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4 sin cos 4 cosec2
1. If tan A = , 0 A 90, then find the ;fn rks dk eku
3 sin cos 5 2 – cosec2
value of sin A. Kkr djsa
4 SSC CHSL 09/08/2023 (Shift-01)
; fn tan A = gS] rks sin A
an
, 0 A 90,
3 16 40
d k eku D;k gksxk\ (a)
25
(b)
41
SSC CPO 03/10/2023 (Shift-01)
a m
nj
41 31
3 7 (c) (d)
40 30
(a) (b)
ty ra
5 5 5. If 7 sin2 A + 3 cos2 A = 4, then find cot
ra
A:
4 4
(c) (d) ; fn 7 sin2 A + 3 cos2 A = 4 gS] rks
cot A dk
5 7 eku Kkr dhft,A
di g 2 SSC CPO 04/10/2023 (Shift-02)
A le
2 sec A
2. If tan A = find the value of .
5 cosec2 A 1 1
(a) (b)
e
3 2
2 sec2 A
;fn tan A = gS] rks dk eku Kkr
by T
5 cosec2 A (c) 3 (d) 2
dhft,A 6. If k(tan 45° sin 60°) = cos 60° cot 30°,
then the value of k is:
n
SSC CPO 03/10/2023 (Shift-03)
; fn k(tan 45° sin 60°) = cos 60° cot 30°
3 4 gS] rks
k d k eku D;k gS\
h s oi
(a) (b)
5 25 SSC CPO 04/10/2023 (Shift-01)
at J
2 9 1
(c) (d) (a) 1 (b)
5 25 3
4 (c) 3 (d) 3
3. If cot ,0 and 5pcos2 sin =
3 2 1
cot2, then the value of p is: 7. If tan(A + B) = 3 and tan (A – B) = ;
3
0º < (A + B) < 90º; A > B, then the values
; fn cot 4 , 0 vkSj5pcos2 sin = of A and B are _____.respectively.
3 2
rksp dk eku D;k gksxk%
M
2
cot , 1
; fn tan(A + B) = 3 v kSjtan (A – B) = ;
SSC CPO 03/10/2023 (Shift-02) 3
7 5
0º < (A + B) < 90º; A > B, rks A v kSjB ds
(a)
27
(b)
27
eku Øe'k%____ gS
SSC CGL MAINS 02/03/2023
25 125 (a) 45º & 15º (b) 15º & 45º
(c) (d)
27 27 (c) 30º & 30º (d) 60º & 30º
8. Evaluate the following:
sin cos 4
4. If , then the value of fuEufyf
[kr dk eku Kkr dhft,A
sin cos 5
cos(36° + A).cos(36° – A) + cos(54° +
cosec2 A).cos(54° – A)
is: SSC CGL 14/07/2023 (Shift-02)
2 – cosec2
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(a) sin 2A (b) cos A 15. Find the value of sin (50° + A) – cos (40° – A).
(c) sin A (d) cos 2A sin (50° + A) – cos (40° – A) dk eku Kkr
9. What is the value of
cos(50º A) sin(40º A)
? dhft,A
cos 40º sec 40º
SSC CPO 05/10/2023 (Shift-02)
cos(50º A) sin(40º A) (a) 1 (b) 0
dk eku D;k gS\
cos 40º sec 40º (c) –1 (d) 2
SSC CHSL 10/03/2023 (Shift-01) 16. If sin (5x – 25°) = cos (5y + 25°), where 5x
– 25° and 5y + 25° are acute angles, then
(a) 1 (b) 2
the value of (x + y) is:
(c) 0.5 (d) 0
10. cosec 2910° + sec 4260° + tan 2565° + cot
;fn sin (5x – 25°) = cos (5y + 25°) gS] tgk¡
1755° = ? 5x – 25° vkSj 5y + 25° U;wudks.k gS](xrks+ y)
dk eku D;k gksxkA
an
SSC CGL 20/07/2023 (Shift-01)
(a) 3 (b) 1 SSC CGL 18/07/2023 (Shift-03)
(c) 4 (d) 0 (a) 50° (b) 40°
a m
(c) 18° (d) 16°
nj
cos 37
11. Find the value of – cos 47° cosec 17. If cos(3) sin( 22) , where 3 90 ,
sin 53
ty ra
43°. then what is the value of ?
ra
cos 37 ;fn cos(3) sin( 22) , dgk¡ 3 90 , rks
– cos 47° cosec 43° dk eku Kkr dk eku D;k gksxk\
sin 53 di g
dhft,A SSC CHSL 02/08/2023 (Shift-03)
A le
SSC CPO 04/10/2023 (Shift-01) (a) 29° (b) 26°
(a) –1 (b) 1 (c) 27° (d) 28°
e
(c) 2 (d) 0 18. Find the value of (tan2 + tan4).
(tan2 + tan4) dk eku Kkr dhft,A
by T
12. 2(sin 1° × sec 89° ) + 3 (cos 11° × cosec
79°) + 5 (tan 21° × tan 69°) = ? SSC CPO 03/10/2023 (Shift-01)
SSC CPO 04/10/2023 (Shift-02) (a) Sec2 – sec4
n
(a) 20 (b) 12 (b) Cot2 – tan2
(c) 11 (d) 10 (c) Sec4 + sec2
h s oi
13. Find the value of/ dk eku Kkr djksa% (d) Sec4 – sec2
19. The value of (1+ sin4A - cos4A) cosec2A
at J
2 2
sin 39 sin 90 – 39 is:
cos 2 35 cos 2 90 – 35 + 3tan15° tan75°: (1+ sin4A - cos4A) cosec2A dk eku D;k gksxk\
SSC CPO 05/10/2023 (Shift-2) SSC CPO 03/10/2023 (Shift-02)
(a) 1 (b) 3 (a) –1 (b) 1
(c) 4 (d) 2 (c) –2 (d) 2
14. What is the value of sec254° – cot236° + 20. If sec A - tan A = p, then find the value
2 of secA.
3
2
sin237° × sec253° +
3
tan60° ? ;fn sec A - tan A = p, gS] rkssecA dk eku
M
Kkr dhft,A
3 SSC CPO 04/10/2023 (Shift-02)
sec254° – cot236° + sin237° × sec253° +
2
p2 1 p2 1
2 (a) (b)
tan60° dk eku D;k gS\ p2 1 p2 1
3
SSC CGL 14/07/2023 (Shift-02) p2 1 p2 1
(c) (d)
2p p
5 9
(a) (b) 21. (sin + cosec)² + (cos + sec)² is equal
2 2
to ?
3 7 (sin + cosec)² + (cos + sec)² fdlds cjkcj gS\
(c) (d)
2 2 SSC CGL MAINS 07/03/2023
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(a) 5 + tan² + cot² 26. If cos – sin = 2 sin, then (cos + sin)
(b) 7 + tan² – cot² is:
(c) 7 + tan² + cot²
(d) 5 + tan² – cot² ;fn cos – sin = 2 sin, gS] rks(cos +
22. Which of the following is equal to sin) dk eku Kkr dhft,A
tan sec – 1 SSC CPO 03/10/2023 (Shift-01)
tan – sec 1 ?
(a) – 2 sin (b) 2 tan
tan sec – 1 (c) 2 cos (d) – 2 cos
fuEufyf[kr esa ls dkSu&lk fodYi
tan – sec 1
27. If a cot + b cosec = p and b cot + a
ds cjkcj gS\ cosec = q then p² – q² is equal to _______.
;fn a cot + b cosec = p vkSjb cot + a cosec
an
SSC CGL MAINS (08/08/2022)
= q gS] rks
p² – q² fuEu esa ls fdlds cjkcj gS\
1 cos 1 cot
(a) (b) SSC CGL MAINS 07/03/2023
sin tan
a m
nj
(a) b² – a² (b) a² – b²
1 sin 1 tan (c) b – a (d) a² + b²
(c) (d)
ty ra
cos cot
ra
28. If a cot + b cosec = p and b cot + a cosec
= q then p² - q² is equal to
cos cos
23. If 10 and 11 , then the value
di g ;fn a cot + b cosec = p vkSjb cot + a cosec
sin cos
of cos2 is: = q gS rks
p² - q² fuEu esa ls fdlosQ cjkcj gS\
A le
SSC CGL 17/07/2023 (Shift-03)
cos cos
;fn sin 10 vkSj 11 gS] rks
cos2dk 2
(a) a + b 2
(b) a2 – b2
e
cos
(c) b2 – a2 (d) a – b
eku D;k gksxk\
by T
29. Find the least value of 16cosec 2 +
SSC CGL 21/07/2023 (Shift-03)
25sin2.
n
(a)
121
(b)
100 16cosec2 + 25sin2 dk U;wure eku Kkr dhft,A
132 221
h s oi
SSC CPO 04/10/2023 (Shift-3)
88 221 (a) 35
(c) (d)
108 121
at J
(b) 38
24. If cosec + cot = m, find the value of (c) 42
2
m 1 (d) 40
.
m2 1 30. If sin + cos = 2 cos then find
2
m 1 sinθ – cosθ
;fn cosec + cot = m gS] rks dk eku :
m2 1 sinθ
Kkr dhft,A
sinθ – cosθ
M
SSC CGL 24/07/2023 (Shift-02) ;fn sin + cos = 2 cos gS] rks
sinθ
(a) 1.1 (b) 2.0 dk eku gS%
(c) cos (d) – sin SSC CPO 05/10/2023 (Shift-03)
25. What is the value of (1 + tan – sec)(1 +
cot + cosec)? (a) – 2
(1 + tan – sec)(1 + cot + cosec) dk eku D;k gS\ (b) –1
SSC CHSL 09/03/2023 (Shift-03) (c) 1
(a) – 1 (b) 0 (d) 2
(c) 1 (d) 2
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ANSWER KEY
1.(c) 2.(b) 3.(c) 4.(c) 5.(c) 6.(a) 7.(a) 8.(d) 9.(d) 10.(c)
11.(d) 12.(d) 13.(c) 14.(b) 15.(b) 16.(c) 17.(d) 18.(d) 19.(d) 20.(c)
21.(c) 22.(c) 23.(b) 24.(c) 25.(d) 26.(c) 27.(a) 28.(b) 29.(d) 30.(a)
an
a m
nj
ty ra
ra
di g
A le
e
by T
n
h s oi
at J
M
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