Class 10 Maths Extra Questions Chapter 8 Introduction to Trigonometry

SHORT ANSWER TYPE QUESTIONS (1)
Prove that :
20. sec4  – sec2 = tan4 + tan2 

1 + sin 
21. = tan + Sec 
1 – sin 

22. If x = p sec + q tan  & y = p tan  + q sec then prove that x2 – y2 = p2 – q2

1
23. If 7 sin2 + 3 cos2 = 4 then show that tan =
3
1 1
24. If sin (A – B) = , cos (A + B) = then find the value of A and B.
2 2

cos 2 20° + cos 2 70°
25. Find the value of .
sin 2 59° + sin 2 31°

cosec 2 A + 1
26. If 3 cot A = 4, find the value of .
cosec 2 A – 1

27. If tan (3x – 15°) = 1 then find the value of x.
28. If A, B, C are interior angles of ABC, the prove that

 A  B  C
cos ec   sec   .
 2   2

(CBSE 2011)
29. In ABC, right angled at B, AB = 5 cm and ACB = 30°. Find BC and AC.
1  sin 60
30. Show that : = 2  3. (CBSE, 2014)
cos 60
cos  cos 
31. Find the value of , if   4 ,  90°. (CBSE, 2014)
1  sin  1  sin 

Mathematics-X 121

SHORT ANSWER TYPE QUESTIONS
Prove that :
tan A + Sec A – 1 1+Sin A
32. tan A – Sec A+1 = Cos A
1 1 1 1
33. –  –
sec x – tan x cos x cos x sec x  tan x
tan  cot 
34. + =1+ tan  + cot  = sec  cosec  +1 (CBSE 2019)
1– cot  1 – tan 
35. (sin + cosec )2 + (cos + sec )2 = 7 + tan2 + cot2 
36. sec A (1 – sin A) (sec A + tan A) = 1
1 1
37. If sec = x + , prove that sec + tan = 2x or
4x 2x
38. If sin + sin2 = 1, prove that cos2 + cos4 = 1
39. Without using trigonometric table, the value of
cot tan (90° – ) – sec (90° – ) cosec + sin2 65° + sin2 25° + 3 tan 5°
tan 85°.
cot (90° – ) cosec(90° – ) sin 
40. Prove that : + = sec2 
tan  tan (90° – )
41. Find the value of :

cos 2 20° + cos 2 70°
+ 2 Cosec2 58° – 2 Cot 58° tan 32° – 4 tan 13° tan 37°
sec 2 50° – cot 2 40
tan 77° tan 45° tan 53°.

42. If A, B, C are the angles of ABC then prove that cosec2

sin 15°cos 75° +cos 15° sin 75°
43. Find the value of sec2 10° – cot2 80° + .
cos  sin (90° – ) + sin  cos (90° – )
44. If cos  + sin  = 2 cos , then show that cos  – sin  = 2 sin .
tan 2 60  4cos 2 45  3sec2 30  5cos 2 90
45. Evaulate :
cosec 30°  sec 60  cot 2 30
46. If a cos  + b sin  = m and a sin  – b cos  = n (CBSE, 2001 C)
Prove that : a2 + b2 = m2 + n2
122 Mathematics-X

LONG ANSWER TYPE QUESTIONS
Prove That:

 1  1  1
47. 1 +  1 +  = sin 2  – sin 4 
 tan   
2
cot  
2

48. 2 (sin6 + cos6 ) – 3 (sin4 + cos4 ) + 1 = 0
49. (1 + cot A + tan A) (sin A – cos A) = sin A tan A – cot A cos A
50. If sin + cos = m and sec + cosec = n then show that n(m2 – 1) = 2m
51. find the value of :

cot (90° – ) tan  – cosec (90° – ) sec  sin 2 (50° + ) + sin 2 (40° – )
+
sin 12° Cos 15° sec 78° cosec75° tan 15° tan 37° tan 53° tan 75°

52. Prove that :
1 1 1 1
– = –
cosec  + cot  sin  sin  cosec  – cot 

cos  cos 
53. If = m and = n , then prove that (m2 + n2) Cos2  = n2
cos  sin 

54. Prove that :

2
sin 2  – 2sin 4 
sec  – =1
2cos 4  – cos 2 

55. cot tan (90° – ) – sec (90° – ) cosec + 3 tan 12° tan 60° tan 78° find its
value.
56. Find the value of —

sec (90° – ) cosec  – tan (90° – ) cot  + cos 2 25° + cos 2 65°
3 tan 27° tan 63°

57. If sin  + cos  = 3 , then prove that tan  + cot  = 1 (CBSE 2020)

cotA – cosA
58. Prove = sec2A + tan2A – 2secA tanA (CBSE 2020 Basic)
cotA+ cosA

Mathematics-X 123

sin   2sin 3 
59. Prove = tan  (CBSE 2020 Basic)
2 cos3   cos 

1
60. If cos(A + B) = sin(A – B) = , 0 < A + B < 90° and A > B then find the value
2
of A and B. (CBSE 2020 Basic)
61. If tan  + sin  = m, tan  – sin  = n, then prove that m2 – n2 = 4 mn .
(CBSE 2020 Standard)

sec 2 (90 – )  cot 2  2 cos 2 60 tan 2 28 tan 2 62
62. Find 
2(sin 2 25  sin 2 65) 3(sec 2 43  cot 2 47)
(CBSE 2020 Standard)
1  sec   tan  1  sin 
63. Prove = (CBSE 2020 Standard)
1  sec   tan  cos 
2
 3sin 43  cos 37 cosec 53
64. Evaluate   –
 cos 47  tan 5 tan 25 tan 45 tan 65 tan 85
(CBSE 2019)
sin  sin 
65. Prove that = 2 (CBSE 2019)
cot   cosec  cot   cosec 

4 sin   cos   1
66. If 4 tan  = 3 then find the value of (CBSE 2018)
4 sin   cos   1

sinA – 2sin 3A
67. Prove = tan A (CBSE 2018)
2 cos3A – cosA

ANSWERS AND HINTS

1. 45° 2. 30°
3. 48° 4. 7
1
5. 6. a2
3
7. 0 8. 9

124 Mathematics-X

1  cot 2 
9. 10. 0°
cot 
11. 40° 12. tan2 
1
13. 14. sec 42° + cot 2°
2
15. (c) 16. (d)
17. (a) 18. (a)
19. (a)
20. LHS = sec2  (sec2 – 1)
RHS = tan2  (tan2  + 1)
Use 1 + tan2  = sec2 
21. Relationalise and proceed in LHS
22. Squaring both sides of x and y and subtracting.
23. Divide both sides by cos2
24. A = 45°, B = 15°
25. 1
17
26.
8
27. 20°
28. Use (A + B + C = 180°)
29. AC = 10, BC = 5 3 , use Pythagoras theorem
30. Substitute values of sin 60° and cos 60° and solve
31. 60°
Note : 32 to 38 use trigonometric identities and prove (based on Ex. 8.4 of
NCERT)
39. 3
40. Use cot (90 – ) = tan , cosec (90 – ) = sec , tan (90 – ) = cot 
41. –1
42. Use A + B + C = 180°
43. 2

Mathematics-X 125

44. cos  + sin = 2 cos 
Square both sides and get 1 + 2 cos  sin  = 2 cos2 
 2 cos  sin  = 2 cos2  – 1 ...(1)
Now square (cos  – sin )2 and get
(cos – sin )2 = 1 – 2 cos  sin  ...(2)
Substitute (1) in (2)
45. 9.
46. Find m2 and n2 and add
Note : Q47 to Q50 Use identities to prove
51. 0
1 1
52. Rationalise in LHS and proceed, use = cosec .
cosec   cot  sin 

1 1
Rationalise on RHS and proceed, use = cosec .
cosec  – cot  sin 
53. Find m2 and n2 and substitute in LHS.
54. Take common sin2  in Numerator and cos2  in Denominator of 2nd term
on LHS and replace 1 by sin2  + cos2 .
55. 0
2
56.
3

57. (sin  + cos ) = 3

1
square both sides and get value of
sin   cos 
Change tan  + cot  into sin  and cos  proceed.
cos A
58. Change cot A = sin A , take cos A common from Numerator and

Denominator, Rationalise remaining term and change into sec A and tan
A.

126 Mathematics-X

sin (1  2sin 2 )
59. LHS = , write 1 = sin2  + cos2  and proceed.
cos (2 cos 2   1)

1
60. cos (A + B) = = cos 60°
2 A  B  60
 Solve these equations
A – B  30 
1
sin (A – B) = = sin 30°
2
A = 45°, B = 15°
61. Find m2 and n2 substitute in m2 – n2 and substitute m and n in 4 mn

2
62.
3
63. Refer NCERT
64. (Use complementary form), 8
65. Convert cot  and cosec  into sin  and cos 
and use sin2 = 1 – cos2

66. Divide Numerator and Denominator by cos , and use sec  = 1  tan 2 
or use pythagoras theorem and trigonometeric ratios,
13
Ans.
11
67. Same as Q 59.

Mathematics-X 127

PRACTICE-TEST
Introduction to Trigonometry
Time : 1 Hrs. M.M.: 20
SECTION-A
4
1. If sin  = what is the value of cos  
5
2. Write the value of sin (45° + ) – cos (45° – ). 1
3. If cos 9 = sin  and 9 < 90°, then the value of tan 5 is 1
1
(a) (b) 3 (c) 1 (d) 0
3
4. If sin A + sin2 A = 1, then the value of (cos2 A + cos4 A) is 1
1
(a) 1 (b) (c) 2 (d) 3
2
SECTION-B
5 sin  – 3cos 
5. If 5 tan = 4 then find the value of 2
5 sin  + 2cos 

6. Find the value of tan 35° tan 40° tan 45° tan 50° tan 55° 2
7. Prove that (sin  + cos ) (tan  + cot ) = sec  + cosec  2

SECTION-C

sin  1 + cos 
8. Prove that + = 2 cosec  3
1 + cos  sin 

9. Prove that cos A sin 2 A 3
– = sin A+ cos A
1 – tan A cos A – sin A

SECTION-D

tan  + sec  – 1 cos 
10. Prove that = . 4
tan  – sec  + 1 1 – sin 

128 Mathematics-X