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Trigonometry MCQ test 1

Trigonometry 11 CBSE - MCQ test 1

Question 1.

Determine the quadrant that contains the terminal side of an angle measuring −7π/6.


1st Quadrent

2nd Quadrent

3rd Quadrent

4th Quadrent

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Solution:

−7π/6
Convert into degree measure =−7π/6*180/π
=-210°
Terminal side of an angle measuring −7π/6 Will be in 2nd Quadrent,


Question 2.

Determine the quadrant that contains the terminal side of an angle x, such that tan x < 0 and sin x > 0.


1st Quadrent

2nd Quadrent

3rd Quadrent

4th Quadrent

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sin x > 0 ,tan x < 0
x is in 2nd Quadrent


Question 3.

Determine the quadrant that contains the terminal side of an angle x, such that cos x < 0 and sin x > 0.


1st Quadrent

2nd Quadrent

3rd Quadrent

4th Quadrent

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cos x < 0 and sin x > 0
x is in 2nd Quadrent


Question 4.

Determine the quadrant that contains the terminal side of an angle x, such that tan x > 0 and sin x > 0.


1st Quadrent

2nd Quadrent

3rd Quadrent

4th Quadrent

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tan x > 0 and sin x > 0
x is in 1st Quadrent


Question 5.

Determine the quadrant that contains the terminal side of an angle measuring −10π/3.


1st Quadrent

2nd Quadrent

3rd Quadrent

4th Quadrent

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−10π/3
Convert into degree measure =>−10π/3 *180/π
=>-600°
=>-600°/360°
=>Remainder =-240°
Terminal side of an angle measuring −10π/3 Will be in 2nd Quadrent,


Question:6

If sin θ and cos θ are the roots of ax² – bx + c = 0, then the relation between a, b and c will be


a2 – b2 + 2ac = 0

a2 + b2 + 2ac = 0

a2 – b2 - 2ac = 0

None

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sin θ and cos θ are the roots of ax² – bx + c = 0.

Sum of roots= sin θ +cos θ = b/a

Product of roots= sin θ.cos θ = c/a

We know that,

(sin θ +cos θ )²=sin²θ +cos²θ+2sin θ.cos θ=1+2sin θ.cos

(b/a )²=1+2c/a

(b²/a² )=1+2c/a

b²=a² +2ca

a2 – b2 + 2ac = 0


Question:7

If tan A = 1/2 and tan B = 1/3, then the value of A + B is :


π/6

π/4

π/5

π/3

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tan(A+B)=(tanA+tanB)/(1+tanA.tanB)

tan(A+B)=(1/2+1/3)/(1-1/2*1/3)

tan(A+B)=(5/6)/(5/6)

tan(A+B)=(5/5)=tan45°

(A+B)=45°=π/4


Question:8

The value of cos 1° cos 2° cos 3° … cos 179° is


1/2

1 object

0

None

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cos 1° cos 2° cos 3° ....... cos 179°

=> cos 1° cos 2° cos 3° ......... cos 90° cos 91° ......... cos 179°

=> cos 1° × cos 2° × cos 3° ..... × 0 × cos 91° .......... cos 179° (∵ cos 90° = 0)

=> 0 × finite

=> 0

Hence, cos 1° cos 2° cos 3° ......... cos 179° = 0


Question:9

What would be the value of : sin 50° – sin 70° + sin 10°


1

0

-1

None of these

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Question:10

What would the value of (1 + tan α) (1 + tan β) , If α + β = π/4


1

-1

2

None

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Question:11

If A lies in the second quadrant and 3 tan A + 4 = 0, then the value of (2 cot A – 5 cos A + sin A) is equal to


48/7

10/23

23/10

None

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Question:12

Which of following is correct cos θ = x + (1/x), when ,If x is real.


θ is an acute angle.

θ is right angle.

No value of θ is possible.

None of these

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Question:13

If tanA + cot A = 2 and 0°< A < 90° then what is the value of tan^n A + cot^n A?


1

2

-1

None

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Question:14

If the value of α + β = 90°, and α : β = 2 : 1, then what is the ratio of cos α to cos β ?


√3

√3:1

1:√3

1:3

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Given,α : β = 2 : 1

Let α=2k,β=k

Therefore ,

2K+K=90°

K=30°

α=2k=60°

β=30°

cos60°=1/2

cos 30°=√3/2

cos α : cos β=1:√3


Question:15

If θ is said to be an acute angle, and 7 sin²θ + 3 cos²θ = 4, then what would be the value of tan θ?


1

√3

1/√3

None

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Question:16

If A is said to be an acute angle, and 4 cos² A - 1 = 0, then what is the value of tan (A - 15°)?


1

-1

√3

None of these

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Question:17

What would be be the value of 1 - 2sin² θ, if cos^4 θ - sin^4 θ = 2/3?


Becomes zero

3/2

2/3

None

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(cos²θ)² - (sin²θ)² = 2/3

(cos²θ - sin²θ)(cos²θ + sin²θ) = 2/3

(cos²θ - sin²θ) = 2/3

(1-sin²θ - sin²θ) = 2/3

(1-2sin²θ ) = 2/3


Question:18

If sin (θ + 18°) = cos 60°, then Which of following is the value of cos5θ, where 0°< θ < 90°?


2:1

1:4

3:1

1:2

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sin (θ + 18°) = cos 60°

=> cos (90° – 30°) = sin 30°

=> θ + 18° = 30°

=> θ = 30° – 18° = 12°

∴ cos5θ = cos 60° = 1/2


Question:19

If y = cosθ, then what is the maximum value of y?


2

0

-1

1

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maximum value of cosθ=1

Therefore,maximum value of y=1


Question:20

If tan θ - cot θ = 0, what will be the value of sin θ + cos θ?


1

1/√2

2

√2

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tan θ - cot θ = 0

tan θ =cot θ

We know that, tan45° = cot45° = 1

so θ=45°

sinθ+cosθ

=>sin45°+cos45°

=>1/√2+1/√2

=>2(1/√2)

=>√2