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

**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.**

**Class 10 MCQ’s questions with solutions-Polynomial**

**Class 10 Maths Sample Paper(Standard) for 2021 CBSE Board Exam -Term 1 with Solutions**

**CBSE Class 10-Question paper of maths 2021 with solutions**

**CBSE Class 10-Half yearly question paper of maths 2020 with solutions**

**CBSE Class 10 -Question paper of maths 2020 with solutions**

**CBSE Class 10-Question paper of maths 2019 with solutions**

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

**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°

**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 +Sin ^{4}A 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 +sin^{4}A

cos A + (sin²A)²

cos A + (cos A)²

cos A + cos²A =1

**Q12.2(sin ^{6}A + cos^{6}A) -3(sin^{4}A + cos^{4}A) is equal to**

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

Ans. (c) -1

2(sin^{6}A + cos^{6}A) -3(sin^{4}A + cos^{4}A)

2[ (sin^{2}A)³ + (cos^{2}A)³] -3[(sin^{2}A)² + (cos^{2}A)²]

2(sin^{2}A + cos^{2}A)[(sin^{2}A)² + (cos^{2}A)²- sin^{2}A.cos^{2}A] – 3[(sin²+cos²A)²-2sin²A.cos²A ]

2[( sin ²A + cos^{2}A)² -2sin^{2}A.cos^{2}A -sin^{2}A.cos^{2}A] – 3( 1 – 2sin^{2}A. 2cos^{2}A )

2( 1- 3sin^{2}A.cos^{2}A) -3( 1 – 2sin^{2}A. 2cos^{2}A )

2 -6 sin^{2}A.cos^{2}A – 3 +6sin^{2}A. 2cos^{2}A = -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 System | Chapter 9-Areas of parallelogram and triangles |

Chapter 2-Polynomial | Chapter 10-Circles |

Chapter 3- Coordinate Geometry | Chapter 11-Construction |

Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |

Chapter 5- Introduction to Euclid’s Geometry | Chapter 13-Surface Areas and Volumes |

Chapter 6-Lines and Angles | Chapter 14-Statistics |

Chapter 7-Triangles | Chapter 15-Probability |

Chapter 8- Quadrilateral |

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**CBSE Class 10-Half yearly question paper of maths 2020 with solutions**

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**CBSE Class 10-Question paper of maths 2019 with solutions**

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Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

**CBSE Class 11-Question paper of maths 2015**

**CBSE Class 11 – Second unit test of maths 2021 with solutions**

**NCERT solutions of class 12 maths**

Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |

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