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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.

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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)

$\fn_cm =\frac{(12-4)/5}{(12+4)/5)}$

= 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)

$\fn_cm =\frac{1+sin^{2}A/cos^{2}A}{1+cos^{2}A/sin^{2}A}$

$\fn_cm =\frac{\left ( cos^{2}A+sin^{2}A \right )/cos^{2}A}{\left ( cos^{2}A+sin^{2}A \right )/sin^{2}A}$

$\fn_cm =\frac{sin^{2}A}{cos^{2}A}=tan^{2}A$

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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 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