =1
BC 1
3 − 3 −1+ 3
tan A − tan B
3 −1
3 − 1 ( 3 − 1)(1 − 3)
Now
=
=
=
=
1− 3
1 + tan A tan B 1 + 3 × 1 1 + 3 (1 + 3)(1 − 3)
2 3 − 4 2( 3 − 2)
=
=
= −( 3 − 2) = 2 − 3.
−2
−2
Q.17. Use the adjoining figure and write the values of :
(i) sin x° (ii) cos y° (iii) 3 tan x° − 2 sin y° + 4 cos y°
Ans. In ∆ABC, ∠B = 90°
AC 2 = AB2 + BC 2 ⇒ (17) 2 = AB2 + (8) 2
⇒ 289 = AB2 + 64 ⇒ AB2 = 289 − 64 = 225 = (15)2
∴ AB = 15 cm
and in ∆BCD, ∠B = 90°
CD 2 = BD 2 + BC2
Math Class IX
(Using Pythagoras Theorem)
10
Question Bank
= (6)2 + (8)2 = 36 + 64 = 100 = (10)2
∴ CD = 10 cm
BC 8
BD 6
(i) sin x° =
=
(ii) cos y ° =
=
AC 17
CD 10
BC 8
BC 8 4
= , sin y° =
=
=
(iii) tan x° =
AB 15
CD 10 5
8
4
3 8 8 12 12
2
∴ 3tan x° − 2sin y ° + 4 cos y ° = 3 × − 2 × + 4 × = − +
=
=2 .
15
5
5 5 5 5
5
5
Q.18. Using the adjoining figure, calculate the values of :
(i) cos θ (ii) tan φ (iii) cosec φ
Ans. Draw CE ⊥ AD
∴ EC = AB and AE = BC = 5 units
and DE = AD − AE = 14 − 5 = 9 units
In ∆ABC, ∠B = 90°
AC 2 = AB2 + BC 2 ⇒ (13)2 = AB2 + (5)2
⇒ 169 = AB2 + 25 ⇒ AB2 = 169 − 25 = 144 = (12)2
∴ AB = 12 units . Hence, EC = 12 units
(Using Pythagoras Theorem)
In ∆DEC, DC 2 = DE 2 + EC2
= (9)2 + (12)2 = 81 + 144 = 225 = (15)2
∴ DC = 15 units
AB 12
Now (i) cosθ =
=
AC 13
CE 12 4
(ii) tan φ =
= =
ED 9 3
CD 15 5
(iii) cosec φ =
=
=
CE 12 4
Q.19. If (tan θ + cot θ) = 5, find the value of (tan 2θ + cot 2θ).
Ans. tan θ + cot θ = 5
Squaring both sides, we get
(tan θ + cot θ )2 = (5) 2
⇒ tan 2 θ + cot 2 θ + 2 tan θ cot θ = 25
{∵ (a + b)2 = a 2 + b 2 + 2ab}
1
= 25 ⇒ tan 2 θ + cot 2 θ + 2 = 25
⇒ tan 2 θ + cot 2 θ + 2 tan θ ×
tan θ
Math Class IX
11
Question Bank
⇒ tan 2 θ + cot 2 θ = 25 − 2 = 23
5
Q.20. If (cos θ + sec θ) = , find the value of (cos 2θ + sec 2θ).
2
5
Ans. (cos θ + secθ ) =
2
Squaring both sides, we get
5
(cosθ + sec θ ) =  
2
2
25
{∵ (a + b)2 = a 2 + b 2 + 2ab}
4
1
25
=
⇒ cos2 θ + sec 2 θ + 2cosθ ×
cosθ
4
25
17
⇒ cos2 θ + sec 2 θ + 2 =
⇒ cos2 θ + sec 2 θ = .
4
4
Q.21. In the given figure, ∆ABC is right angled at B. If
3
AC = 20 cm and tan A = , find the lengths of AB and BC.
4
Ans. In ∆ABC, ∠B = 90°
3
AC = 20 cm, tan A =
4
BC 3 3 x
But tan A =
= =
∴ BC = 3 x cm and AB = 4 x cm
AB 4 4 x
AC 2 = AB2 + BC 2 ⇒ (20)2 = (4 x )2 + (3 x )2
400
= 16 = (4)2
⇒ 400 = 16 x 2 + 9 x 2 = 25 x 2 ∴ x2 =
25
∴ x=4
Hence, AB = 4 x = 4 × 4 = 16 cm and BC = 3 x = 3 × 4 = 12 cm.
2x
Q.22. If cos θ =
, find the values of sin θ and tan θ in terms of x.
1 + x2
2x
Base
AB
Ans. cosθ =
=
=
2
Hypotenuse AC
1+ x
cos2 θ + sec2 θ + 2cos θ secθ =
∴
AB = 2x and AC = 1 + x 2
Now in ∆ABC,
Math Class IX
12
Question Bank
(Using Pythagoras Theorem)
AC 2 = AB2 + BC 2
2 2
2
2
⇒ (1 + x ) = (2 x) + BC ⇒ 1 + 2 x 2 + x 4 = 4 x 2 + BC2
⇒ BC 2 = 1 + 2 x 2 + x 4 − 4 x 2 = 1 − 2x 2 + x 4
⇒ BC 2 = (1 − x 2 )2 ⇒ BC2 = ( x2 − 1) 2
BC = x 2 − 1
x 2 − 1 x2 − 1
( x 2 − 1)
Now sin θ =
, tan θ =
=
2x
1 + x2 x 2 + 1
1
Q.23. If tan x = 1 , find the value of 4 sin 2 x − 3 cos2 x + 2.
3
1 4 Perpendicular BC
=
Ans. tan x = 1 = =
3 3
Base
AB
∴ BC = 4, AB = 3
∴
AC 2 = AB2 + BC 2
∴
∴
∴
(Using Pythagoras Theorem)
= (3)2 + (4)2 = 9 + 16 = 25 = (5)2
AC = 5
BC 4
AB 3
sin x =
= and cos x =
=
AC 5
AC 5
2
2
4
3
4sin 2 x − 3cos 2 x + 2 = 4   − 3   + 2
5
5
16
9
= 4 × − 3× + 2
25
25
64 27
=
− +2
25 25
64 − 27 + 50 87
12
=
=
=3 .
25
25
25
Q.24. If cosec θ = 5, find the value of :
(i) 2 − sin 2θ − cos 2θ
Ans. cosec θ =
∴
(ii) 2 +
1
sin 2θ
−
cos2θ
sin 2θ
.
5 Hyp. AC
=
=
1
Perp. BC
AC = 5, BC = 1
Math Class IX
13
Question Bank
But AC 2 = AB2 + BC 2
(Using Pythagoras Theorem)
2
2
2
2
⇒ ( 5) = AB + (1) ⇒ 5 = AB + 1
⇒ AB2 = 5 − 1 = 4 = (2)2 ∴ AB = 2
BC
1
AB
2
Now sin θ =
, cos θ =
=
=
AC
AC
5
5
2
2
1 4 10 − 1 − 4 5
 1   2 
= = 1.
(i) 2 − sin θ − cos θ = 2 − 
 −
 = 2− 5 − 5 =
5
5
 5  5
2
2
2
 2 
4
2


1
1
cos θ
1
 5  = 2+ − 5 = 2+ 5 − 4×5
(ii) 2 + 2 −
=
2
+
−
2
2
1 1
1 5 1
sin θ sin 2 θ
 1 
 1 




5 5
 5
 5
= 2 + 5 − 4 = 7 − 4 = 3.
Q.25. If sec A = 2, find the value of
3 cos2 A + 5 tan 2 A
4 tan 2 A − sin 2 A
.
2
1
Hyp. AC
2
=
=
Base AB
1
AC = 2, AB = 1
Ans. sec A = 2 =
∴
But AC 2 = AB2 + BC 2
(Using Pythagoras Theorem)
⇒ ( 2)2 = (1)2 + (BC)2 ⇒ 2 = 1 + BC 2
⇒ BC2 = 2 − 1 = (1) 2 ∴ BC = 1
BC 1
1
1
BC
1
, sin A =
, tan A =
= =1
cos A =
=
=
AB 1
sec A
AC
2
2
2
∴
 1 
3
+ 5 × (1)2
2
2

3cos A + 5 tan A
2
= 
2
2
2
4 tan A − sin A
2  1 
4 × (1) − 

 2
Math Class IX
14
Question Bank
1
3
13
3 × + 5 ×1
+5
13 2 13
6
2
=
= 2
= 2 = × = =1 .
1
1
7
2 7 7
7
4 ×1 −
4−
2
2
2
x
y cos θ − x sin θ
Q.26. If tan θ = , find the value of
.
y
x sin θ − y cos θ
x Perp. BC
Ans. tan θ = =
=
y Base AB
∴
BC = x, AB = y . But AC 2 = AB2 + BC 2
⇒ AC2 = y 2 + x 2 ⇒ AC = x 2 + y 2
BC
x
AB
y
, cosθ =
Now sin θ =
=
=
AC
AC
x2 + y 2
x2 + y 2
y
x
y·
+ x·
x2 + y 2
x2 + y2
y cos θ + x sin θ
=
x
y
x sin θ − y cosθ x·
− y·
x2 + y 2
x2 + y 2
y2
=
x2 + y2
x2
x2 + y 2
=
y 2 + x2
x2 − y2
+
−
=
x2
x2 + y 2
y2
=
y 2 + x2
x2 + y2
×
x2 + y 2
x2 − y2
x2 + y 2
x2 + y 2
x2 − y2
.
Q.27. If cot θ = 1, find the value of : 5 tan 2θ + 2 sin 2θ − 3.
Base AB 1
Ans. cot θ = 1 =
=
=
Perp. BC 1
∴ AB = 1, BC = 1
But AC 2 = AB2 + BC 2
(Using Pythagoras Theorem)
2
2
2
AC = (1) + (1) = 1 + 1 = 2
∴
AC = 2
Math Class IX
15
Question Bank
Now tan θ =
1
1
BC
1
= = 1 , sin θ =
=
cot θ 1
AC
2
2
1
 1 
5 tan θ + 2sin θ − 3 = 5(1) + 2 
 − 3 = 5 × 1 + 2 × 2 − 3 = 5 + 1 − 3 = 3.
 2
Q.28. In the following figure, AD ⊥ BC, AC = 26, CD = 10, BC = 42,
∠DAC = x, ∠B = y, find the value of :
2
2
(i) cot x
2
1
(ii)
2
sin y
−
1
tan 2 y
Ans. In ∆ABC, AD ⊥ BC
AC = 26, CD = 10, BC = 42
∴ BD = BC − CD = 42 − 10 = 32
In ∆ACD,
(Using Pythagoras Theorem)
AC 2 = CD 2 + AD 2
⇒ (26)2 = (10)2 + AD 2 ⇒ 676 = 100 + AD2 ⇒ AD 2 = 576
∴ AD = 24
AD 24
AD 24 12
Now cot x =
=
= 2.4, cos x =
=
=
CD 10
AC 26 13
In ∆ADB,
AB2 = AD 2 + BD 2
= (24)2 + (32)2 = 576 + 1024 = 1600 = (40)2
∴ AB = 40
AD 24 3
BD 32 4
AD 24 3
sin y =
=
= , cos y =
=
= , tan y =
=
=
AB 40 5
AB 40 5
BD 32 4
(i) cot x = 2.4
1
1
1
1
1
1 25 16 9
(ii)
−
=
−
=
−
=
− = = 1.
2
2
2
2
9
9
9 9 9
sin y tan y  3 
 3
 
 
25 16
5
 4
Q.29. In the following figure, BD = 18, DC = 24, AC = 50, ∠A = x, ∠BCD = y
and ∠ABC = 90° = ∠BDC find :
(i) 2 tan x − sin y
(ii) 3 − 2cos x + 3cot y
(iii) 5 − 3tan 2 x + 3sec 2 x
Ans. In ∆BCD, ∠D = 90°
Math Class IX
(iv) 2 cot 2 y − 2cosec 2 y + 3
16
Question Bank
BC 2 = BD 2 + CD 2
= (18)2 + (24)2 = 324 + 576 = 900 = (30)2
∴ BC = 30
In ∆ABC, ∠B = 90°
AC 2 = AB2 + BC 2 ⇒ (50)2 = AB2 + (30)2
⇒ 2500 = AB2 + 900 ⇒ AB2 = 2500 − 900 = 1600 = (40)2
∴ AB = 40
BC 30 3
AB 40 4
sin x =
=
= , cos x =
=
=
AC 50 5
AC 50 5
BC 30 3
1
5
tan x =
=
= , sec x =
=
AB 40 4
cos x 4
BD 18 3
CD 24 4
sin y =
=
= , cot y =
=
=
BC 30 5
BD 18 3
1
5
cosec y =
=
sin y 3
3 3 3 3 15 − 6 9
(i) 2 tan x − sin y = 2 × − = − =
=
4 5 2 5
10
10
4
4
8
8 35 − 8 27
2
=
=5
(ii) 3 − 2cos x + 3cot y = 3 − 2 × + 3 × = 3 − + 4 = 7 − =
5
3
5
5
5
5
3
2
2
9
25
3
5
(iii) 5 − 3tan 2 x + 3sec 2 x = 5 − 3 ×   +3   = 5 − 3 × + 3 ×
16
16
4
4
27 75 80 − 27 + 75 155 − 27 128
=5−
+
=
=
=
=8
16 16
16
16
16
2
2
4
5
(iv) 2 cot y − 2 cosec y + 3 = 2   − 2   + 3
3
 3
16
25
32 50
= 2× − 2× + 3 =
− +3
9
9
9
9
32 − 50 + 27 59 − 50 9
=
=
= = 1.
9
9
9
2
Math Class IX
2
17
Question Bank
Q.30. In an isosceles triangle ABC; AB = AC = 10 cm and BC = 18 cm.
Find the value of : (i) sin 2B + cos2C
Ans. Draw AD ⊥ BC which bisects BC
18
∴ BD = DC = = 9 cm
2
In ∆ABD,
(ii) tan 2C − sec 2B + 2
AB2 = BD 2 + AD 2
⇒ (10) 2 = (9)2 + AD2
⇒ 100 = 81 + AD 2
⇒ AD 2 = 100 − 81 = 19 ⇒ AD = 19
AD
19
AB 10
sin B =
=
, sec B =
=
AB 10
BD 9
CD 9
AD
19
cos C =
= , tan C =
=
AC 10
DC
9
2
 19   9  2 19
81 100
+
=
= 1.
(i) sin B + cos C = 
 +  =
10
10
100
100
100




 19   10  2
19 100
19 − 100 + 162
2
2
(ii) tan C − sec B + 2 = 
+2 =
−  +2 = −
81 81
81
 9  9
181 − 100 81
=
= = 1.
81
81
Q.31. In rhombus ABCD, diagonals AC and BD intersect each other at point O.
If cosine of angle CAB is 0.6 and OB = 8 cm, find the length of the side and
the diagonals of the rhombus.
Ans. cos ∠CAB = 0.6
6 3 OA
=
= =
10 5 AB
Let OA = 3x, AB = 5 x
In ∆AOB,
2
2
AB2 = AO 2 + OB2 ⇒ (5 x) 2 = (3x)2 + OB2
⇒ 25 x2 = 9 x 2 + OB2
⇒ OB2 = 25 x 2 − 9 x 2 = 16 x 2 = (4 x)2 ∴ OB = 4x
Math Class IX
18
Question Bank
But, OB = 8 cm ⇒ 4 x = 8 cm ⇒ x =
8
8
cm ∴ OA = × 3 = 6 cm
4
4
8
× 5 = 10 cm
4
Hence, each side = 10 cm and diagonal AC = 2OA = 2 × 6 = 12 cm and diagonal
BD = 2OB = 2 × 8 = 16 cm.
and AB =
Q.32. If sin A = cos A, find the value of 2 tan 2 A − 2sec 2 A + 5.
sin A
Ans. sin A = cos A ⇒
= 1 ⇒ tan A = 1
cos A
∴ 2 tan 2 A − 2sec2 A + 5 = 2 tan 2 A − 2 (1 + tan 2 A) + 5 (∵ sec2 A = 1 + tan 2 A)
= 2 tan 2 A − 2 − 2 tan 2 A + 5 = −2 + 5 = 3.
Q.33. In rectangle ABCD, diagonal BD = 26 cm and cotangent of angle ABD =
1.5. Find the area and the perimeter of the rectangle ABCD.
Ans. In rectangle ABCD, diagonal BD = 26 cm
15 3
cot (∠ABD) = 1.5 =
=
10 2
3 AB
⇒ cot θ = =
2 DA
∴ AB = 3, DA = 2
In ∆ABD,
BD 2 = AB2 + AD 2 = (3)2 + (2)2 = 9 + 4 = 13
∴ BD = 13
If BD = 13, then AB = 3
3 × 26 78
If BD = 26, then AB =
=
cm
13
13
2
52
and AD =
× 26 =
cm
13
13
Area of ABCD = AB × AD
78
52 78 × 52
=
×
=
= 312 cm 2
13
13
13
52 
 78
Perimeter of rectangle ABCD, = 2 (l + b) = 2 
+

13 
 13
Math Class IX
19
Question Bank
= 2×
130 260 × 13 260 13
=
=
= 20 13 cm
13
13
13 × 13
Q.34. If 2 sin x = 3, evaluate: (i) 4 sin 3 x − 3 sin x
(ii) 3 cos x − 4 cos 3 x
Ans. 2sin x = 3
2
 3
3
3
⇒ sin x =
⇒ cos x = 1 − sin 2 x = 1 − 
 = 1− =
2
4
 2 
4−3
1 1
=
=
4
4 2
(i) 4sin 3 x − 3sin x = sin x (4sin 2 x − 3)
2


3   3
 = 3  4 × 3 − 3  = 3 (3 − 3) = 3 × 0 = 0.
=
4
−
3

2
2   2 
 2 
4
 2


(ii) 3cos x − 4cos3 x = cos x (3 − 4 cos 2 x)
1 
1
3 − 4 ×  
2 
 2
2
1
1 1
 1
 =  3 − 4 ×  = (3 − 1) = × 2 = 1.
2
4 2
 2 
Q.35. Use the informations given in the following figure to evaluate :
10
6
+
− 6cot y
sin x sin y
Ans. In ∆ADC, we get
=
(Using Pythagoras Theorem)
AC 2 = AD 2 + DC2
2
2
2
⇒ 20 = 12 + DC
⇒ 400 = 144 + DC 2
⇒ DC 2 = 400 − 144 = 256 = (16)2
∴ DC = 16
But BC = 21
∴ BD = BC − DC = 21 − 16 = 5
In ∆ADB,
∴
AB2 = AD 2 + BD 2
= (12)2 + (5)2 = 144 + 25 = 169 = (13)2
AB = 13
BD 5
AD 12 3
CD 16 4
sin x =
= , sin y =
=
= , cot y =
= =
AB 13
AC 20 5
AD 12 3
Math Class IX
20
Question Bank
10
6
10 6
4
 4  10 × 13 6 × 5
+
− 6cot y =
+ −6  =
+
− 6×
5 3
sin x sin y
5
3
3
3
13 5
= 26 + 10 − 8 = 36 − 8 = 28.
Q.36. Without using tables; find the value of :
Now
(i) tan 2 30° + tan 2 45° + tan 2 60°
tan 45°
sec 60° 5 sin 90°
(ii)
+
−
cosec 30° cot 45° 2 cos 0°
(iii)
sin 30° − sin 90° + 2 cos 0°
tan 30° × tan 60°
(iv)
5 sin 2 30° + cos2 45° − 4 tan 2 30°
2 sin 30° cos 30° + tan 45°
(v)
tan 2 60° + 4 cos 2 45° + sec2 30° + 5 cos 2 90°
cosec 30° + sec 60° − cot 2 30°
(vi) 4 (sin4 30° + cos4 60°) − 3 (cos2 45° − sin2 90°)
Ans. (i) tan 2 30° + tan 2 45° + tan 2 60° = (tan 30°)2 + (tan 45°)2 + (tan 60°)2
2
1 + 3 + 9 13
1
 1 
2
2 1
= =4
=
 + (1) + ( 3) = 3 + 1 + 3 =
3
3
3
 3
tan 45°
sec 60° 5sin 90° 1 2 5 × 1 1 2 5 1 + 4 − 5 0
(ii)
= + −
= + − =
= = 0.
+
−
2
2
cosec 30° cot 45° 2cos 0° 2 1 2 × 1 2 1 2
1
1
−
1
+
2
×
1
−1+ 2 1
sin 30° − sin 90° + 2cos 0° 2
3
2
(iii)
=
=
= +1 = .
1
tan 30°× tan 60°
1
2
2
× 3
3
2
2
 1 
1  1 
−4
5×  + 

2
2
2

5sin 30° + cos 45° − 4 tan 30°
 2  2 
 3
(iv)
=
2sin 30° cos 30° + tan 45°
1
3
2× ×
+1
2 2
Math Class IX
21
2
Question Bank
1 1
1 5 1 4
5× + − 4×
+ −
4 2
3 =4 2 3
=
1
3
3
2× ×
+1
+1
2 2
2
15 + 6 − 16
5
2
5
12
=
= ×
=
.
12 2 + 3 6 (2 + 3)
3+2
2
2
2
 1   2 
( 3)2 + 4 
 +
2
2
2
2
 + 5× 0
tan 60° + 4 cos 45° + sec 30° + 5cos 90°
2  3

(v)
=
cosec 30° + sec 60° − cot 2 30°
2 + 2 − ( 3) 2
1 4
4
3 + 4× + + 0 3 + 2 +
2 3
3
=
=
4−3
1
4 15 + 4 19
1
= 5+ =
=
=6 .
3
3
3
3
4
4
2
2
(iv) 4 (sin 30° + cos 60°) − 3 (cos 45° − sin 90°)
 1  2

 1  4  1 4 
2
= 4   +    − 3 
−
(1)


 2 

 2   2  
1
1 1
1 
 1 1 3 4
= 4  +  − 3  − 1 = 4 × − 3  −  = + = = 2.
8
16 16 
2 
 2 2 2 2
Q.37. Verify each of the following :
(i) cos 60° cos 30° − sin 60° sin 30° = 0.
(ii) cos 60° = (1 − 2 sin 2 30°) = (2 cos 2 30° − 1)
 tan 60° − tan 30° 
(iii) tan 30° = 

 1 + tan 60° tan 30° 
2
1 + cos 30°
 tan 60° + 1 
(iv) 
 =
1 − cos 30°
 tan 60° − 1 
Ans. (i) LHS = cos60° cos30° − sin 60° sin 30°
1
3
3 1
= ×
−
×
2 2
2 2
Math Class IX
22
Question Bank
3
3
−
= 0 = RHS
4
4
1
(ii) LHS = cos 60° =
2
=
2
1
1 1
1
1 − 2sin 30° = 1 − 2 ×   = 1 − 2 × = 1 − =
4
2 2
2
2
2
 3
3
3
1
2 cos 30° − 1 = 2 × 
 −1 = 2 × −1 = − 1 =
4
2
2
 2 
1
(iii) LHS = tan 30° =
3
1
3−
tan 60° − tan 30°
3 = 3 −1
RHS =
=
1 + tan 60° tan 30° 1 + 3 × 1
3
3 1+1
2
2 1
1
= 3 =
× =
. Hence, LHS = RHS
2
3 2
3
2
 tan 60° + 1 
(iv) LHS = 

 tan 60° − 1 
2
2
 3 +1
3 + 1 + 2 3 4 + 2 3 2 (2 + 3)
=
=
=
 =
3
−
1
3
+
1
−
2
3
4
−
2
3
2 (2 − 3)


2+ 3
=
2− 3
1 + cos30°
RHS =
1 − cos 30°
3 2+ 3
1+
2 = 2 = 2+ 3 × 2 = 2+ 3
=
2
3 2− 3
2− 3 2− 3
1−
2
2
From eqn. (i) and (ii), we get
∴ LHS = RHS
Math Class IX
23
...(i)
...(ii)
Question Bank
Q.38. Without using tables, prove that :
(i) sin (2 × 30°) =
2 tan 30°
(ii) cos (2 × 30°) =
1 + tan2 30°
2 tan30°
(iii) tan (2 × 30°) =
1 − tan 2 30°
2 tan 30°
Ans. (i) sin (2 × 30°) =
1 + tan 2 30°
1 − tan 2 30°
1 + tan 2 30°
3
2
LHS = sin (2 × 30°) = sin 60° =
1
3
2
2
2
2 tan 30°
3 = 3 = 3 = 2 ×3
=
=
RHS =
2
1 3 +1
4
3 4
1 + tan 2 30°
 1 
1+
1+ 

3
3
3
 3
3
3× 3
3
=
=
=
2
2 3 2 3× 3
∴ LHS = RHS
1 − tan 2 30°
(ii) cos (2 × 30°) =
1 + tan 2 30°
1
LHS = cos (2 × 30°) = cos 60° =
2
2×
2
 1 
1 3 −1 2
1− 
1−
2

1 − tan 30°
3

 =
3 = 3 = 3 = 2×3 = 2 = 1
RHS =
=
2
1 3 +1 4 3 4 4 2
1 + tan 2 30°
 1 
1+
1+ 

3
3
3
 3
LHS = RHS
∴
cos (2 × 30°) =
(iii) tan (2 × 30°) =
1 − tan 2 30°
1 + tan 2 30°
2 tan 30°
.
1 − tan 2 30°
LHS = tan (2 × 30°) = tan 60° = 3
Math Class IX
24
Question Bank
2
2 tan 30°
RHS =
=
= 3
2
2
1
1 − tan 30°
 1 
1−
1− 

3
 3
2
2
2 3
3
3
= 3 = 3 =
× =
×
= 3
3 −1
2
3 2
3
3
3
3
∴ LHS = RHS
2 tan 30°
.
∴ tan (2 × 30°) =
1 − tan 2 30°
Q.39. If A = 45°, verify that
2×
1
3
(i) sin 2A = 2 sin A cos A
(ii) cos 2A = (2cos2 A − 1) = (1 − 2sin 2 A)
Ans. (i) LHS = sin 2 A = sin (2 × 45°) = sin 90° = 1
RHS = 2sin A cos A
1
1
1
= 2 × sin 45° cos 45° = 2 ×
×
= 2× =1
2
2
2
Hence, LHS = RHS
(ii) cos 2 A = cos (2 × 45°) = cos 90° = 0 and 2 cos 2 A − 1 = 2cos 2 45° − 1
2
1
 1 
=2
 −1 = 2 × 2 −1 = 1−1 = 0
 2
2
1
 1 
Again, 1 − 2sin A = 1 − 2sin 45° = 1 − 2 
 = 1− 2 × 2 = 1−1 = 0
 2
2
2
Hence, cos 2 A = (2cos2 A − 1) = (1 − 2sin 2 A)
Q.40. If A = 30°, prove that
 1 − tan 2 A 
(ii)
cos
2A
=
.

2 

(1 + tan 2 A)
1
+
tan
A


3
Ans. (i) LHS = sin 2 A = sin (2 × 30°) = sin 60° =
2
(i) sin 2A =
Math Class IX
2 tan A
25
Question Bank
2
2
2 tan A
2 tan 30°
2 3
3
RHS =
=
=
= 3 = 3 =
× =
.
2
2
2
1
4
4
2
3
1 + tan A 1 + tan 30°
 1 
1+
1+ 

3
3
 3
Hence, LHS = RHS
1
(ii) LHS = cos 2 A = cos (2 × 30°) = cos 60° =
2
2×
1
3
2
 1 
1 2
1− 
1−
2
2

1 − tan A 1 − tan 30°
 3 =
3 = 3 = 2× 3 = 1.
RHS =
=
=
2
2
2
1 4 3 4 2
1 + tan A 1 + tan 30°
 1 
1+
1+ 

3 3
 3
Hence, LHS = RHS.
Q.41. If A = B = 45°, show that :
(i) sin (A − B) = sin A cos B − cos A sin B.
(ii) cos (A + B) = cos A cos B − sin A sin B.
Ans. Given A = B = 45°
(i) LHS = sin ( A − B ) = sin (45° − 45°) = sin 0° = 0 (zero)
RHS = sin A cos B − cos A sin B
= sin 45° cos 45° − cos 45° sin 45°
1
1
1
1
1 1
=
×
−
×
= − =0
2
2
2
2 2 2
∴ LHS = RHS
(ii) LHS = cos ( A + B ) = cos (45° + 45°) = cos90° = 0
RHS = cos A cos B − sin A sin B
1
1
1
1
1 1
= cos 45° cos 45° − sin 45° sin 45° =
×
−
×
= − =0
2
2
2
2 2 2
∴ LHS = RHS
Q.42. If A = 60° and B = 30°, show that :
(sin A cos B + cos A sin B)2 + (cos A cos B − sin A sin B) 2 = 1.
Ans. We are given A = 60° and B = 30°
LHS = (sin A cos B + cos A sin B )2 + (cos A cos B − sin A sin B )2
= (sin 60° cos30° + cos 60° sin 30°)2 − (cos60° cos30° − sin 60° sin 30°) 2
Math Class IX
26
Question Bank
2
2
2
 3
3 1 1 1
3
3 1 3 1  3
3
=
×
+ ×  − ×
−
×  =  +  −
−

2 2 2  2 2
2 2  4 4  4
4 
 2
2
= (1)2 − (0) 2 = 12 = 1 = RHS.
Q.43. If A = 60° and B = 30°, prove that :
(i) sin (A + B) = sin A cos B + cos A sin B.
(ii) cos (A + B) = cos A cos B − sin A sin B.
(iii) cos (A − B) = cos A cos B + sin A sin B.
tan A − tan B
.
1 + tan A tan B
Ans. We are given A = 60° and B = 30°
(i) LHS = sin ( A + B ) = sin (60° + 30°) = sin 90° = 1
RHS = sin A cos B + cos A sin B = sin 60° cos 30° + cos 60° sin 30°
3
3 1 1 3 1
=
×
+ × = + =1
2
2 2 2 4 4
∴ LHS = RHS
(ii) LHS = cos ( A + B ) = cos (60° + 30°) = cos 90° = 0
RHS = cos A cos B − sin A sin B = cos 60° cos30° − sin 60° sin 30°
1
3
3 1
3
3
= ×
−
× =
−
=0
2 2
2 2
4
4
∴ LHS = RHS
3
(iii) LHS = cos ( A − B ) = cos (60° − 30°) = cos 30° =
2
RHS = cos A cos B + sin A sin B
1
3
3 1
3
3
+
× =
+
= cos60° cos30° + sin 60° sin 30° = ×
2 2
2 2
4
4
2 3
3
=
=
.
4
2
∴ LHS = RHS
1
(iv) LHS = tan ( A − B ) = tan (60° − 30°) ⇒ tan 30° =
3
tan A − tan B
tan 60° − tan 30°
RHS =
=
1 + tan A tan B 1 + tan 60° tan 30°
(iv) tan (A − B) =
Math Class IX
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Question Bank
1
3 −1
2
3 = 3 = 3 = 2 = 1
=
1
1 +1
2
3×2
3
1+ 3 ×
3
∴ LHS = RHS
cos 3A + 2 cos 4A
Q.44. Evaluate :
, when A = 15°.
sin 3A + 2 sin 4A
Ans. A = 15°
cos 3 A + 2cos 4 A cos (3 × 15°) + 2cos (4 × 15°) cos 45° + 2cos 60°
=
=
sin 3 A + 2sin 4 A sin (3 × 15°) + 2sin (4 × 15°)
sin 45° + 2sin 60°
1
1
1
+ 2×
+1
2
2
2
=
=
1
3
1
3
+ 2×
+
2
2
2 1
3−
1+ 2
2
1+ 2
2 = 1+ 2 ×
=
=
.
1+ 6
2
1+ 6 1+ 6
2
3 sin 3A + 2 cos (2A + 5°)
Q.45. Evaluate :
, when A = 20°.
2 cos 3A − sin (2A − 10°)
3sin 3 A + 2cos (2 A + 5°) 3sin (3 × 20°) + 2cos (2 × 20° + 5°)
Ans.
=
[∵ A = 20°]
2cos3 A − sin (2 A − 10°) 2 cos (3 × 20°) − sin (2 × 20° − 10°)
3sin 60 + 2cos (40° + 5°) 3sin 60° + 2cos 45°
=
=
2cos 60° − sin 30°
2cos 60° − sin (40° − 10°)
3× 3
1
3 3
2
3 6 +4
+ 2×
+
2
2 = 2
2 = 2 2
=
1 1
1
1
2× −
1−
2 2
2
2
2 (3 6 + 4) (3 6 + 4) 2 3 6 + 4 (3 6 + 4) × 2
=
=
=
=
2 2
2 2
2
2× 2
3 6 × 2 + 4 2 3 12 + 4 2 3 × 4 × 3 + 4 2 3 × 2 3 + 4 2
=
=
=
=
2
2
2
2× 2
Math Class IX
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Question Bank
=
2 (3 3 + 2 2)
= 3 3 + 2 2.
2
Q.46. Show that 4 (sin4 30° + cos4 60°) − 3 (cos2 45° − sin2 90°) = 2.
Ans. LHS = 4 (sin 4 30° + cos4 60°) − 3 (cos 2 45° − sin 2 90°)
 1  2

 1  4  1 4 
2
= 4   +    − 3 
−
(1)


 2 

 2   2  
2
1 1
1 
 1
= 4  +  − 3  − 1 = 4 × − 3 ×  − 
16
16 16 
2 
 2
1 3 4
= + = = 2 = RHS
2 2 2
Q.47. If 0° ≤ x ≤ 90°, state the numerical value of x for which sin x° = cos x°.
Ans. ∵ sin x° = cos x°
sin x° cos x°
=
⇒
⇒ tan x° = 1 = tan 45°
cos x° cos x°
⇒ x° = 45° or x = 45.
Q.48. (i) If sin x = cos x and x is acute, state the value of x .
(ii) If sec A = cosec A and 0° ≤ A ≤ 90°, state the value of A.
(iii) If tan θ = cot θ and 0° ≤ θ ≤ 90°, state the value of θ.
(iv) If sin x = cos y; write the relation between x and y, if both the angles x
and y are acute.
Ans. (i) sin x = cos x
sin x
= 1 ⇒ tan x = 1 ∴ tan x = tan 45°
⇒
cos x
or x = 45°
(ii) sec A = cosec A
1
1
sin A
=
⇒
= 1 ⇒ tan A = 1 ∴ 0° ≤ A ≤ 90°
cos A sin A
cos A
∴ tan A = tan 45° or A = 45°
(iii) tan θ = cot θ
1
⇒ tan θ =
⇒ tan 2 θ = 1
tan θ
tan θ = ± 1 i.e. tan θ = ±1
Math Class IX
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Question Bank
But, 0° ≤ θ ≤ 90° ⇒ tan θ = 1
⇒ tan θ = tan 45° ∴ θ = 45°
(iv) sin x = cos y ⇒ sin x = sin (90° − y )
∴ x = 90° − y or x + y = 90°
Q.49. (i) If sin x = cos y, then x + y = 45°; write true or false.
(ii) sec θ cot θ = cosec θ; write true or false.
(iii) For any angle θ , state the value of : sin 2θ + cos 2θ.
Ans. (i) ∵ sin x = cos y
sin x = sin (90° − y ) ⇒ x = 90° − y
x + y = 90°
Hence, given statement is false.
(ii) secθ ·cot θ = cosec θ
1 cos θ
1
LHS = secθ ·cot θ =
=
= cosec θ =RHS
·
cosθ sin θ sin θ
Hence, given statement is true.
(iii) Consider the right angled ∆OMP in which ∠POM = θ
PM
OM
∴ sin θ =
, cosθ =
OP
OP
2
2
 PM   OM 
∴ sin θ + cos θ = 
 +

 OP   OP 
PM 2 OM 2 PM 2 + OM 2 OP 2
=
+
=
=
[Using Pythagoras Theorem]
OP 2 OP 2
OP 2
OP 2
=1
2
2
∴ sin θ + cos θ = 1
Q.50. State for any acute angle θ whether :
(i) sin θ increases or decreases as θ increases.
(ii) cos θ increases or decreases as θ increases.
(iii) tan θ increases or decreases as θ decreases.
Ans. (i) when θ = 0° , then sin 0° = 0
when θ = 90° , then sin 90° = 1
Hence, value of sin θ increases from 0 to 1 as θ increases from 0° to 90°.
(ii) For θ = 0° ⇒ cos 0° = 1
For θ = 90° ⇒ cos 90° = 0
Hence, value of cosθ decreases from 1 to 0 as θ increases from 0° to 90°.
2
Math Class IX
2
30
Question Bank
(iii) For θ = 90° ⇒ tan 90° = ∞
For θ = 0 ⇒ tan 0° = 0
Hence, the value of tan θ decreases from ∞ to 0 as θ decreases fro 90° to 0°.
Hence, (i) increasing, (ii) decreasing and (iii) decreasing.
Q.51. If 3 = 1.732, find (correct to two decimal places) the value of each the
following:
2
(i) sin 60°
(ii)
tan 30°
3 1.732
Ans. (i) sin 60° =
(correct to 2 decimal places)
=
= 0.866 = 0.87
2
2
2
2
=
= 2 3 = 2 × 1.732 = 3.464 = 3.46 (correct to 2 decimal places)
(ii)
1
tan 30°
3
Q.52. If x = 20°, evaluate : 12 sin
3x
cos 3x + 5 tan (2x + 5°) − 3cot 2 3x.
2
Ans. x = 20°
3x
cos 3 x + 5 tan (2 x + 5°) − 3cot 2 3 x
2
3 × 20°
= 12sin
cos 3 × 20° + 5 tan (2 × 20° + 5°) − 3cot 2 (3 × 20°)
2
= 12sin 30° cos 60° + 5 tan 45° − 3cot 2 60°
Now 12sin
2
1 1
 1 
= 12 × × + 5 × 1 − 3 
 = 3 + 5 − 1 = 7.
2 2
 3
Q.53. Without using trigonometric tables, evaluate the following:
2
 sin 49°   cos 41° 
(i) 
 +

 cos 41°   sin 49° 
2
2
 cot 40°  1  cos 35° 
(ii) 
− 

 tan 50°  2  sin 55° 
2
2
 sin 49°   cos 41° 
 sin (90° − 41°)   cos (90° − 49°) 
Ans. (i) 
 +
 =
 + 

cos 41°
sin 49°
 cos 41°   sin 49° 

2
2
2
 cos 41°   sin 49° 
2
2
=
 +
 = (1) + (1) = 1 + 1 = 2.
cos
41
°
sin
49
°

 

cot 40° 1  cos 35°  cot (90° − 50°) 1 cos (90° − 55°)
(ii)
−
− 
=
tan 50°
2
sin 55°
tan 50° 2  sin 35° 
Math Class IX
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Question Bank
tan 50° 1 sin 55°
1
1 1
−
= 1 − ×1 = 1 − = .
tan 50° 2 sin 55°
2
2 2
Q.54. Without using trigonometric tables, prove that :
=
(ii) sin 2 56° − cos2 34° = 0
(i) cos 72° − sin 18° = 0
(iii) sin 2 48° + sin2 42° = 1
(iv) cos 2 75° + cos 2 15° = 1
Ans. (i) LHS = cos 72° − sin18°
= cos (90° − 18°) − sin18°
= sin18° − sin18° = 0 = RHS
(ii) LHS = sin 2 56° − cos 2 34° = sin 2 (90° − 34°) − cos2 34°
= cos2 34° − cos2 34° = 0 = RHS
(iii) LHS = sin 2 48° + sin 2 42° = sin 2 (90° − 42°) + sin 2 42°
= cos2 42° + sin 2 42° = 1 = RHS
(iv) LHS = cos2 75° + cos 2 15° = cos2 (90° − 15°) + cos2 15°
= sin 2 15° + cos2 15° = 1 = RHS
Q.55. Without using trigonometric tables, prove that :
(i) sin 20° cos 70° + cos 20° sin 70° = 1
(ii) cos 64° cos 26° − sin 64° sin 26° = 0
Ans. (i) LHS = sin 20° cos 70° + cos 20° sin 70°
= sin 20° cos (90° − 20°) + cos 20° sin (90° − 20°)
= sin 20° sin 20° + cos 20° cos 20°
= sin 2 20° + cos 2 20° = 1 = RHS
(ii) LHS = cos64° cos 26° − sin 64° sin 26°
= cos (90° − 26°) cos 26° − sin (90° − 26°)sin 26°
= sin 26° cos 26° − cos 26° sin 26° = 0 = RHS
Q.56. Without using the trigonometric table, prove that :
cos 81° cos 14°
(i)
+
=2
sin 9° sin 76°
2
2
 sin 47°   cos 43° 
2
(ii) 
 +
 − 4 cos 45° = 0
 cos 43°   sin 47° 
(iii) sin 2 20° + sin 2 70° − tan 2 45° = 0
cos81° cos14° cos (90° − 9°) cos (90° − 76°)
Ans. (i) LHS =
+
=
+
sin 9° sin 76°
sin 9°
sin 76°
Math Class IX
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Question Bank
=
sin 9° sin 76°
+
= 1 + 1 = 2 = RHS
sin 9° sin 76°
2
2
 sin 47°   cos 43° 
2
(ii) LHS = 
 +
 − 4cos 45°
 cos 43°   sin 47° 
2
2
2
1 
 1  
 sin (90° − 43°)   cos (90° − 47°) 
4
=
+
−
∵
cos
45
°
=



 

cos 43°
sin 47°
2 

 2 
2
2
1
 cos 43°   sin 47° 
=
 +
 − 4×
2
 cos 43°   sin 47° 
2
2
= (1) + (1) − 2 = 2 − 2 = 0 = RHS
(iii) LHS = sin 2 20° + sin 2 70° − tan 2 45°
= sin 2 20° + sin 2 (90° − 20°) − tan 2 45°
= sin 2 20° + cos 2 20° − tan 2 45° = 1 − 1 = 0 =RHS
Q.57. Prove that :
cosθ
sinθ
(i)
+
=2
sin (90° − θ) cos (90° − θ)
(ii) sin θ cos (90° − θ) + cos θ sin (90° − θ) = 1.
cosθ sin θ
cosθ
sin θ
=
+
= 1 + 1 = 2 = RHS
Ans. (i) LHS =
+
sin (90° − θ ) cos (90° − θ ) cosθ sin θ
(ii) LHS = sin θ cos (90° − θ ) + cosθ sin (90° − θ )
= sin θ sin θ + cos θ cosθ = sin 2 θ + cos2 θ = 1 = RHS
cos 35° sin 11°
Q.58. Prove that :
+
− cos 28° cosec 62° = 1.
sin 55° cos 79°
cos35° sin11°
Ans. LHS =
+
− cos 28° cosec 62°
sin 55° cos 79°
cos (90° − 55°) sin (90° − 79°)
=
+
− cos (90° − 62°) cosec 62°
sin 55°
cos79°
sin 55° cos 79°
=
+
− sin 62° cosec 62°
sin 55° cos 79°
= 1 + 1 − 1 = 2 − 1 = 1 = RHS
Math Class IX
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Question Bank
Q.59. Find the magnitude of angle A, if:
(i) 2 sin A cos A − cos A − 2 sin A + 1 = 0
(ii) tan A − 2 cos A tan A + 2 cos A − 1 = 0
(iii) 2 cos2 A − 3 cos A + 1 = 0
(iv) 2 tan 3A cos 3A − tan 3A + 1 = 2 cos 3A
Ans. (i) 2sin A cos A − cos A − 2sin A + 1 = 0 ⇒ cos A (2sin A − 1) − 1 (2sin A − 1) = 0
(2sin A − 1) (cos A − 1) = 0
If 2sin A − 1 = 0, then 2sin A = 1
1
⇒ sin A = ⇒ sin A = sin 30° ⇒ A = 30°
2
If cos A − 1 = 0, then cos A = 1
⇒ cos A = cos 0°
⇒ A = 0°
(ii) tan A − 2 cos A tan A + 2 cos A − 1 = 0 ⇒ tan A (1 − 2cos A) − 1 + 2cos A = 0
tan A (1 − 2cos A) − 1 (1 − 2cos A) = 0 ⇒ (1 − 2 cos A) (tan A − 1) = 0
1
If 1 − 2 cos A = 0, then −2cos A = −1 ⇒ 2cos A = 1 ⇒ cos A =
2
⇒ cos A = cos 60° ⇒ A = 60°
If tan A − 1 = 0, then tan A = 1 ⇒ tan A = tan 45° ⇒ A = 45°
(iii) 2 cos 2 A − 3cos A + 1 = 0 . Putting, cos A = x
2 x 2 − 3x + 1 = 0 ⇒ 2 x 2 − 2 x − x + 1 = 0
⇒ 2 x ( x − 1) − 1 ( x − 1) = 0 ⇒ ( x − 1) (2 x − 1) = 0
If ( x − 1) = 0, then x = 1
1
If (2 x − 1) = 0, then 2 x = 1 ⇒ x =
2
Putting, x = cos A in eqn. (i), we get
∴ cos A = 1
⇒ cos A = cos 0° ⇒ A = 0°
Putting, x − = cos A in eqn. (ii), we get cos A =
⇒
⇒
(iv)
⇒
⇒
...(i)
...(ii)
1
2
cos A = cos 60°
A = 60°
2 tan 3 A cos3 A − tan 3 A + 1 = 2cos 3 A
2 tan 3 A cos3 A − tan 3 A + 1 − 2cos 3 A = 0
tan 3 A (2 cos 3 A − 1) − 1 (2cos 3 A − 1) = 0
Math Class IX
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Question Bank
⇒ (2cos 3 A − 1) (tan 3 A − 1) = 0
If (2 cos 3 A − 1) = 0, then 2 cos3 A = 1 ⇒ cos3 A =
1
⇒ cos 3 A = cos 60°
2
60°
⇒ A = 20°
3
If (tan 3 A − 1) = 0, then tan 3 A = 1 ⇒ tan 3 A = tan 45°
45°
⇒ 3 A = 45° ⇒ A =
⇒ A = 15°
3
⇒ 3 A = 60° ⇒ A =
Q.60. Solve for x :
x
(ii) cos − 1 = 0
3
(i) 2 cos 3x − 1 = 0
(iii) sin (x + 10°) =
1
2
(iv) cos (2x − 30°) = 0
(v) 2 cos (3x − 15°) = 1
(vi) tan 2 (x − 5°) = 3
(vii) 3 tan 2 (2x − 20°) = 1
3
x

(viii) cos  + 10°  =
2
2

(ix) sin 2 x + sin 2 30° = 1
(x) cos 2 30° + cos 2 x = 1
(xi) cos 2 30° + sin 2 2x = 1
Ans. (i) 2 cos 3x − 1 = 0
⇒ 2 cos 3x = 1 ⇒ cos3x =
(xii) sin 2 60° + cos2 (3x − 9) = 1
1
⇒ cos3x = cos 60° ⇒ 3x = 60°
2
60°
∴ x = 20°
3
x
x
x
x
cos − 1 = 0 ⇒ cos = 1 ⇒ cos = cos 0° ⇒ = 0°
3
3
3
3
x = 0°
1
sin ( x + 10°) = ⇒ sin ( x + 10°) = sin 30°
2
Comparing both sides, we get
x + 10° = 30° ⇒ x = 30° − 10°
x = 20°
cos (2 x − 30°) = 0 ⇒ cos (2 x − 30°) = cos 90°
⇒ x=
(ii)
∴
(iii)
∴
(iv)
Math Class IX
35
Question Bank
Comparing both sides, we get
2 x − 30° = 90° ⇒ 2 x = 90° + 30°
⇒ 2 x = 120° ⇒ x = 60°
(v) 2 cos (3 x − 15°) = 1 ⇒ cos (3 x − 15°) =
1
2
cos (3 x − 15°) = cos 60°
Comparing both sides, we get
3 x − 15° = 60° ⇒ 3 x = 60° + 15° ⇒ 3 x = 75° or x = 25°
(vi) tan 2 ( x − 5°) = 3
⇒ tan ( x − 5°) = ± 3 . Taking +ve sign only
tan ( x − 5°) = tan 60°
⇒ tan ( x − 5°) = 3
⇒
Comparing both sides, we get
x − 5° = 60° ⇒ x = 60° + 5°
or, x = 65°
1
(vii) 3tan 2 (2 x − 20°) = 1 ⇒ tan 2 (2 x − 20°) =
3
1
⇒ tan (2 x − 20°) = ±
3
Taking +ve sign only
1
⇒ tan (2 x − 20°) = +
⇒
tan (2 x − 20°) = tan 30°
3
Comparing both sides, we get
∴ 2 x − 20° = 30° ⇒ 2 x = 30° + 20°
⇒ 2 x = 50° or x = 25°
3
x

(viii) cos  + 10°  =
2
 2
x

cos  + 10°  = cos 30°
2

Comparing both sides, we get
x
x
+ 10° = 30° ⇒ = 30° − 10°
2
2
x
= 20° or x = 2 × 20°
⇒
2
or, x = 40°
Math Class IX
36
Question Bank
(ix) sin 2 x + sin 2 30° = 1
2
⇒
⇒
⇒
⇒
(x)
1
sin x +   = 1
⇒
 2
1
sin 2 x = 1 −
⇒
4
3
sin x =
⇒
4
sin x = sin 60° ∴ x = 60°
cos2 30° + cos 2 x = 1
2
2
 3
2
⇒ 
 + cos x = 1
 2 
3
⇒ cos2 x = 1 −
4
1
⇒ cos x =
4
⇒ cos x = cos 60°
(xi) cos2 30° + sin 2 2 x = 1
1
=1
4
3
sin 2 x =
4
3
sin x =
2
sin 2 x +
⇒
3
+ cos 2 x = 1
4
⇒
cos2 x =
⇒
⇒
2
 3
2
⇒ 
 + sin 2 x = 1 ⇒
 2 
3
⇒ sin 2 2 x = 1 −
⇒
4
1
⇒ sin 2 x =
⇒
4
⇒ sin 2 x = sin 30°
⇒
⇒ x = 15°.
(xii) sin 2 60° + cos2 (3x − 9°) = 1
1
4
1
cos x =
2
x = 60°
3
+ sin 2 2 x = 1
4
sin 2 2 x =
1
2
2 x = 30°
sin 2 x =
2
 3
2
⇒ 
 + cos (3 x − 9°) = 1 ⇒
 2 
3
⇒ cos2 (3 x − 9°) = 1 −
⇒
4
Math Class IX
1
4
3
+ cos2 (3 x − 9°) = 1
4
cos2 (3 x − 9°) =
37
1
4
Question Bank
1
4
⇒ cos (3x − 9°) = cos 60°
⇒ 3x = 60° + 9°
69°
⇒ x=
3
∴ x = 23°
⇒ cos (3x − 9°) =
⇒
⇒
⇒
cos (3 x − 9°) =
1
2
3 x − 9° = 60°
3 x = 69°
Q.61. If 4 cos2 x = 3 and x is an acute angle find the value of :
(ii) cos 2 x + cot 2 x
(iv) sin 2x
(i) x
(iii) cos 3x
Ans. (i) 4 cos2 x = 3 ⇒
cos2 x =
3
4
3
3
=
= cos 30° ∴ x = 30°
4
2
(ii) cos2 x + cot 2 x = cos 2 30° + cot 2 30°
3
3
3
= + ( 3)2 = + 3 = 3
4
4
4
(iii) cos3 x = cos (3 × 30°) = cos 90° = 0
⇒ cos x =
(iv) sin 2 x = sin (2 × 30°) = sin 60° =
3
.
2
Q.62. In ∆ABC, ∠B = 90°, AB = y units, BC = 3 units, AC = 2 units and angle
A = x°, find:
(i) sin x°
(ii) x°
(iii) tan x°
(iv) Use cos x° to find the value of y.
Ans. In ∆ABC, ∠B = 90° , ∠A = x° , AB = y units, BC = 3 units and
AC = 2 units
(Using Pythagoras Theorem)
AC 2 = AB2 + BC 2
(2)2 = y 2 + ( 3)2 ⇒ 4 = y 2 + 3
⇒ y2 = 4 − 3 = 1 ⇒ y = 1
Now,
BC
3
(i) sin x° =
=
= sin 60° (ii) x° = 60°
AC
2
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(iii) tan x° = tan 60° = 3
(iv) cos x° = cos 60° =
1 AB
=
2 AC
AB = 1 ⇒ y = 1.
Q.63. If A, B, C are the interior angles of a triangle, prove that :
∴
A
B+C
tan 
 = cot .
2
 2 
Ans. In ∆ABC,
∠A ∠B ∠C
+
+
= 90°
2
2
2
∠B ∠C
∠A
B+C
A
+
= 90° −
= 90° −
⇒
⇒
2
2
2
2
2
A
A
 B+C 

Now LHS = tan 
 = tan  90° −  = cot 2 = RHS
2
 2 

∴
∠A + ∠B + ∠C = 180° ⇒
Q.64. A balloon is connected to a meteorological station by a cable of length 200
metres, inclined at 60° to the horizontal. Determine the height of the balloon
from the ground, assuming that there is no slack in the string. (Take
3 = 1.73)
Ans. In ∆ABC,
String = AC = 200 m
Height of ballon from the ground = BC = h m
∠BAC = 60° (Given)
h
3
h
⇒
=
⇒ sin 60° =
200
2
200
200 × 3 200 × 1.73
=
= 173 m
⇒ h=
2
2
Hence, height of balloon from the ground is 173 m.
Q.65. A ladder leaning against a wall, makes an angle of 60° with the horizontal
and the foot of the ladder is 9.5 metres away from the wall. Find the length
of the ladder.
Ans. In ∆ACB,
Length of ladder = AB = x m
Height of wall = BC = h m
AC = 9.5 m (Given)
∠BAC = 60° (Given)
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Question Bank
9.5
1 9.5
⇒ =
x
2
x
⇒ x = 9.5 × 2 = 19
Hence, length of ladder is 19 m.
Q.66. A kite is flying with a thread 150 m long. If the thread is assumed stretched
straight and makes an angle of 60° with the horizontal, find the height of the
kite above the ground. (Take 3 = 1.73 )
Ans. In ∆ABC,
Length of thread = AC = 150 m (Given)
Height of kite = h m
(Given)
∠BAC = 60°
Perpendicular
∴ sin θ =
Hypotenuse
BC
h
=
⇒ sin 60° =
AC 150
3
h
150 × 3
=
⇒
⇒ h=
2 150
2
⇒ h = 75 × 3 = 75 × 1.32 = 129.75 m
Hence, height of the kite is 129.75 m.
Q.67. A kite is flying at a height of 120 m from the level ground. It is attached to
a string inclined 60° to the horizontal. Find the length of the string. (Take
3 = 1.73)
Ans. In ∆ABC,
Length of string = AC = l m
Height = BC = 120 m
(Given)
(Given)
∠BAC = 60°
Perpendicular
∴ sin θ =
Hypotenuse
⇒ cos 60° =
BC 120
3 120
=
=
⇒
AC
l
2
l
120 × 2 120 × 2 × 3
⇒ l=
=
3
3× 3
120 × 2 × 3
=
= 40 × 2 × 3 m
3
⇒ sin 60° =
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Question Bank
= 80 (1.73) m = 138.40 m
Hence, length of string is 138.4 m.
Q.68. In a ∆ABC, right angled at B, if ∠A = 30° and BC = 8 cm, find the
remaining angles and sides.
Ans. In ∆ABC,
∠B = 90°, ∠A = 30° (Given)
(Given)
BC = 8 cm
∴ ∠C = 180° − (∠B + ∠A)
= 180° − (90° + 30°)
= 180° − 120° = 60°
Perpendicular BC
sin θ =
=
Hypotenuse
AC
8
1
8
=
⇒ sin 30° =
⇒
AC
2 AC
⇒ AC = 16 cm
Perpendicular BC
tan θ =
=
Base
AB
8
1
8
⇒ tan 30° =
⇒
=
AB
3 AB
Hence, AB = 8 3 cm, AC = 16 cm.
Q.69. In a rectangle ABCD, AB = 12 cm and ∠BAC = 30°, Calculate the lengths
of side BC and diagonal AC.
Ans. In rectangle ABCD
(Given)
AB = 12 cm
∠BAC = 30°, ∠B = 90°
In ∆BAC,
Perpindecular
tan θ =
Base
BC
1
BC
⇒ tan 30° =
⇒
=
AB
3 12
⇒ BC =
12
3
cosθ =
Math Class IX
⇒
BC =
12 × 3 12 3
=
= 4 3 cm
3
3× 3
Base
Hypotenuse
41
Question Bank
AB
3 12
=
⇒
AC
2
AC
⇒
3AC = 12 × 2
24 24 × 3 24 3
=
= 8 3 cm
=
⇒ AC =
3
3
3× 3
Hence, BC = 4 3 cm and AC = 8 3 cm.
Q.70. Find the length of AB, from the adjoining
figure.
Ans. In ∆ACF,
FC
20
tan 45° =
1=
⇒
AC
AC
⇒ AC = 20
In ∆BDE,
BD
1
30
= cot 60° =
∴ BD =
= 10 3
30
3
3
Draw FM ⊥ DE
∴ FM = CD
Also, ME = DE − DM = DE − CF = 30 − 20 = 10
FM
Now,
= tan 60° = 3 ⇒ FM = ME 3 = 10 3
ME
∴ CD = 10 3
Now,
AB = AC + CD + DB = 20 + 10 3 + 10 3
= 20 + 20 3 = 20 ( 3 + 1)
= 20 (2.732) = 54.640 = 54.64.
⇒ cos30° =
Q.71. In trapezium ABCD, as shown, AB || DC, AD = DC = 20 cm and ∠A = 60°.
Find :
(i) Length of AB
(ii) distance between AB and DC.
Ans. Draw DL ⊥ AB, CM ⊥ AB
DL
3
= sin 60° =
⇒ DL = 10 3 cm
20
2
Thus,
distance
between
AB
and
= 10 3 = 10 (1.732) = 17.32 cm.
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DC
Question Bank
Again, In ∆ADL,
AL
1
= cos 60° =
20
2
20
= 10
2
Now In ∆BCM,
MB
1
20
= cos 60° = ∴ MB =
= 10
20
2
2
AB = AL + LM + MD = 10 + LM + 10 = 20 + LM
= 20 + 20 = 40 cm.
Hence, length of AB is 40 cm.
Q.72. Use the information given to find the length of AB.
Ans. In ∆APQ,
AP
= cot 30° = 3
10
∴ AP = 10 3 = 10 (1.732) = 17.32 cm
In ∆PRB,
PB
tan 45° =
=1
8
∴ PB = 8 cm
Hence, AB = AP + PB = 17.32 + 8 = 25.32 cm
Q.73. Find the length of AB.
Ans. In ∆DEA,
AE
tan 45° =
= 1 ∴ AE = DE = CB = 30 cm
DE
∴ AE = 30 cm
In ∆BDE,
BE
tan 60° =
= 3
30
∴ DE = 30 3 = 30 (1.732) = 51.960 = 51.96
∴ BE = DC = 51.96 ⇒ AE + BE = 30 + 51.96 = 81.96
Hence, length of AB is 81.96.
∴
AL =
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