RRB Group-D Trigonometry Questions PDF

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RRB Group-D Trigonometry Questions PDF
RRB Group-D Trigonometry Questions PDF

RRB Group-D Trigonometry Questions PDF

Download Top 20 RRB Group-D Trigonometry Questions and Answers PDF. RRB Group-D Maths questions based on asked questions in previous exam papers very important for the Railway Group-D exam.

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Question 1: 5tanθ=4, then the value of (5sinθ3cosθ5sinθ+3cosθ) is

a) 17

b) 27

c) 57

d) 25

Question 2: The least value of (4sec2θ+9cosec2θ) is

a) 1

b) 19

c) 25

d) 7

Question 3: If x=cosecθsinθ and y=secθcosθ, then the value of $x2y2(x2 + y2 + 3)$

a) 0

b) 1

c) 2

d) 3

Question 4: If 0θπ2, 2ycosθ=sinθ and x2cosecθ=y, then the value of x24y2 is

a) 1

b) 2

c) 3

d) 4

Question 5: If sinθ+sin2θ=1, then the value of cos12θ + 3cos10θ + cos6θ +  3cos8θ  – 1 is

a) 0

b) 1

c) -1

d) 2

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Question 6: The value of 1cosecθcotθ1sinθ

a) cotθ

b) cosecθ

c) tanθ

d) 1

Question 7: If cosθ+sinθ=2cosθ, then cosθsinθ is

a) -2cosθ

b) -2sinθ

c) 2sinθ

d) 2tanθ

Question 8: If cos4θsin4θ=23, then the value of 12sin2θ is,

a) 0

b) 23

c) 13

d) 43

Question 9: If tanθ = 3/4 and θ is acute, then cosecθ is equal to

a) 53

b) 2

c) 12

d) 4

Question 10: The value of 11+tan2θ + 11+cot2θ is

a) 1

b) 2

c) 12

d) 14

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Question 11: Maximum value of (2sinθ+3cosθ) is

a) 2

b) 13

c) 15

d) 1

Question 12: If CosΘ+SecΘ=2, find Cos2017Θ+Sec2017Θ

a) 2

b) 3

c) 1018

d) 2017

Question 13: If sinx+1sinx=2 then cos5x+cot5x=?

a) 0

b) -1

c) 1

d) 2

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Question 14: If cos2x+(1+sinx)2=3, then find the value of cosec x + sin(60+x), where 0º < x < 90º?

a) 3

b) 2

c) 212

d) 312

Question 15: If x and y are acute angles such that their sum is less than 90 . If cos(2x20)=sin(2y+20) , then the value of tan(x+y) is

a) 0

b) 1

c) 13

d) 3

Question 16: If x is an acute angle such that
Tan (4x – 50°) = Cot ( 50° – x), then the value of x would be?

a) 60

b) 45

c) 50

d) 30

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Question 17: If cosec 9x= sec 9x (0 <x<10), what is the value of x?

a) 9°

b) 3°

c) 5°

d) 6°

Question 18: If sinx+1=2sinx then cos3x+cosec3x=?

a) 0

b) -1

c) 1

d) 2

Question 19: If sin2xsinx+cos2x+sin2x=0, find the value of x, if 0x180.

a) 45º

b) 90º

c) 180º

d) More than one value is possible

Question 20: If sin2Θ4=cos2Θ9. Find the value of tan2Θcot2Θ.

a) 1/6

b) 13/27

c) 1/18

d) None of these

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Answers & Solutions:

1) Answer (A)

Taking cosθ outside in numerator and in denominator and making tanθ
hence eq will be  (5tanθ35tanθ+3)
As it is given that 5tanθ = 4
after putting values and solving we will get the equation reduced to 1/7.

2) Answer (A)

4sec2θ+9cosec2θ
or 4+4tan2θ+9+9cot2θ
or 13+4tan2θ+9cot2θ
or 13+4tan2θ+9tan2θ
or 1312+(2tanθ+3tanθ)2    (eq. (1) )
or now above expression to be minimum, equation (2tanθ+3tanθ)2 should be minimum.
So applying A.M.G.M.
(2tanθ+3tanθ)26
or (2tanθ+3tanθ)=26 ( for value to be minimum)
After putting above value in eq.(1) , we will get least value of expression as 25.

3) Answer (B)

x=cosecθsinθ=cos2θsinθ=cotθcosθ
Similarly y=tanθsinθ
xy=sinθcosθ
x2+y2+3=(sec2θ+cosec2θ)
Now putting above values in given equation, and after solving it will be reduced to 1

4) Answer (A)

2y=tanθ
x=2ycosecθ
Hence value of x24y2 = 4y2(cosec2θ1)
or tan2θcot2θ = 1

5) Answer (A)

Given equation can be written as (cos4θ+cos2θ)31
as sinθ+sin2θ=1
or sinθ=cos2θ
putting above value in given equation it will be
(sin2θ+sinθ)31=0

6) Answer (A)

sinθ1cosθ1sinθ

or cosθcos2θ(1cosθ)sinθ = cotθ

7) Answer (C)

sin2θ+cos2θ=1
So, sin2θ+cos2θ+2sinθcosθ=2cos2θ
Hence, cos2θsin2θ=2sinθcosθ
So, cosθsinθ=2sinθ

8) Answer (B)

cos4θsin4θ=(cos2θsin2θ)(cos2θ+sin2θ)=cos2θsin2θ=23
cos2θsin2θ=12sin2θ=23

9) Answer (A)

sinθcosθ=34
So, sin2θcos2θ=916
Hence, sin2θ=99+16=925
So, cosecθ=53

10) Answer (A)

1+tan2θ=sec2θ
1+cot2θ=csc2θ
So, the given fraction becomes,

1sec2θ+1csc2θ=sin2θ+cos2θ=1

11) Answer (B)

Maximum Value of asinθ+bcosθ=a2+b2
Maximum Value of 2sinθ+3cosθ=22+32
=13
Hence, Correct option is B.

12) Answer (A)

CosΘ+SecΘ=2
(CosΘ+SecΘ)2=22
==> Cos2Θ+Sec2Θ+2CosΘSecΘ=4 (Note :Cos \Theta * Sec \Theta = 1)
==> Cos2Θ+Sec2Θ+2=4
==> Cos2Θ+Sec2Θ = 2
(CosΘ+SecΘ)3=23
==> (Cos3Θ+Sec3Θ+3CosΘSecΘ(CosΘ+SecΘ)=23
==> (Cos3Θ+Sec3Θ = 2
Therefore, irrespective of the power, the expression always equals to 2.

13) Answer (A)

sinx+1sinx=2
=>sin2x+1=2sinx
=>(sinx1)2=0
=>sinx=1
=>cosx=0
The given expression would be,
cos5x+cot5x
cos5x+cos5xsin5x
=0

14) Answer (A)

cos2x+(1+sinx)2=3
=> cos2x+1+sin2x+2sinx=3 (Since cos2x+sin2x=1)
=> 2sinx=1 or sin x = ½ which means x=30 degrees [Solution in the first quadrant]
=> cosec x = 1/sin x = 2 and sin (60+x) = sin 90 = 1
=> The value of the expression is 2+1 = 3

15) Answer (B)

Given that , x+y<90(1)
cos(2x20)=cos(90(2y+20))
Using the given property in (1)
2x20=702y
2x+2y=90
x+y=45
tan(x+y)=1
Hence , the correct option is B

16) Answer (D)

We know that Tan ( 90 – x) = Cot x and Cot (90 – x) = Tanx
So 4x – 50 = 90 – y
Then 50 – x will be y
=> y = 50 – x
x + y = 50
Moreover,
90 – y = 4x – 50
=> 4x + y = 140
Subtracting the two equations, we get
3x = 90
=> x = 30
Hence the correct answer is option D.

17) Answer (C)

If cosec 9x= sec 9x,
Then sin9x=cos9x
In the first quadrant cos and sin are equal when 9x=45°
Hence, x = 5°

18) Answer (C)

sinx+1=2sinx
=>sin2x+sinx=2
=>(sinx1)(sinx+2)=0
As sinx can only take values between -1 and 1,
=>sinx=1
=>cosx=0
The given expression would be,
cos3x+cosec3x
cos3x+sin3x
=1

19) Answer (B)

sin2xsinx+cos2x+sin2x=0
We know that sin2x=2sinxcosx and cos2x=cos2xsin2x
=> 2cosx+cos2x=0
=> cos x =0 or cos x =2
Now, cos x =2 is not possible.
=> cos x = 0
In the range, 0x180, cos x =0 at only 90º
Thus, B is the correct answer.

20) Answer (D)

sin2Θ4=cos2Θ9
==> sin2Θcos2Θ = 49
==> tan2Θ=49
==> cot2Θ=94
==> tan2Θcot2Θ = 4/9 – 9/4 = -65/36
Since there is no such option, the correct option to choose is D.

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