CBSE Class 10 Introduction to Trigonometry MCQ New Pattern

## CBSE Class 10 Introduction to Trigonometry MCQ By Clarify Knowledge

CBSE Class 10 Introduction to Trigonometry MCQ New Pattern 2022

## CBSE Class 10 Introduction to Trigonometry MCQ Table

1. In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. The value of tan C is:

(a) 12/7

(c) 20/7

(d) 7/24

Explanation: AB = 24 cm and BC = 7 cm

tan C = Opposite side/Adjacent side

2. (Sin 30°+cos 60°)-(sin 60° + cos 30°) is equal to:

(a) 0

(b) 1+2√3

(c) 1-√3

(d) 1+√3

Explanation: sin 30° = ½, sin 60° = √3/2, cos 30° = √3/2 and cos 60° = ½

Putting these values, we get:

(½+½)-(√3/2+√3/2)

= 1 – [(2√3)/2]

= 1 – √3

3. The value of tan 60°/cot 30° is equal to:

(a) 0

(b) 1

(c) 2

(d) 3

Explanation: tan 60° = √3 and cot 30° = √3

Hence, tan 60°/cot 30° = √3/√3 = 1

4. 1 – cos2A is equal to:

(a) sin2A

(b) tan2A

(c) 1 – sin2A

(d) sec2A

Explanation: We know, by trigonometry identities,

sin2A + cos2A = 1

1 – cos2A = sin2A

5. sin (90° – A) and cos A are:

(a) Different

(b) Same

(c) Not related

(d) None of the above

Explanation: By trigonometry identities.

Sin (90°-A) = cos A {since 90°-A comes in the first quadrant of unit circle}

6. If cos X = ⅔ then tan X is equal to:

(a) 5/2

(b) √(5/2)

(c) √5/2

(d) 2/√5

Explanation: By trigonometry identities, we know:

1 + tan2X = sec2X

And sec X = 1/cos X = 1/(⅔) = 3/2

Hence,

1 + tan2X = (3/2)= 9/4

tan2X = (9/4) – 1 = 5/4

tan X = √5/2

7. If cos X = a/b, then sin X is equal to:

(a) (b2-a2)/b

(b) (b-a)/b

(c) √(b2-a2)/b

(d) √(b-a)/b

Explanation: cos X = a/b

By trigonometry identities, we know that:

sin2X + cos2X = 1

sin2X = 1 – cos2X = 1-(a/b)2

sin X = √(b2-a2)/b

8. The value of sin 60° cos 30° + sin 30° cos 60° is:

(a) 0

(b) 1

(c) 2

(d) 4

Explanation: sin 60° = √3/2, sin 30° = ½, cos 60° = ½ and cos 30° = √3/2

Therefore,

(√3/2) x (√3/2) + (½) x (½)

= (3/4) + (1/4)

= 4/4

= 1

9. 2 tan 30°/(1 + tan230°) =

(a) sin 60°

(b) cos 60°

(c) tan 60°

(d) sin 30°

Explanation: tan 30° = 1/√3

Putting this value we get;[2(1/√3)]/[1 + (1/√3)2] = (2/√3)/(4/3) = 6/4√3 = √3/2 = sin 60°

10. sin 2A = 2 sin A is true when A =

(a) 30°

(b) 45°

(c) 0°

(d) 60°

Explanation: sin 2A = sin 0° = 0

2 sin A = 2 sin 0° = 0

11. The value of (sin 45° + cos 45°) is

(a) 1/√2

(b) √2

(c) √3/2

(d) 1

Explanation:

sin 45° + cos 45° = (1/√2) + (1/√2)

= (1 + 1)/√2

= 2/√2

= (√2 . √2)/√2

= √2

12. If sin A = 1/2 , then the value of cot A is

(a) √3

(b) 1/√3

(c) √3/2

(d) 1

Explanation:

Given,

sin A = 1/2

cos2A = 1 – sin2A

= 1 – (1/2)2

= 1 – (1/4)

= (4 – 1)/4

= 3/4

cos A = √(3/4) = √3/2

cot A = cos A/sin A = (√3/2)/(1/2) = √3

13. If ∆ABC is right angled at C, then the value of cos(A+B) is

(a) 0

(b) 1

(c) 1/2

(d) √3/2

Explanation:

Given that in a right triangle ABC, ∠C = 90°.

We know that the sum of the three angles is equal to 180°.

∠A + ∠B + ∠C = 180°

∠A + ∠B + 90° = 180° (∵ ∠C = 90° )

∠A + ∠B = 90°

Now, cos(A+B) = cos 90° = 0

14. The value of (tan 1° tan 2° tan 3° … tan 89°) is

(a) 0

(b) 1

(c) 2

(d) 1/2

Explanation:

tan 1° tan 2° tan 3°…tan 89°

= [tan 1° tan 2°…tan 44°] tan 45° [tan (90° – 44°) tan (90° – 43°)…tan (90° – 1°)]

= [tan 1° tan 2°…tan 44°] [cot 44° cot 43°…cot 1°] × [tan 45°]

= [(tan 1°× cot 1°) (tan 2°× cot 2°)…(tan 44°× cot 44°)] × [tan 45°]

= 1 × 1 × 1 × 1 × …× 1     {since tan A × cot A = 1 and tan 45° = 1}

= 1

15. The value of the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is

(a) -1

(b) 0

(c) 1

(d) 3/2

Explanation:

[cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)]

= cosec [90° – (15° – θ)] – sec (15° – θ) – tan (55° + θ) + cot [90° – (55° + θ)]

We know that cosec (90° – θ) = sec θ and cot(90° – θ) = tan θ.

= sec (15° – θ) – sec (15° – θ) – tan (55° + θ) + tan (55° + θ)

= 0

16. If cos(α + β) = 0, then sin(α – β) can be reduced to

(a) cos β

(b) cos 2β

(c) sin α

(d) sin 2α

Explanation:

Given,

cos(α + β) = 0

cos(α + β) = cos 90°

⇒ α + β = 90°

⇒ α = 90° – β

sin(α – β) = sin(90° – β – β)

= sin(90° – 2β)

= cos 2β    {since sin(90° – A) = cos A}

17. If sin A + sin2A = 1, then the value of the expression (cos2A + cos4A) is

(a) 1

(b) 1/2

(c) 2

(d) 3

Explanation:

Given,

sin A + sin2A = 1

sin A = 1 – sin2A

sin A = cos2A {since sin2θ + cos2θ = 1}

Squaring on both sides,

sin2A = (cos2A)2

1 – cos2A = cos4A

⇒ cos2A + cos4A = 1

18. If cos 9α = sinα and 9α < 90°, then the value of tan 5α is

(a) 1/√3

(b) √3

(c) 1

(d) 0

Explanation:

Given,

cos 9α = sin α and 9α < 90°

That means, 9α is an acute angle.

cos 9α = cos(90° – α)

9α = 90° – α

9α + α = 90°

10α = 90°

α = 9°

tan 5α = tan(5 × 9°) = tan 45° = 1

19. The value of the expressionis

(a) 3

(b) 2

(c) 1

(d) 0

Explanation:

20. The value of the expression sin6θ + cos6θ + 3 sin2θ cos2θ is

(a) 0

(b) 3

(c) 2

(d) 1

Explanation:

We know that, sin2θ + cos2θ = 1

Taking cube on both sides,

(sin2θ + cos2θ)3 = 1

(sin2θ)3 + (cos2θ)3 + 3 sin2θ cos2θ (sin2θ + cos2θ) = 1

sin6θ + cos6θ + 3 sin2θ cos2θ = 1

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