8 INTRODUCTION TO TRIGONOMETRY CHAPTER

CHAPTER
8
J
A
INTRODUCTION TO TRIGONOMETRY
Points to Remember :
1. If ABC is a right triangle right angled at B and BAC = , 0°    90°, we have :
Base = AB, Perpendicular = BC and, Hypotenuse = AC
here,
;
AB
Base
cos =

AC Hypotenuse
BC Perpendicular

AB
Base
;
cosec  
AC Hypotenuse

AB
Base
;
cot  
tan =
sec  
2. We have, cosec  
T
I
3. Values of various Trigonometric ratios :

T-ratio
0°
sin 
0
M
A
cos 
1
tan 
0
cosec 
not defined
sec 
1
cot 
not defined
30°
1
2
3
2
1
3
2
C
AC Hypotenuse
=
BC Perpendicular
AB
Base

BC Perpendicular
B
1
1
1
, sec  
, cot  
sin 
cos 
tan 
Also, tan   sin  , cot   cos 
cos 
sin 
132
J
A
BC Perpendicular
sin =

AC
Hypotenuse
B
90°

A
45°
60°
90°
1
3
2
1
2
1
2
0
1
3
not defined
2
1
2
2
2
3
2
3
1
INTRODUCTION TO TRIGONOMETRY
3
2
1
3
1
not defined
0
MATHEMATICS–X
4. The value of sin  increases from 0 to 1 as  increases from 0 ° to 90°. Also, the value of cos  decreases
from 1 to 0 as  increases from 0° to 90°.
5. If  is an acute angle, then
sin (90° – ) = cos , cos (90° – ) = sin 
tan (90° – ) = cot , cot (90° – ) = tan 
sec (90° – ) = cosec , cosec (90° – ) = sec 
6. Basic trigonometric identities :
J
A
(i) sin 2   cos 2   1 or 1  cos 2   sin 2  or 1  sin 2   cos 2 
(ii) 1  tan 2   sec2  or sec2   tan 2   1
(iii) 1  cot 2   c osec2  or cosec2 – cot2 = 1
J
A
ILLUSTRATIVE EXAMPLES
Example 1. In ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine : (i) sin A, cos A (ii) sin C, cos C
[NCERT]
Solution. In right angled ABC, we have
AC2 = AB2 + BC2
 AC2 = (24)2 + (7)2
 AC2 = 576 + 49
 AC2 = 625  AC = 25
(ii)
sin A 
opposite side BC 7

hypotenuse AC 25
cos A 
adjacent side AB 24

hypotenuse AC 25
T
I
opposite side AB 24
sin C 

hypotenuse AC 25
B
7cm
C
adjacent side BC 7

hypotenuse AC 25
M
A
cosC 
B
A
24 cm
(i)
20
Example 2. Given : cot  
, find all other trigonometric ratios.
21
Solution.
cot  
20
Base
20


21
Perpendicular 21
Let Base (B) = 20 k, then perpendicular (P) = 21 k
H = 29k
Since P2 + B2 = H2 where H = hypotenuse


H2 = (21 k)2 + (20 k)2
2
2
P = 21k
B = 20k
2
= 441 k + 400 k = 841 k

H  841 k 2  29 k
MATHEMATICS–X
INTRODUCTION TO TRIGONOMETRY
133

sin  
P 21k 21
B 20k 20



; cos   
H 29k 29
H 29k 29
tan  
P 21k 21
H 29k 29

 ; cosec   

B 20k 20
P 21k 21
and sec  
Example 3. If 3 cot A = 4, check whether
Solution.
cot A 
1 – tan2 A
= cos 2 A – sin 2 A or not.
1+tan2 A
4
1
3
 tan A 

3
cot A 4
2
3
1   1 9
1

tan
A
16  9 7
4
consider,
   2  16 

2
1  tan A
 3  1  9 16  9 25
1  
16
4
2
B
In right ABC, AC2 = AB2 + BC2
= (4k)2 + (3k)2 = 16k2 + 9k2 = 25k2

AC  25k 2  5k
Now, cos A 
AB 4k
BC 3k


and sin A 
AC 5k
AC 5k
T
I
2

2
2
2
2
2
M
A
[NCERT]
J
A
16k  9k
7k
7
 4k   3k 
cos A  sin A       


2
2
5
k
5
k
25
k
25
k
25
   
2
from above, we conclude that
C
5k
A
4k
3k
B
1  tan 2 A
 cos 2 A  sin 2 A .
1  tan 2 A
2
4sec 2 30° – tan 2 45°
Example 4. Evaluate : 5cos 60° +
sin 2 30° + cos 2 30°
Solution.
J
A
H 29k 29


B 20k 20
[NCERT]
5cos 2 60  4sec 2 30  tan 2 45
sin 2 30  cos 2 30
2
2
 2 
1
2
5    4  
  (1)
3
2



2
2


3
1


  
 2   2 
5 16
15  64  12
 1
67
4
3
12



1 3
4
12

4 4
4
134
Ans.
INTRODUCTION TO TRIGONOMETRY
MATHEMATICS–X
Example 5. If tan (A + B) = 3 and tan (A – B)  1 ; 0° < A + B  90°; A > B, find A and B.
3
Solution.
tan (A + B) = 3 = tan 60°  A + B = 60°
and, tan (A  B) 
1
cos 2 40  cos 2 50
sin 2 40  sin2 50
...(2)
J
A
(i)
cos (40° – ) – sin (50° + ) +
(ii)
2 sin 68 2 cot 15 3 tan 45 tan 20 tan 40 tan 50 tan 70


cos 22 5 tan 75
5
(i) We have, cos (40° – ) – sin (50° + ) +
[CBSE 2002]
B
cos 2 40  sin 2 40
sin 2 40  cos 2 40
T
I
[CBSE 2004]
cos 2 40  cos 2 50
sin 2 40  sin 2 50
cos 2 40  cos 2 (90  40)
 sin[90  (40  )]  sin(50  ) 
sin 2 40  sin 2 (90  40)
 sin (50  )  sin(50  ) 
J
A
...(1)
 tan 30  A  B  30
3
On adding (1) and (2), we get, 2A = 90°  A = 45°
subtracting (2) from (1), we get, 2B = 30°  B = 15°
So, A = 45° and B = 15° Ans.
Example 6. Evaluate :
Solution.
[NCERT]
[ cos   sin (90  )]
1
 0   1 Ans.
[ sin2 + cos2 = 1]
1
2 sin 68 2 cot15 3 tan 45 tan 20 tan 40 tan 50 tan 70


(ii) We have,
cos 22 5 tan 75
5
2sin(90  22)
2 cot15
3tan 45 tan 20 tan 40 tan (90  40) tan (90  20)


cos 22
5 tan(90  15)
5

M
A
2 cos 22 2 cot15 3.1.tan 20 tan 40 cot 40 cot 20


cos 22 5 cot15
5
[ sin (90° – ) = cos , tan (90° – ) = cot ]
1 
2
3 tan 20 tan 40

 cot  
 2 1   1  

tan  
5
5 tan 40 tan 20

2 3
 2    1  2  1  1 Ans.
5 5
Example 7. If tan 2 A = cot (A – 18°), where 2A is an acute angle, find the value of A.
[NCERT]

Solution.
We have, tan 2A = cot (A – 18°) = tan [90° – (A – 18°)]

2A = 108° – A

3A = 108°  A 
MATHEMATICS–X
[ cot   tan (90  )]
= tan (108° – A)
108
 A = 36° Ans.
3
INTRODUCTION TO TRIGONOMETRY
135
A
 B +C 
Example 8. If A, B, C are the interior angles of a triangle ABC, show that sin 
 = cos .
2
2


Solution. We know, in any ABC, A + B + C = 180°
 B + C = 180° – A
[NCERT]
J
A
B  C 180  A
A

 90 
2
2
2
taking sine to both sides, we get

A
 BC

sin 
  sin  90  
2
2



A
 BC 
 sin 
  cos
2
 2 
Hence shown.
Example 9. Prove the following Trigonometric identities :
J
A
[ sin (90  )  cos ]
1  cos 
 cosec   cot 
1  cos 
(ii) (sin  + cosec )2 + (cos  + sec )2 = 7 + tan2 + cot2
(i)
Solution.
B
[CBSE 2000]
(iii)
tan   sec   1 1  sin

tan   sec   1
cos 
(iv)
1
1
1
1
–

–
cosec A – cot A sin A sin A co sec A  cot A
(v)
tan A
cot A

 1  tan A  cot A  1  sec A cosec A
1  cot A 1  tan A
T
I
(i) We have, LHS =
M
A
=
[CBSE 2002]
[CBSE 2002(C)]
[CBSE 2002(C)]
1  cos 
1  cos  1  cos 


1  cos 
1  cos  1  cos 
(1  cos ) 2
(1  cos ) 2

2
1  cos 
sin 2 
1  cos 
1
cos 


sin 
sin  sin 
= cosec  + cot  = RHS
(ii) We have, LHS = (sin  + cosec )2 + (cos  + sec )2
= sin2  + cosec2 + 2 sin  cosec  + cos2 + sec2 + 2 cos sec 

2
2
2
2
= (sin   cos )  cosec   sec   2 sin 
1
1
 2 cos 
sin 
cos 
 1  cosec2   sec2   2  2
136
1
1 

2
2
 sin   cos   1, cosec   sin  , sec   cos  
INTRODUCTION TO TRIGONOMETRY
MATHEMATICS–X
= 5 + (1 + cot2) + (1 + tan2)
= 7 + tan2 + cot2
= RHS
[ cosec2 = 1 + cot2, sec2 = 1 + tan2]
J
A
(iii) We have, LHS  tan   sec   1
tan   sec   1

(tan   sec )  1
(tan   sec )  1

(sec   tan )  (sec 2   tan 2 )
tan   sec   1

(sec   tan )  (sec   tan )(sec   tan )
tan   sec   1

(sec   tan )[1  (sec   tan )]
tan   sec   1

(sec   tan )(1  sec   tan )
tan   sec   1
 sec   tan  

1  sin 
 RHS
cos 
T
I
(iv) We have, LHS 

M
A

B
1
sin 

cos  cos 
1
1

cos ec A  cot A sin A
1
co sec A  cot A
1
.

cos ecA  cot A cos ec A  cot A sin A
cosec A  cot A
 cosec A
cosec 2 A  cot 2 A
= cosec A + cot A – cosec A
= cot A
RHS 
[ cosec2 – cot2  = 1]
...(1)
1
1

sin A cosec A  cot A

1
1
cos ec A  cot A

.
sin A co sec A  cot A cosec A  cot A

1
cosec A  cot A

sin A cosec 2  cot 2 A
 cosec A  (cosec A  cot A)
= cot A
from (1) and (2), it follows that LHS = RHS.
MATHEMATICS–X
J
A
[ sec2 – tan2 = 1]
[ cosec 2   cot 2   1]
...(2)
INTRODUCTION TO TRIGONOMETRY
137
(v) We have, LHS 
tan A
cot A

1  cot A 1  tan A
1
tan A

 tan A
1
1  tan A
1
tan A

tan A
1

tan A  1 tan A(1  tan A)
tan A
2

tan A
1

tan A  1 tan A(1  tan A)

tan 2 A
1

tan A  1 tan A (tan A  1)

tan 3 A  1
tan A(tan A  1)

(tan A  1)(tan 2 A  tan A  1)
tan A (tan A  1)

tan 2 A  tan A  1
tan A
T
I
J
A
1 

 cot   tan  


B
J
A
[ a3 – b3 = (a – b) (a2 + ab + b2)]
tan 2 A tan A
1


tan A tan A tan A
= tan A + 1 + cot A = 1 + tan A + cot A

 1
sin A cos A

cos A sin A
 1
sin 2 A  cos 2 A
cos A sin A
 1
1
 1  cosec A sec A = RHS
sin A cos A
M
A
Example 10. (i) If tan  + sin  = m and tan  – sin  = n, show that m2  n 2  4 mn .
(ii) If sec  + tan  = p, show that
Solution.
138
[CBSE 2000]
p2  1
 sin 
p2  1
(i) We have, LHS = m2 – n2
= (tan  + sin )2 – (tan  – sin )2
= (tan2 + sin2 + 2tan  . sin ) – (tan2  + sin2 – 2 tan  sin )
= 4 tan  sin 
INTRODUCTION TO TRIGONOMETRY
...(1)
MATHEMATICS–X
RHS  4 mn
 4 (tan   sin )(tan   sin )
J
A
 4 tan 2   sin 2 
4
sin 2 
sin 2   sin 2  cos 2 
2

sin


4
cos 2 
cos 2 
4
sin 2 (1  cos 2 )
sin 2 .sin 2 

4
cos 2 
cos 2 
sin 4 
sin 2 
sin 
 4.
 4.sin .
2
cos 
cos 
cos 
= 4 sin  tan  = 4 tan  . sin 
from (1) and (2), it follows LHS = RHS.
 4.
p 2  1 (sec   tan ) 2  1

p 2  1 (sec   tan ) 2  1
(ii) We have, LHS 

sec 2   tan 2   2sec  tan   1
sec 2   tan 2   2sec  tan   1

(sec 2   1)  tan 2   2 sec  tan 
sec 2   2 sec  tan   (1  tan 2 )

tan 2   tan 2   2sec  tan 
sec 2   2sec  tan   sec 2 
T
I


M
A
2 tan 2   2sec  tan 
2sec 2   2sec  tan 
2 tan  (tan   sec )
2sec  (sec   tan )
J
A
...(2)
B
[ 1  tan 2   sec 2 ]
tan 
sin 
sin 


sec  cos .sec 
1
= sin  = RHS

PRACTICE EXERCISE
Question based on Trigonometric Ratios :
1. In ABC, right angled at B, if AB = 12 cm and BC = 5 cm, find (i) sin A and tan A (ii) sin C and cot C.
2. In each of the following, one of the six trigonometric ratios is given. Find the values of the other
trigonometric ratios.
(i) sin  
3
4
15
17
(ii) cot  
13
5
(v) cosec  
(iv) sec  
MATHEMATICS–X
(iii) cos  
41
9
(vi) tan  
7
25
5
12
INTRODUCTION TO TRIGONOMETRY
139
3. If tan A  2  1, show that sin A.cos A 
2
4
4. If cos A 
12
35
, verify that sin A(1  tan A) 
.
13
156
5. If sin A 
1
, verify that 3 cos A – 4 cos3 A = 0.
2
6. If sin  
5
, verify that tan2 – sin2 = tan2 . sin2.
13
1  cos  1
 .
1  cos  7
7. If 7 cot   24, prove that
5
tan 
sin 
8. If sec   , verify that

2
4
1  tan  sec 
9. If 4 cot   5, show that
5sin   3cos  7

5sin   2 cos  2
B
10. If 2 tan  = 1, find the value of 3cos   2 sin  .
2 cos   sin 
11. If cot  
2 sin  cos 
13
, find the value of
cos 2   sin 2 
12
T
I
a
cos   sin 
12. If tan   , find the value of
b
cos   sin 
13. If sec  
13
3cos   2sin 
, show that
3
5
9cos   4sin 
M
A
J
A
J
A
14. In ABC, right angled at B, AC + BC = 25 cm and AB = 5 cm, find the value of sin2A + cos2A.
15. If 21 cosec  = 29, find the value of :
2
(i)
2
cos   sin 
1  2sin 2 
(ii)
2 cos 2   1
cos 2   sin 2 
16. If tan  
1
1
 2; show that tan 2  
2
tan 
tan 2 
17. If sin A 
1
1
 3, find the value of sin 2 A 
sin A
sin 2 A
18. If cot   3, show that :
(i)
140
2  cos 2  11

2  sin 2  7
(ii) cosec2– cot2 = 1
INTRODUCTION TO TRIGONOMETRY
MATHEMATICS–X
19. If sec  
3sin   4sin 3  3tan   tan 3 
5

verify that
4
4cos 3   3cos 
1  3tan 2 
J
A
20. If B and Q are acute angles such that cos B = cos Q, show that B = Q.
Questions based on Trigonometric ratios of some specific angles
21. Evaluate each of the following :
(i) sin 60° cos 30° + cos 60° sin 30°
(iii) 2 cos2 60 cot 30° + 6 sin2 30° cosec2 60°
(v)
cot 2 30
cos 30  sin 2 30
(ii) cos 60° cos 30° – sin 60° sin 30°
(iv) 4 cot2 45° – sec2 60° + sin2 60° + cos2 90°
(vi)
2
(vii) 2 (cos2 45° + tan2 60°) – 6 (sin245° – tan230°)
(viii)
tan 2 60  3sec 2 30  4 cos 2 45  5 cos 2 90
cosec 30  sec 60  cot 2 30
22. Verify each of the following :
(i) sin 60° = 2 sin 30° cos 30° 
(iii) tan 60 
2 tan 30
1  tan 2 30
2 tan 30
1  tan 2 30
23. Show that :
T
I
(i) cosec2 30° sin2 45° – sec2 60° + 2 = 0
(iii)
cos 2 60  3cos 60  2
1
sin 2 60
24. If  = 30°, verify that :
2 tan 
1  tan 2 
2 tan 
(iii) tan 2 
1  tan 2 
25. If A = 45°, verify that :
(i) sin 2 A = 2 sin A . cos A
M
A
(i) sin 2 
(iii) sin A 
1  cos 2A
2
(v) tan(A  B) 
MATHEMATICS–X
tan A  tan B
1  tan A.tan B
B
(ii) cos 60  cos 2 30  sin 2 30 
1  tan 2 30
1  tan 2 30
2
2
2
(ii) 4 tan 30  sec 30  sin 45 
13
6
(iv) 4 (sin2 30° + cos2 60°) – 3 (cos2 45° – tan245°) =
(ii) cos2 
7
2
1  tan 2 
1  tan 2 
(iv) sin 3  = 3 sin  – 4 sin3 
(ii) cos 2 A = 2 cos2 A – 1
(iv) cos A 
26. If A= 60° and B = 30°, verify that :
(i) sin (A + B) = sin A cos B + cos A sin B
(iii) cos (A + B) = cos A cos B – sin A sin B
J
A
5sin 2 30  cos 2 45  4 tan 2 30
2sin 30 cos 30  tan 45
1  cos 2A
2
(ii) sin (A – B) = sin A cos B – cos A sin B
(iv) cos (A – B) = cos A cos B + sin A sin B
(vi) tan(A  B) 
tan A  tan B
1  tan A.tan B
INTRODUCTION TO TRIGONOMETRY
141
27. Given that sin (A + B) = sin A cos B + cos A sin B, find the value of sin 75°.
28. Given that cos (A – B) = cos A cos B + sin A sin B, find the value of cos 15°.
29. Find x in each of the following, if :
(i) 2 cos x = 1
(ii) 2 sin 2x = 1
(iii) tan 3x  3
(iv) 2sin 3x  3
(v)
(vi)
3 cot 3x  1
30. Find acute angle  in each of the following cases; if,
(i) sin (3 – 15°) = 1
2
(iii) sin  
J
A
(ii) 2 sin (3 – 15°)  3
1
4
(iv) 3 tan 2   1  0
(vi) cot 2 (2  5)  3
(v) cos (40° + ) = sin 30°
31. If sin (A  2B) 
3
and cos (A + 4B) = 0, find the values of angles A and B.
2
32. If sin (A + B) = 1 and cos (A  B) 
33. If sin (A – B) 
J
A
 x
3.sec    2
2
B
3
; 0  A  B  90, and A > B; find A and B.
2
1
1
and cos (A + B)  ; 0 < A + B  90°, and A > B, find A and B.
2
2
T
I
M
A
12 cm
34. ABC is a right triangle, right angled at C. If A = 30°, and AB = 40 units, find the remaining two sides and
B of ABC.
A
E
B
35. ABCD is a rectangle with AD =12 cm and DC = 20 cm
as shown. The line segment DE is drawn making an
angle of 30° with AD, intersecting AB in E. Find the
30°
lengths of DE and AE.
D
20 cm
C
Questions based on Trigonometric Identitities of Complementary Angles
36. Evaluate each of the following :
(i)
cos 49
sin 41
(ii)
sec 32
cosec 58
37. Evaluate each of the following :
(i) sin 54° – cos 36°
(iii) cosec 47° – sec 43°
(v) sin2 29° + sin2 61
38. Evaluate each of the following :
(i)
142
cos10
 cos 59cosec 31
sin 80
(iii)
tan 21
cot 69
(ii) tan 62° – cot 28°
(iv) sec2 31° – cot2 59°
(vi) tan2 48° – cosec242°
(ii)
tan 53 cot 79

cot 37 tan11
INTRODUCTION TO TRIGONOMETRY
MATHEMATICS–X
3sin 62
sec 42
(iii) cos 28  cosec 48
(iv)
2 cos 67 tan 40

 sin 90
sin 23 cot 50
J
A
39. Evaluate each of the following :
(i)
2
cos 2 20  cos 2 70
sin 2 57  sin 2 33
2
 sin 27   cos 63 
(ii) 
 

 cos 63   sin 27 
 tan 36   cot 54 
(iii) 
 

 cot 54   tan 36 
2
2
2
 cosec 27   sec 63 
(iv) 

 
 sec 63   c osec 27 
2
J
A
40. Evaluate each of the following :
(i) tan 5° tan 25° tan 30° tan 45° tan 65° tan 85° (ii) cot 12° cot 38° cot 52° cot 60° cot 78°
(iii) sec (35° + A) – cosec (55° – A)
(iv) cos (36° + A) – sin (54° – A)
41. Prove that :
(i)
cos (90  )
sin 

2
sin 
cos (90  )
(ii) cos  . cos (90° – ) + cos  . sin (90° – ) = 1
(iii)
(iv)
B
sin .cos (90   cos  cos  sin (90  ).sin 

1
sin (90  )
cos (90  )
cos (90  ) . sec (90  ) . tan 
tan (90  )

2
cos ec (90  ) sin (90  ) .cot (90  )
cot 
T
I
42. Without using trigonometric tables, find the value of each of the following :
(i) sin (50° + ) – cos (40° – ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89°
(ii) cos (40  )  sin (50  ) 
2
cos 2 40  cos 2 50
sin 2 40  sin 2 50
[CBSE 2002]
[CBSE 2002]
2
 sin 47   cos 43 
2
2
2
(iii) 
 
  4 cos 45  cosec 67  tan 23
 cos 43   sin 47 
M
A
sin15 cos 75  cos15 sin 75
(iv) sec 2 10  cot 2 80 
cos  sin(90  )  sin  . cos (90  )
[CBSE 2002]
(v) cot  tan (90° – ) – sec (90° – ) cosec  + sin2 25° + sin2 65° + 3 (tan 5° tan 45° tan 85°)
(vi)
(vii)
3tan 25 tan 40 tan 50 tan 65 1
 tan 2 60
2
2
4 (cos 29  cos 61)
2
 tan  cot (90  )  sec  cosec (90  )  sin 2 35  sin 2 55
tan10 tan 20 tan 45 tan 70 tan 80
cosec 2 (90  )  tan 2  2 tan 2 30 sec 2 52 sin 2 32
(viii) 4 (cos 2 48  cos 2 42)  cosec 2 70  tan 2 20
MATHEMATICS–X
INTRODUCTION TO TRIGONOMETRY
[CBSE 2004]
[CBSE 2002 (C), 2005]
[CBSE 2006]
143
2
2
 tan 20   cot 20 
(ix) 
 
  2 tan15 tan 37 tan 53 tan 60 tan 75
 cosec70   sec 70 
[CBSE 2003]
sec 39
2

.tan17 tan 38 tan 60 tan 52 tan 73  3(sin 2 31  sin 2 59)
cosec 51
3
(x)
J
A
[CBSE 2006 (C)]
43. Express each of the following in terms of trigonometric ratios of angles between 0° and 45° :
(i) sin 79° + cosec 83°
(ii) tan 58° + sec 46°
(iii) cosec 57° + cot 57°
(iv) sin 82° + tan 82°
44. In any ABC, prove that :
C
AB
(i) sin 
  cos
2
2


45.
46.
47.
48.
49.
J
A
A
 BC
(ii) tan 
  cot
2
2


If sin 3 = cos ( – 6°), where 3 and  – 6°) are acute angles, find the value of .
If tan 2 = cot ( + 6°), where 2 and  + 6° are acute angles, find the value of .
If sec 5  = cosec ( – 36°), where 5 is an acute angle, find the value of .
If sin A = cos B, prove that A + B = 90°.
Prove that :
(i) tan 1° tan 2° tan 3° ..... tan 89° = 1
B
19
.
2
50. If A and B are acute angles and sin (A + B) = cos (A – B), show that A = 45°.
(ii) sin2 5 + sin2 10° + sin2 15° + ..... + sin2 85° + sin2 90° 
T
I
Questions based on Trigonometric Identities :
51. Using trigonometric identities, write the following expressions as an integer :
(i) 5 cot2 A – 5 cosec2 A
(ii) 4 tan2  – 4 sec2 
(iii) 7 cos2 + 7 sin2
(iv) 3 sec2 – 3 tan2
52. Simplify the following expressions :
M
A
(i) (1 + cos ) (cosec  – cot )
(iii)
(v)
1
1

1  sin A 1  sin A
1  tan 2 
cot 2   1
sin 3   cos3 
(ii)
sin   cos 
(iv)
sin 4 A  cos 4 A
sin 2 A  cos 2 A
(vi) cosec  (1+ cos ) (cosec  – cot )
Prove that following identities (53–100) :
53. cos2  (1 + tan2) = 1
55.
144
sin 2   cos 2 
 tan   cot 
sin  cos 
54. cosec2 + sec2 = cosec2 . sec2
2
56. (cot   cosec ) 
INTRODUCTION TO TRIGONOMETRY
[CBSE 2001]
1  cos 
1  cos 
MATHEMATICS–X
57. (1 + tan2 ) (1 + sin ) (1 – sin ) = 1
2sin 2   1
sin  cos 
58.
sin A
 cosec A  cot A
1  cos A
60.
1
1

 2cosec 2 
1  cos  1  cos 
59.
tan   cot  
61.
tan   sin  sec   1

tan   sin  sec   1
62. sec A (1 – sin A) (sec A + tan A) = 1
63.
1  sin 
cos 

 2 sec 
cos  1  sin 
1  tan 2   1  tan  

64.

1  cot 2   1  cot  
65.
tan 
cot 

 1  tan   cot   1  sec  cosec 
1  cot  1  tan 
66.
 1  cos   1  cos 

 
 sin   1  cos 
67.
68.
1  cot 2 
 sin 2 
1  cot 4 
69.
2
J
A
[CBSE 2002 (C)]
1  cot A sin A  cos A

1  cot A sin A  cos A
1  sin A
 sec A  tan A
1  sin A
B
70.
1  sin A
cos A

1  sin A 1  sin A
71.
72.
1  cos 
 cosec   cot 
1  cos 
1
73. cosec   cot   cosec   cot 
74.
1  cos 
1  cos 

 2 cosec 
1  cos 
1  cos 
T
I
J
A
2
1  cos 
 cosec   cot 
1  cos 
75. (sin  + sec )2 + (cos  + cosec )2 = (1 + sec  cosec )2
76.
sin A  cos A sin A  cos A
2
2



2
2
2
sin A  cos A sin A  cos A sin A  cos A 2 sin A  1
77.
sin   2sin 3 
 tan 
2cos3   cos 
M
A
[CBSE 2000]
sin 
sin 
78. cot   cosec   2  cot   c osec  [CBSE 2000]
79. (1 + cot  – cosec ) (1 + tan  + sec ) = 2
80.
81.
82.
83.
[CBSE 2000 (C)]
[CBSE 2000]
cosec 
cosec 

 2sec2 
cosec   1 cosec   1
1 
1 
1

1 
1 

2
2
2
4
 tan A  cot A  sin A  sin A
1
1
1
1



sec A  tan A cos A cos A sec A  tan A
1
1
1
1



cosec A  cot A sin A sin A c osec A  cot A
MATHEMATICS–X
INTRODUCTION TO TRIGONOMETRY
[CBSE 2006(C), 2007]
[CBSE 2005]
[CBSE 2002C, 2006]
145
84.
sec A  tan A  1 1  sin A

tan A  sec A  1
cos A
86.
sin   cos   1 1  sin 

sin   cos   1
cos 
cot A  cosec A  1 1  cos A
85. cot A – cosec A + 1  sin A
87.
sec   tan 
1

 sec   tan 
sec   tan  sec   tan 
88.
sec   tan 
1
 sec   tan  
sec   tan 
sec   tan 
90.
cos 2 
sin 3 

 1  sin  cos 
1  tan  sin   cos 
91.
1  cos   sin 2 
 cot 
sin (1  cos )
92.
tan 
tan   1

1  cot  2  cosec 2 
tan 2 A  tan 2 B 
[CBSE 2000(C)]
[CBSE 2003]
B
93. sec2 A  cosec2 A  tan A  cot A
cos 2 B  cos 2 A sin 2 A  sin 2 B

cos 2 B.cos 2 A
cos 2 A cos 2 B
T
I
96. 2 (sin6  + cos6 ) – 3 (sin4  + cos4) + 1 = 0
sec   1
sec   1

 2 cosec 
sec   1
sec   1
97.
98.
sin   sin  cos   cos 

0
cos   cos  sin   sin 
M
A
100.
J
A
89. sec4 – sec2 = tan4 + tan2
94. 2 sec2  – sec4  – 2 cosec2  + cosec4  = cot4 – tan4
95.
J
A
[CBSE 2001(C)]
99.
[CBSE 2000]
[CBSE 2005]
[CBSE 2001, 2006(C)]
sin 2 
cos3 

 1  sin  cos 
1  cot  cos   sin 
sin A
cos A

1
sec A  tan A  1 cosec A  cot A  1
101. If cos  + sin  =
2 cos , show that cos – sin  =
2 sin 
[CBSE 2002(C)]
102. If sin  + cos  = p and sec  + cosec  = q, show that q ( p 2  1)  2 p
103. If x = a sec  + b tan  and y = a tan  + b sec , prove that x2 – y2 = a2 – b2.
2000 (C)]
104. If
[CBSE 2001,
cos 
cos 
 m and
 n, show that (m2 + n2) cos2  = n2
cos 
sin 
105. If a cos  + b sin  = m and a sin  – b cos  = n prove that a2 + b2 = m2 + n2.
146
INTRODUCTION TO TRIGONOMETRY
MATHEMATICS–X
106. If a cos  = x and b cot  = y, show that
107. If
a2 b2

1
x2 y 2
J
A
x2 y2
x
y
x
y
2
2
cos   sin   m and sin   cos   n, prove that 2  2  m  n .
a
b
a
b
a
b
108. If sec   x 
1
1
, prove that sec  + tan  = 2x or
2x
4x
109. If sec   tan   p, prove that
[CBSE 2001]
p 2 1
 sin 
p2  1
J
A
[CBSE 2004]
110. If tan  + sin  = m and tan  – sin  = n, prove that m2  n2  4 mn . [CBSE 2000, 2002(C), 2004(C)]
111. If a cos  – b sin  = c, prove that a sin   b cos    a 2  b 2  c 2
112. If 3sin   5cos   5, prove that 5sin   3cos   3
113. If sin  + sin2 = 1, prove that cos2+ cos4 = 1
[CBSE 2001(C)]
[CBSE 2002(C)]
B
114. If (sec A + tan A) (sec B + tan B) (sec C + tan C) = (sec A – tan A) (sec B – tan B) . (sec C – tan C), prove
that each side is equal to ± 1.
115. If x = r sin  cos , y = r sin  sin  and z = r cos , then show that x2 + y2 + z2 = r2.
HINTS TO SELECTED QUESTIONS
5sin   3cos 
5
9. here, cot   . Consider LHS =
5sin   2 cos 
4
T
I
dividing numerator and denominator by sin , we get
5  3cot 
LHS 

5  2 cot 
5
5  3 
4  7
5 2
5  2 
4
M
A
16.
tan  
1
 2. Squaring both sides,
tan 
tan 2  

1
1
 2.tan .
4
2
tan 
tan 
tan 2  
1
 4  2  2.
tan 2 
20. Consider two right triangles ACB and PRQ.
BC
QR
We have, cos B 
, cos Q 
AB
PQ
Since, cos B  cos Q 
MATHEMATICS–X
BC QR

 k , say ...(1)
AB PQ
A
C
P
B
INTRODUCTION TO TRIGONOMETRY
R
Q
147
Now,
AC

PR
AB2  BC2
2
PQ  QR
2
AB2  k 2 AB2

2
2
PQ  k PQ
2

AB
PQ
J
A
Thus, In ACB and PRQ, we have
AC AB BC


 ACB  PRQ  B  Q.
PR PQ QR
27. sin 75° = sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30°
31.
3
 sin 60  A  2B  60
2
and, cos (A + 4B) = 0 = cos 90°  A + 4B = 90°
sin (A  2B) 
...(1)
J
A
...(2)
solve (1) and (2) now.
49. (ii) sin2 5° + sin2 10° + sin2 15° + .... + sin2 85° + sin2 90°
 (sin 2 5  sin 2 85)  (sin 2 10  sin 2 80)  ...  (sin 2 40  sin 2 50)  sin 2 45  sin 2 90
 (sin 2 5  cos 2 5)  (sin 2 10  cos 2 10)  ...  (sin 2 40  cos 2 40)  sin 2 45  sin 2 90
2
 1 
1
19
2
 1
 1  .....  1  
  (1)  8   1 
2
2
2


8 times
1  cos  1  cos 
1  cos  1  cos 



1  cos  1  cos 
1  cos  1  cos 
74. LHS 
T
I
B
2
(1  cos )2
(1  cos ) 2 1  cos  1  cos 

 2cosec   RHS



2
2
sin 
1  cos 
1  cos 
sin 
sin 

86. LHS  sin   cos   1 dividing numerator and denominator by cos ,
sin   cos   1
tan   1  sec  tan   sec   1

tan   1  sec  tan   sec   1
M
A
LHS 
90. LHS =
148
2

(tan   sec )  (sec   tan 2 )
tan   sec   1

(tan   sec )  (sec   tan )(sec   tan )
tan   sec   1

(tan   sec )(1  sec   tan )
= tan  + sec   sin   1  1  sin   RHS
tan   sec   1
cos  cos 
cos 
cos 2 
sin 3 

1  tan  sin   cos 
INTRODUCTION TO TRIGONOMETRY
MATHEMATICS–X


cos 2 
sin 3 
cos 2 
sin 3 



sin  sin   cos  cos   sin  cos   sin 
1
cos 
J
A
cos3   sin 3  (cos   sin )(cos 2   sin 2   cos  sin )

cos   sin 
cos   sin 
= 1 + sin  cos  = RHS
96.
sin 6   cos6   (sin 2 ) 3  (cos 2 ) 3
J
A
 (sin 2   cos 2 ) [(sin 2 ) 2  (cos 2 ) 2  (sin 2 )(cos 2 )]
[ a 3  b 3  ( a  b)( a 2  b 2  ab)]
 1.[(sin 2 ) 2  (cos 2 ) 2  2 sin 2  cos 2   3sin 2  cos 2 ]
 (sin 2   cos 2 ) 2  3sin 2  cos 2   1  3sin 2  cos 2 
also, sin4  + cos4  = (sin2 )2 + (cos2 )2 + 2 sin2  cos2  – 2 sin2  cos2 
B
= (sin2  + cos2 )2 – 2 sin2 cos2 = 1 – 2 sin2  cos2 
Now, put these values in LHS and simplify.
101. cos  + sin  =
2 cos   sin  = ( 2  1) cos   cos  
 cos   sin .

1
2 1
T
I

2 1
2 1
 cos   ( 2  1).sin 
1
1 
1 


 tan     x    1    x  
4x
4x 
4x 


M
A
sec   x 

.sin 
cos   2 sin   sin   cos   sin   2 sin 
2
108.
1
2 1
1 

tan     x  
4x 

2
1
1
1
Now, when tan     x   , then sec  + tan   x 
 x
 2x
4x
4x
4x 

1 
1  
1  1


and, when tan     x   , then sec   tan    x     x   
4x 
4x  
4x  2 x


111. Consider, (a cos  – b sin )2 + (a sin  + b cos )2
 a 2 cos2   b 2 sin 2   2ab sin  cos   a 2 sin 2   b2 cos2   2ab sin  cos 
 a 2 (cos 2   sin 2 )  b 2 (cos 2   sin 2 )  a 2  b 2
MATHEMATICS–X
INTRODUCTION TO TRIGONOMETRY
149
 c 2  (a sin   b cos ) 2  a 2  b 2
J
A
 a sin   b cos    a 2  b 2  c 2
114. We have, (sec A + tan A) (sec B + tan B) (sec C + tan C) = (sec A – tan A) (sec B – tan B) .
(sec C – tan C). Multiplying both sides by (sec A – tan A) (sec B – tan B) (sec C – tan C), we get
(sec2A – tan2A) (sec2B – tan2B) (sec2C – tan2C) = [(sec A – tan A) (sec B – tan B) (sec C – tan C)]2

(sec A – tan A) (sec B – tan B) (sec C – tan C) = ± 1
Similarly, multiplying by (sec A + tan A) (sec B + tan B) (sec C + tan C), we get
(sec A + tan A) (sec B + tan B) (sec C + tan C) = ± 1.
2
2
2
115. LHS = x + y + z
= r2 sin2 cos2 + r2 sin2  sin2 + r2cos2
= r 2 sin 2  (cos 2   sin 2 )  r 2 cos 2 
 r 2 sin 2   r 2 cos 2   r 2 (sin 2   cos 2 )  r 2  RHS.
B
J
A
MULTIPLE CHOICE QUESTIONS
Mark the correct alternative in each of the following :
1. If 3 tan  = 4, then the value of
(a) 1
(b)
1  sin 
is :
1  sin 
T
I
1
2
2. If 5 sin  – 3 = 0, then the value of
(a)
1
4
(b)
1
8
M
A
3. If tan  
(a)
a2  b2
a 2  b2
2
3
5. If sin  
(a) 1
150
1
3
(d)
2
3
cosec   cot 
is :
2 cot 
(c)
3
4
(d)
7
8
(c)
ab
a b
(d)
a–b
ab
(c)
1
6
(d)
5
2
(c)
7
4
(d)
28
29
a
a sin   b cos 
, then the value of
is
b
a sin   b cos 
(b)
a 2  b2
a2  b2
4. If 5 tan   4  0 , then the value of
(a)
(c)
(b)
5sin   3cos 
is
5sin   2 cos 
1
3
4
4 tan   5 cos 
, then the value of
is
5
sec   cot 
(b)
5
8
INTRODUCTION TO TRIGONOMETRY
MATHEMATICS–X
6. The value of 4 cot245° – sec260° + sin260° + cos290° is
1
2
3
(b)
(c)
4
4
4
7. If tan 3x = sin 45° cos 45° + sin 30°, then the value of x is :
(a)
(a) 15°
(b) 30°
4
(d) 1
(c) 45°
4
2
J
A
(d) 60°
2
8. The value of 4 (sin 30° + cos 30°) – 3 (sin 45° – 2 cos 45°) is
(a) 0
9. The value of
(a) 0
(b) 1
(c) 4
sin 2 63  sin 2 27
is
cos 2 17  cos 2 73
(b) 1
(c) 2
10. The value of tan 5° tan 10° tan 15° tan 75° tan 80° tan 85° is :
(a) –1
11. The value of
(a) –1
(b) 0
(c) 1
sin 75 sin12

 cos18cosec 72 is :
cos15 cos 78
(b) 0
(c) 1
2
2
B
 sin 35   cos55 
12. The value of 
 
  2 cos 60 is :
 cos 55   sin 35 
(a) 0
(b) 1
(c) –1
13. The value of
(a) 0
sin 75  sin15
is :
cos15  cos 75
(b) –1
T
I
14. If sin   cos 4, then the value of  is :
(a) 18°
15. The value of
(b) 36°
J
A
(c) 45°
(d) none of these
(d) none of these
(d) none of these
(d) 2
(d) none of these
(d) none of these
sec  cosec (90  )  tan  cot (90  )  (sin 2 35  sin 2 55)
is
tan10 tan 20 tan 45 tan 70 tan 80
(b) –1
(c) 2
(d) 3
M
A
(a) 1
(c) 1
(d) none of these
16. (cosec A – sin A) (sec A – cos A) (tan A + cot A) is equal to :
(a) 1
17.
(b) –1
(c) 2
(d) –2
cosec   1
cosec   1

is equal to :
cosec   1
cosec   1
1
cos 
18. If x = r sin  cos , y = r sin  sin  and z = r cos , then
(a) cos 
(b) 2 cos 
(c)
(a) x 2  y 2  z 2  r 2
(b) x 2  y 2  z 2  r 2
(c) x 2  y 2  z 2  r 2
(d) z 2  y 2  x 2  r 2
(d)
2
cos 
19. If a cot  + b cosec  = p and b cot  + a cosec  = q, then p2 – q2 is equal to :
(a) a2 – b2
(b) b2 – a2
(c) a2 + b2
(d) a + b
MATHEMATICS–X
INTRODUCTION TO TRIGONOMETRY
151
20. If (1 + cos ) (1 + cos ) (1 + cos ) = (1 – cos ) (1 – cos ) (1 – cos ), then each of the expression is equal
to :
(a) sin  sin  sin 
(b) cos  cos  cos 
(c) ± sin  sin  sin 
(d) ± cos  cos  cos 
J
A
VERY SHORT ANSWER TYPE QUESTIONS (1 MARK QUESTIONS)
1. What is the maximum value of
2. If sin 2 
1
?
cosec 
3
, find the value of cos .
2
3. If sin   cos   1, the find the value of sin  cos .
J
A
15
and     90 , find the value of sec .
7
5. What is the value cos 1° cos 2° cos 3° .... cos 89° cos 90° ...... cos 180°?
4. If cosec  =
6. If 7 sin2  + 3 cos2 = 4, find the value of tan  if 0°    90°.
7. Write the value of sin2 68° + sin2 22° – 1.
8. If A and B are acute angles and sin B = cos A, then write the value of A + B.
9. Write the value of sin (55° + ) – cos (35° – ).
10. Express tan 87° + sin 63° in terms of trigonometric ratios of angles between 0° and 45°.
11. Write the value of tan 5° tan 35° tan 45° tan 55° tan 85°.
T
I
B
sin 
cos 
12. Write the simplest form of sec (90  )  cosec (90  ) .
13. If tan ( – 36°) = cot 2; 2 and ( – 36°) are acute angles, then find .
14. If cosec2  (1 + cos ) (1 – cos ) = k, then find the value of k.
15. If sin 2 = cos 3, then find the value of .
16. Find the value of
cos   cos3 
.
sin   sin 3 
M
A
17. If tan A 
1
and sin B 
1
, find the value of A + B.
3
2
18. If cos 3 = 1, then find the value of tan .
 BC 
19. If A, B and C are the angles of a triangle, then find the value of cot 
 in terms of A.
 2 
20. If tan2  – 5 tan  + 1 = 0, find the value of tan  + cot .
PRACTICE TEST
M.M : 30
Time : 1 hour
General Instructions :
Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each.
152
INTRODUCTION TO TRIGONOMETRY
MATHEMATICS–X
4 sin   3cos 
.
2 sin   6 cos 
2. Evaluate : 4 cot2 45° + sin2 60° + cos2 90° – sec2 60°.
1. If 3 cot  = 2, find the value of
J
A
3. If sin ( + 36°) = cos , where  + 36° and  are acute angles, find the value of .
4. If cosec  + cot  = a and cosec  – cot  = b, prove that ab = 1.
5. In OPQ, right angled at P, OP = 7 cm, OQ – PQ = 1 cm. Determine the values of sin Q and cos Q.
1  tan 2 
.
1  tan 2 
7. Without using trigonometric tables, evaluate :
6. If  = 30°, verify that cos 2 
2
2
sin 20  sin 70 sin (90  )sin  cos (90  ).cos 


cos 2 20  cos 2 70
tan 
cot 
8. Prove that :
J
A
tan A
cot A

 1  sec A . cosec A
1  cot A 1  tan A
9. A contractor plans to instal two slides for the children to play in a park. For the children below the age
of 6 years, he prefers to have a slide, which is inclined to an angle of 30° to the ground, whereas for elder
children, he wants to have a steep slide, inclined at an angle of 60° to the ground. The length of the slide
in both cases is 3m. What should be the required length of the ladder, in each case, to reach the top of
the slide and at what distance should it be placed from the base of the slide?
B
10. If tan A + sin A = m and tan A – sin A = n, prove that (m2 – n2)2 = 16 mn.
T
I
ANSWERS OF PRACTICE EXERCISE
1. (i)
5 5
,
13 12
(ii)
12 5
,
13 12
8
15
17
17
8
, tan   , cosec   , sec   , cot  
17
8
15
8
15
2. (i) cos  
M
A
(ii) sin  
4
3
4
5
5
, cos   , tan   , cosec   , sec  
5
5
3
4
3
(iii) sin  
24
24
25
25
7
, tan  
, sec  
, cosec  
, cot  
25
7
7
24
24
(iv) sin  
12
5
12
13
5
, cos   , tan   , cosec   , cot  
13
13
5
12
12
(v) sin  
9
40
9
41
40
, cos  
, tan  
, sec  
, cot  
41
41
40
40
9
(vi) sin  
5
12
13
13
12
, cos   , cosec   , sec   , cot  
13
13
5
12
5
MATHEMATICS–X
INTRODUCTION TO TRIGONOMETRY
153
10.
8
3
312
25
11.
17. 11
21. (i) 1 (ii) 0 (iii)
3 1
27.
12.
28.
2 2
b a
ba
3
34
(iv)
4
2
3 1
2 2
14. 1
(v) 6 (vi)
5
(2  3) (vii) 6
6
33. A = 45°, B = 15°
31. A = 30°, B = 15°
J
A
34. AC = 20 3 units, BC = 20 units and B = 60°
35. DE = 8 3 cm, AE = 4 3 cm
36. (i) 1 (ii) 1 (iii) 1
37. (i) 0 (ii) 0 (iii) 0 (iv) 1 (v) 1 (vi) –1
38. (i) 2 (ii) 0 (iii) 2
39. (i) 1 (ii) 2 (iii) 2 (iv) 0
40. (i)
1
3
(ii)
1
B
3
(iv) 0
(iii) 0 (iv) 0
3
5
3 (vi) 4 (vii) 2 (viii) 12 (ix) 1  2 3 (x) 0
42. (i) 1 (ii) 1 (iii) 1 (iv) 2 (v)
J
A
(viii) 9
29. (i) 60° (ii) 15° (iii) 20° (iv) 20° (v) 20° (vi) 60°
30. (i) 35° (ii) 25° (iii) 30° (iv) 30° (v) 20° (vi) 17.5°
32. A = 60°, B = 30°
15. (i) 1 (ii) 1
43. (i) cos 11° + sec 9° (ii) cot 32° + cosec 44° (iii) sec 33° + tan 33° (iv) cos 8° + cot 8°
45. 24°
46. 28°
47. 21°
51. (i) –5 (ii) –4 (iii) 7 (iv) 3
2
52. (i) sin  (ii) 1 – sin  cos  (iii) 2 sec A (iv) 1 (v) tan2  (vi) 1
T
I
ANSWERS OF MULTIPLE CHOICE QUESTIONS
1.
6.
11.
16.
(c)
(c)
(c)
(a)
2. (b)
7. (a)
12. (b)
17. (d)
M
A
3. (b)
8. (c)
13. (c)
18. (a)
4. (c)
9. (b)
14. (a)
19. (b)
5. (d)
10. (c)
15. (c)
20. (c)
ANSWERS OF VERY SHORT ANSWER TYPE QUESTIONS
1. 1
2.
3
2
15
7
3. 0
4.
7. 0
8. 90°
9. 0
10. cot 3° + cos 27°
11. 1
12. 1
13. 42°
14. 1
15. 18°
16. tan 
17. 75°
18. 0
19. tan
1
6.
3
5. 0
A
2
20. 5
ANSWERS OF PRACTICE TEST
154
1.
1
3
7. 2
3
7
24
INTRODUCTION
TO
TRIGONOMETRY
2.
3. 27°
5. sin   , cos   MATHEMATICS–X
4
25
25
9. 2 3 m, 3 m