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Trigonometry MCQ test 2

Trigonometry 11 CBSE - MCQ test 2

Question 1.

Which of following is equal to cos²x+cos²y-2cosx.cosy.cos(x+y):

sin(x+y)

sin²(x+y)

sin³(x+y)

None of these

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Solution:

cos² x + cos² y - 2cos x cos y cos (x + y)

=> cos² x + cos² y - 2cos x cos y [cos x cos y - sin x sin y]

=> cos² x + cos² y - 2cos² x cos² y + 2sin x sin y cos x cos y

=> cos² y - cos² x cos² y + cos² x - cos²y cos² x + 2sin x sin y cos x cos y

=> (1 - cos²x)cos²y + (1 - cos²y)cos²x + 2sin x sin y cos x cos y

=> sin²x cos²y + cos²x sin²y + 2sin x sin y cos x cos y

=> [ sin x cos y + cos x sin y ]²

=> [sin (x + y)]² = sin² (x + y)


Question 2.

Determine the value of (asinX+bcosX)² if acosX-bsinX=c :

a²+b²+c²

a²+b²-c²

a²-b²-c²

a²-b²+c²

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Given, acosX+bcosY=C

Squaring both sides, (acosX-bsinX)²=c²

=> (a²cos²X+b²sin²X-2acosX.bsinX)=c²

=> a²(1-sin²X)+b²(1-cos²X)+2acosX.bsinX=c²

=> a²-a²sin²X+b²-b²cos²X+2acosX.bsinX=c²

=> a²+b²-c²=a²sin²X+b²cos²X-2acosX.bsinX

=> a²+b²-c²=(asinX+bcosX)


Question 3.

If in △ABC, cosA+2cosB+cosC=2,What is relation between a,b and c.

a-c=2b

a+c=2b

a+b=2c

2a+c=b

Click Here For Solution

A,B,C are the angles of a △ABC.

Therefore, A+B+C=180°

Given, cosA+2cosB+cosC=2

cosA+cosC=2(1−cosB)

2cos(A+C)/2.cos(A-C)/2=2.2sin²B/2

2.sinB/2.cos(A-C)/2=2.2sin²B/2

[ since,cos(A+C)/2=cos(180°-B)/2= sinB/2 ]

cos(A-C)/2=2sinB/2

Multiplying both sides by 2cosB/2

cos(A-C)/2*2cosB/2 =2sinB/2 *2cosB/2

[ since,sin(A+C)/2=sin(180°-B)/2= cosB/2 ]

[cos(A-C)/2]*[2sin(A+C)/2] =2sinB/2 *2cosB/2

2*[cos(A-C)/2]*[sin(A+C)/2] =2sinB

2*[sin(A+C)/2]*[cos(A-C)/2] =2sinB

sinA+sinC=2sinB

a+c=2b


Question 4.

A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

12π

10π

8π

6π

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360 revolutions in 1 minute.

360 revolutions in 60 second.

6 revolutions in 1 second.

6 *2π radian in 1 second.

12π radian in 1 second.


Question 5.

What will be the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm (Use π = 22/7).

10°36'

12°36'

14°36'

11°36'

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Angle subtended at the centre of a circle=22/100 radian

Angle subtended at the centre of a circle=22/100*180/π degree

Angle subtended at the centre of a circle=(22/100)*(180*7/22) degree

Angle subtended at the centre of a circle=(180*7/100)=12.6°=12°36'


Question:6

If tanA=(cos9°-sin9°)/(cos9°+sin9°) what will be vaule of A?

36°

54°

45°

None

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tanA=(cos9°-sin9°)/(cos9°+sin9°)

Dividing RHS part by cos9° in numerator and denumarator

tanA=(1-tan9°)/(1+tan9°)

tanA=(tan45°-tan9°)/(1+tan45°tan9°)

tanA=tan(45°-9°)=tan36°


Question:7

In a triangle ABC, cosec A (sin B cos C + cos B sin C) is equal to :

a/c

1

c/a

None of these

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cosec A (sin B cos C + cos B sin C)

cosec A *sin (B + C)

cosec A *sin (180°-A)

cosec A *sin A

1


Question:8

The perimeter of a triangle ABC is 6 times the arithmetic mean of the sines of its angles. If the side b is 2, then the angle B is

π/3

2π/3

π/2

None

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a+b+c=6(sinA+sinB+sinC)/3

a+b+c=2(sinA+sinB+sinC)

(a+b+c)/2=(sinA+sinB+sinC)

(sinA+sinB+sinC)/(a+b+c)=1/2----Eqn 1

We know that:

sinA/a=sinB/b=sinC/c

sinA/a=sinB/b=sinC/c=(sinA+sinB+sinC)/(a+b+c)

sinA/a=sinB/b=sinC/c=1/2

sinB/b=1/2

sinB=b/2=2/2=1=sin(π/2)

B=π/2


Question:9

If tan A – tan B = x and cot B – cot A = y, then What will be the value of cot (A – B):

x + y

1/x + 1/y

1/x + y

None of these

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tan A – tan B = x ----Eqn1

cot B – cot A = y ----Eqn2

Eqn1

tan A – tan B = x

( 1/cotA ) – ( 1/cotB ) = x

( cotB - cotA )/cotA*cotB = x

putting (cot B – cot A )= y

y/cotA*cotB = x

tanA*tanB = x/y ----Eqn3

cot (A – B)=1/tan (A – B)

(1+tanA*tanB)/(tan A – tan B)

From Eqn1 and Eqn3 we get,

[1+(x/y)]/x

1/x + 1/y

(y+x)/xy

1/x + 1/y


Question:10

If the angles of a triangle be in the ratio 1 : 4 : 5, then the ratio of the greatest side to the smallest side is

(√5 – 1):4

3 : (√5 – 1)

4 : (√5 – 1)

None

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Calculation of angles of triangle:

Calculation of value of sin18°:
Let θ = 18°….(i)
We can write 90° as:
90° = 36° + 54°
90 = 2 × 18° + 3 × 18°
Substituting (i) in the above equation,
90° = 2θ + 3θ
3θ = 90° – 2θ
Taking “sin” on both sides,
sin 3θ = sin(90° – 2θ)
Using the identity sin 3A = 3 sin A – 4 sin³A,
3 sin θ – 4 sin³θ = cos 2θ
Now, using the identity cos 2θ = 1 – 2 sin²θ,
3 sin θ – 4 sin³θ = 1 – 2 sin²θ = 0
4 sin³θ – 2 sin²θ – 3 sin θ + 1 = 0
Let sin θ = x.
4x³ – 2x² – 3x + 1 = 0
(x – 1)(4x² + 2x – 1) = 0
x – 1 = 0, 4x² + 2x – 1 = 0
x = 1, 4x² + 2x – 1 = 0
Here x can not be equal to 1 as x=sin18° which must be less tha 1.
4x² + 2x – 1 = 0
Using quadratic formula,
x = [-2 ± √{4 – 4(-4)}]/2(4)
= [-2 ± √(4 + 16)]/8
= [-2 ± √20]/8
= [-2 ± 2√5]/8
= 2[-1 ± √5]/8
= (-1 ± √5)/4
The value of sin θ is positive as x=sin18°.
x=sin18°=(-1 + √5)/4
Let greatest side =c in front of angle 90° and the smallest side = a infront of angle 18°.
sin90°/c=sin18°/a
sin90°/sin18°=c/a
1/[(-1 + √5)/4]=c/a
4/(-1 + √5)=c/a


Question:11

If cosecA-cotA=1/5 then what will be sinA:

48/7

10/23

5/13

None

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cosecA-cotA=1/5 ------Eqn 1

so, cosecA+cotA=5 ------Eqn 2

Adding Eqn 1 & Eqn 2,We get

2cosecA=5+1/5=26/5

cosecA=13/5

sinA=5/13


Question:12

If secA-tanA=5 then what will be cosA:

5/3

10/3

3/5

None of these

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secA-tanA=3 ------Eqn 1

secA+tanA=1/3 ------Eqn 2

Adding Eqn 1 & Eqn 2,We get

2secA=3+1/3=10/3

secA=5/3

cosA=3/5


Question:13

If sinx + cosecx =2, then which of following is equal to (sin³x + cosec³x) :

10

2

2³

None

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sinx+cosecx=2

sinx+1/sinx=2

sin²x+1=2sinx

sin²x+1-2sinx=0

(sinx-1)²=0

(sinx-1)=0

sinx=1

so,cosecx =1

(sin³x + cosec³x)=1³+1³=2


Question:14

If sinx + sin²x = 1 then the value ofcos^8 x + 2cos^6 x+ cos^4 x

√3/2

√3

1

1:3

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Given,

sinx + sin²x = 1

sinx = 1- sin²x =cos²x

We have to find value of :

cos^8 x + 2cos^6 x+ cos^4 x

(cos^2 x)^4 + 2(cos^2)^3 x+ (cos^2 x)2

(cos^2 x)^4 + 2(cos^2)^3 x+ (cos^2 x)2

(sinx)^4 + 2(sinx)^3 x+ (sinx)2

(sin²x)² + 2(sin²x)(sinx)+ (sinx)2

(sinx + sin²x )²=1


Question:15

Which of following is equal to sin420°.cos390°+cos(-300°)sin(-330°)

11/√3

√3

1

None

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sin420°.cos390°+cos(-300°)sin(330°)

sin(360°+60°).cos(360°+30°)-cos(360°-60°)sin(360°-30°)

sin(60°).cos(30°)+cos(60°)sin(30°)

sin(60°+30°)=1


Question:16

If tan x=1/3–√ What will equal to cos2x?

1/2

-1/2

√3/2

None of these

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We know, cos2x = cos²x – sin²x

=> (cos²x – sin²x)/(cos²x+sin²x) {1 = sin²x + cos²x}

=> (1-tan²x)/(1+tan²x)

=> (1-(1/3–√)2)/(1+(1/3–√)2)

=>(1-1/3)/(1+1/3) = (2/3)/(4/3) = 1/2.


Question:17

Which of following is correct?

cos 3x = 3cosx – 4cos³x

cos 3x = 4cosx – 3cos³x

cos 3x = 3cos³x – 4cosx

cos 3x = 4cos³x – 3cosx

Click Here For Solution

cos 3x = cos (2x +x)

=> cos 2x cos x – sin 2x sin x

=> (2cos² x – 1) cos x – 2sin x cos x sin x

=> (2cos² x – 1) cos x – 2cos x (1 – cos² x)

=> 2cos³ x – cos x – 2cos x + 2 cos³ x

=> 4cos³ x – 3cos x.


Question:18

Which of following is equal to 5 tan² A – 5 sec² A + 1

-2

1

3

-4

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5 tan² A – 5 sec² A + 1

-5(sec² A-tan² A)+1

-5*1+1

-4


Question:19

If sinX/sinY=1/2 and cosX/cosY=3/2 where x,y∈(0,π/2), then the value of tan(x+y) is equal to:

√12

0

√13

√15

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Question:20

lf in a ΔABC,∠C=90° , then What will be the maximum value of sinAsinB :

1

1/√2

2

1/2

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