Class 10 Maths Trigonometry MCQ's With Solutions for Term-1 CBSE Board Exam 2021 - Future Study Point

Class 10 Maths Trigonometry MCQ’s With Solutions for Term-1 CBSE Board Exam 2021

class 10 trigonometry MCQ for term-1

Class 10 Maths Trigonometry MCQ’s With Solutions for Term-1 CBSE Board Exam 2021

class 10 trigonometry MCQ for term-1

Class 10 Maths Trigonometry MCQ’s With Solutions for Term-1 CBSE Board Exam 2021 are created  for boosting your preparation of Maths subject in  Term-1 CBSE Board exam 2021 which is going to be held in the month of November-December.We have planned to introduce  bunches of Maths MCQ’s with their detail solutions for achieving excellent marks for our readers so the students who didn’t subscribe our website and Utube channel,those are requested to do that as soon as possible.These Maths MCQ’s are created and collected as per the Maths sample paper published by CBSE,hence these Maths MCQ questions of Trigonometry are very important inputs for the Term-1 CBSE exams and  are based on the chapter 8-introduction to Trigonometry.

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Class 10 Maths Trigonometry MCQ’s With Solutions for Term-1 CBSE Board Exam 2021

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Q1. Given that sin α = 1/2 and cos β = 1/2,then the value of α + β is

(a) 0°    (b)  30°    (c)  60°   (d) 90°

Ans.(d) 90°

sin α = 1/2 ⇒sin α = sin 30° ⇒α = 30°

cos β = 1/2 ⇒cos β = cos 60° ⇒β = 60°

α + β = 30° +60° = 90°

Q2.If 4tan θ = 3, then (4 sin θ – cos θ)/(4 sin θ +cos θ) is equal to

(a) 2/3    (b) 1/3    (c) 1/2    (d) 3/4

Ans.(c) 1/2

We are given that 4tan θ = 3

tan θ = 3/4 = perp./Base

Hypotenuse = √(3² +4²) =√25 =5

sin θ = 3/5 and cos θ = 4/5

(4 sin θ – cos θ)/(4 sin θ +cos θ)

( 4×3/5 – 4/5)(4×3/5 + 4/5)

(12/5 – 4/5)(12/5 + 4/5)

= 8/16 = 1/2

Q3.sin (45° + θ) – cos (45° -θ) is equal to

(a) 2 cos θ    (b) 0   (c) 2 sin θ    (d) 1

sin (45° + θ) – sin [90° -(45° -θ)]

sin (45° + θ) -sin (90° -45° + θ)

sin (45° + θ) -sin (45° + θ) =0

Q4.If √2 sin (60° – α) = 1 ,then α is

(a) 45°    (b) 15°    (c) 60°    (d) 30 °

Ans.  (b) 15°

√2 sin (60° – α) = 1

sin (60° – α) = 1/√2

sin (60° – α) = sin 45°

60° – α = 45°

α = 60° – 45° = 15°

Q5. The value of sin²30° – cos² 30° is

(a) -1/2     (b) √3/2    (c) 3/2    (d) -2/3

Ans. (a) -1/2

The given expression is

sin²30° – cos² 30°

Value of sin 30° = 1/2 and of cos 30°= √3/2

(1/2)² -(√3/2)²

1/4 – 3/4 = -2/4 = -1/2

Q6.If √2 sin (60° – α) =1,then α is

(a) 45°    (b) 15°    (c) 60°   (d) 30°

Ans.(b) 15°

We are given

√2 sin (60° – α) =1

sin (60° – α) = 1/√2 = sin 45°

60° – α = 45°

α = 60° – 45° = 15°

Q7. The maximum value of 1/ cosec α is

(a) 0    (b) 1    (c) √3/2    (d) -1/√2

Ans.  (b) 1

The given expression is

1/ cosec α

= sin α

The maximum value of sin α is 1

Class 10 Maths Trigonometry MCQ’s With Solutions for Term-1 CBSE Board Exam 2021

Q8.If cos (40°+A) = sin 30°,then value of A is

(a) 30°    (a) 40°      (a) 60°    (a) 20°

Ans. (a) 20°

The given trigonometric equation is

cos (40°+A) = sin 30°

Applying the formula sin θ = cos (90° -θ)

cos (40°+A) = cos (90°- 30°)

40°+A = 90°- 30°=60°

A = 60° – 40° = 20°

See the Videoes-NCERT Solutions of Exercise 8.4 Introduction to Trigonometry

Q9.If  cosec θ – cot θ = 1/3 , the value of (cosec θ + cot θ) is

(a) 1       (b) 2       (c) 3       (d) 4

Ans.(c) 3

The given trigonometric equation is

cosec θ – cot θ = 1/3

cosec θ – cot θ = (cosec²θ -cot²θ)/3

cosec θ – cot θ = (cosec θ – cot θ)(cosec θ + cot θ)/3

(cosec θ + cot θ) = 3

Q10. (1 + tan²A)/(1 + cot²A) is equal to

(a) sec²A      (b) -1       (c) cot²A     (d) tan²A

Ans.  (d) tan²A

(1 + tan²A)/(1 + cot²A)

Q11.If cos A + cos²A =1,then sin²A +Sin4A is equal to

(a) -1      (b) 0      (c) 1     (d) None of these

Ans. (c) 1

The given trigonometric equation is

cos A + cos²A =1

cos A = 1 – cos²A = sin²A

sin²A +sin4A

cos A + (sin²A)²

cos A + (cos A)²

cos A + cos²A =1

Q12.2(sin6A + cos6A) -3(sin4A + cos4A) is equal to

(a) 0   (b) 6    (c) -1    (d) None of these

Ans. (c) -1

2(sin6A + cos6A) -3(sin4A + cos4A)

2[ (sin2A)³ + (cos2A)³] -3[(sin2A)² + (cos2A)²]

2(sin2A + cos2A)[(sin2A)² + (cos2A)²- sin2A.cos2A] – 3[(sin²+cos²A)²-2sin²A.cos²A ]

2[( sin ²A  + cos2A)² -2sin2A.cos2A -sin2A.cos2A] – 3( 1 – 2sin2A. 2cos2A )

2( 1- 3sin2A.cos2A)  -3( 1 – 2sin2A. 2cos2A )

2 -6 sin2A.cos2A – 3 +6sin2A. 2cos2A = -1

Class 10 Maths Trigonometry MCQ’s With Solutions for Term-1 CBSE Board Exam 2021

Q13.3 sin² 20° – 2 tan² 45° + 3 sin² 70° is equal to

(a) 0      (b) 1      (c) 2      (d) -1

Ans.(b) 1

The given expression is

3 sin² 20° – 2 tan² 45° + 3 sin² 70°

3 sin²(90° – 70°) + 3 sin² 70°-2 tan² 45°

3 cos² 70° + 3 sin² 70°- 2

3(cos² 70° +  sin² 70°) -2

3 -2 = 1

Q14. If sin 2A = 1/2(tan²45°) where A is an acute angle, then the value of A is

(a) 60°     (b) 45°     (c) 30°    (d) 15°

Ans. (d) 15°

The given equation is

sin 2A = 1/2(tan²45°)

sin 2A = 1/2 = sin 30°

2A = 30°

A = 15°

Q15. sin θ/(1 + cos θ) is

(a) cos θ/(1 – sin θ)   (b) (1-sin θ)/sin θ    (c) (1-sin θ)/cos θ  (d) (1-cos θ)/sin θ

Ans. (d) (1-cos θ)/sin θ

The given expression is

sin θ/(1 + cos θ)

Multiplying the numerator and denominator by the conugate of the denominator (1 – cos θ)

sin θ(1-cos θ)/(1 + cos θ)(1-cos θ)

=sin θ(1-cos θ)/(1-cos² θ)

= sin θ(1-cos θ)/sin² θ

=(1-cos θ)/sin θ

Q16.If  A + B = 90°, cot B = 3/4 ,then tan A is equal to

(a) 5/3     (b) 1/3     (c) 3/4     (d) 1/4

Ans.  (c) 3/4

We are given that

A + B = 90°

A = 90° – B

tan A = tan (90° – B)

tan A = cot B

Also given to us ,cot B = 3/4

tan A = 3/4

Class 10 Maths Trigonometry MCQ’s With Solutions for Term-1 CBSE Board Exam 2021

Q17. If A and B are complementary angles then

(a) sin A = sin B     (b) cos A = cos B   (c) tan A = tan B    (d) sec A = cosec B

Ans.    (d) sec A = cosec B

We are given that A and B are complementary angles

Therefore

A + B = 90°

sec A = sec( 90° – B) = cosec B

Q18. The value of tan 1° tan 2° tan 3°……tan 87° tan 88° tan 89°

(a) 1       (b) -1       (c) 0        (d) None of these

Ans.(a) 1

The given expression is

tan 1° tan 2° tan 3°……tan 87° tan 88° tan 89°

=tan(90°-89°) tan (90°-88°) tan (90°-87°)…..tan 87° tan 88° tan 89°

= cot 89°. cot 88°.cot 87°…..tan 87° tan 88° tan 89°

= (1/tan 87°).1/ tan 88°.1/ tan 89°….tan 87° tan 88° tan 89°

= 1

Q19.The value of cos 1° cos 2° cos 3°……  cos 180° 

(a) 1       (b) -1       (c) 0        (d) None of these

Ans.(c) 0

We are given that

cos 1° cos 2° cos 3°……  cos 180°

=cos 1° cos 2° cos 3°..cos 90°cos 91°….  cos 180°

=cos 1° ×cos 2° ×cos 3°..0×cos 91°….  cos 180° = 0

Class 10 Maths Trigonometry MCQ’s With Solutions for Term-1 CBSE Board Exam 2021

Q20. If sec θ + tan θ = x, then sec θ =

(a) (x²+1)/x     (b) (x²+1)/2x     (c) (x²-1)/2x     (d) (x²-1)/x

Ans. (b) (x²+1)/2x

The given equation is

sec θ + tan θ = x….(i)

We have

1 + tan ²θ = sec²θ

sec²θ – tan ²θ = 1

(sec θ – tan θ)(sec θ + tan θ) = 1

Putting the value of x

x (sec θ – tan θ) = 1

sec θ – tan θ = 1/x…..(ii)

Adding both equations (i) and (ii)

2 sec θ = x + 1/x

2 sec θ = (x² +1)/x

sec θ = (x² +1)/2x

Why are these MCQ’S are important to study for the preparation of term-1 Exam 2021 CBSE Board ?

CBSE has changed pattern of the question paper,so according to latest pattern in the maths question paper there will be 50 questions of MCQ each of 1 marks,so this pattern is easier than the previous years question papers,but it is very important to clear the concept of the chapters because maximum questions would be out of the books,hence it is mandatory to study additional questions which are technical and short.

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NCERT Solutions of  Science and Maths for Class 9,10,11 and 12

NCERT Solutions of class 9 maths

Chapter 1- Number SystemChapter 9-Areas of parallelogram and triangles
Chapter 2-PolynomialChapter 10-Circles
Chapter 3- Coordinate GeometryChapter 11-Construction
Chapter 4- Linear equations in two variablesChapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s GeometryChapter 13-Surface Areas and Volumes
Chapter 6-Lines and AnglesChapter 14-Statistics
Chapter 7-TrianglesChapter 15-Probability
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Chapter 1-Matter in our surroundingsChapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?Chapter 10- Gravitation
Chapter3- Atoms and MoleculesChapter 11- Work and Energy
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Chapter 5-Fundamental unit of lifeChapter 13-Why do we fall ill ?
Chapter 6- TissuesChapter 14- Natural Resources
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Chapter 1-Real numberChapter 9-Some application of Trigonometry
Chapter 2-PolynomialChapter 10-Circles
Chapter 3-Linear equationsChapter 11- Construction
Chapter 4- Quadratic equationsChapter 12-Area related to circle
Chapter 5-Arithmetic ProgressionChapter 13-Surface areas and Volume
Chapter 6-TriangleChapter 14-Statistics
Chapter 7- Co-ordinate geometryChapter 15-Probability
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CBSE Class 10-Half yearly question paper of maths 2020 with solutions

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Chapter 1- Chemical reactions and equationsChapter 9- Heredity and Evolution
Chapter 2- Acid, Base and SaltChapter 10- Light reflection and refraction
Chapter 3- Metals and Non-MetalsChapter 11- Human eye and colorful world
Chapter 4- Carbon and its CompoundsChapter 12- Electricity
Chapter 5-Periodic classification of elementsChapter 13-Magnetic effect of electric current
Chapter 6- Life ProcessChapter 14-Sources of Energy
Chapter 7-Control and CoordinationChapter 15-Environment
Chapter 8- How do organisms reproduce?Chapter 16-Management of Natural Resources

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Chapter 1-SetsChapter 9-Sequences and Series
Chapter 2- Relations and functionsChapter 10- Straight Lines
Chapter 3- TrigonometryChapter 11-Conic Sections
Chapter 4-Principle of mathematical inductionChapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbersChapter 13- Limits and Derivatives
Chapter 6- Linear InequalitiesChapter 14-Mathematical Reasoning
Chapter 7- Permutations and CombinationsChapter 15- Statistics
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NCERT solutions of class 12 maths

Chapter 1-Relations and FunctionsChapter 9-Differential Equations
Chapter 2-Inverse Trigonometric FunctionsChapter 10-Vector Algebra
Chapter 3-MatricesChapter 11 – Three Dimensional Geometry
Chapter 4-DeterminantsChapter 12-Linear Programming
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