# CBSE 10th Maths Important MCQs from Chapter 8 Introduction to Trigonometry with Detailed Solutions

MCQ questions for class 10 Maths Chapter 8- Introduction to Trigonometry are provided here. All these MCQs are provided with answers and are important for CBSE Class 10 Maths Exam which will be conducted on 02nd June, 2021. All the questions are also available in PDF. Practice with these questions to revise the important concepts and score good marks in the CBSE Class 10 Maths Exam 2021.

**Note - **Students must note that the following topics have been excluded from the revised CBSE syllabus of Class10 Maths.

**Motivate the ratios whichever are defined at 0 ^{o} and 90^{o}**

**Trigonometric ratios of complementary angles**

So, students should prepare according to the new syllabus only.

**Check below solved MCQs from Class 10 Maths Chapter 8 Introduction to Trigonometry:**

**1. **(sin30° + cos30°) – (sin 60° + cos60°)

(A) – 1

(B) 0

(C) 1

(D) 2

**Answer:**** (B)**

**Explanation: **According to question

**2.** Value of tan30°/cot60° is:

(A) 1/√2

(B) 1/√3

(C) √3

(D) 1

**Answer: ****(D)**

**Explanation:**

**
**

**3. **sec^{2}θ – 1 = ?

(A) tan^{2}θ

(B) tan^{2}θ + 1

(C) cot^{2}θ – 1

(D) cos^{2}θ

**Answer: (A)**

**Explanation: **From trigonometric identity

1+ tan^{2}θ = sec^{2}θ

⇒sec^{2}θ – 1 = tan^{2}θ

**4. **The value of sin θ and cos (90° – θ)

(A) Are same

(B) Are different

(C) No relation

(D) Information insufficient

**Answer: ****(A)**

**Explanation: **Since from trigonometric identities,

cos(90° – θ) = sin θ

So, both represents the same value.

**5. **If cos A = 4/5, then tan A = ?

(A) 3/5

(B) 3/4

(C) 4/3

(D) 4/5

**Answer: (B)**

**Explanation: **From trigonometric identity

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

(A) 1

(B) −1

(C) 0

(D) 1/2

**Answer: (C)**

**Explanation: **Since

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

= cosec (75° + θ) – cosec [90° - (15° - θ)] – tan (55° + θ) + tan [90° - (35° - θ)]

= cosec (75° + θ) – cosec (75° + θ) – tan (55° + θ) + tan (55° + θ)

= 0

**7. **Given that: SinA = a/b, then cosA = ?

(C) b/a

(D) a/b

**Answer:(B)**

**Explanation: **We have

**
**

**8.** The value of (tan1° tan2° tan3° ... tan89°) is

(A) 0

(B) 1

(C) 2

(D)1/2

**Answer: (B)**

**Explanation: **This can be written as,

(tan1° tan2° tan3° ... tan89°)

(tan1° tan2° tan3° ....... tan44° tan45° tan46° ..... tan87°tan88°tan89°)

= [tan1° tan2° tan3° ....... tan44° tan45° tan (90 – 44)° ..... tan(90° - 3) tan (90° - 2) tan (90° - 1)]

= (tan1° tan2° tan3° ....... tan44° tan45° cot 44° ..... cot3° cot2° cot 1°)

= 1

Since tan and cot are reciprocals of each other, so they cancel each other.

**9.** If sin A + sin^{2} A = 1, then cos^{2} A + cos^{4} A = ?

(A) 1

(B) 0

(C) 2

(D) 4

**Answer: (A)**

**Explanation: **We have

sin A + sin ^{2} A = 1

⇒ sin A = 1 – sin^{2} A

⇒ sin A = cos^{2} A ......(i)

Squaring both sides

⇒sin^{2}A = cos^{4}A ......(ii)

From equations (i) and (ii), we have

cos^{2}A + cos^{4}A = sin A + sin^{2}A = 1

**10.** If sin A = 1/2 and cos B = 1/2, then A + B = ?

(A) 0^{0}

(B) 30^{0}

(C) 60^{0}

(D) 90^{0}

**Answer: (D)**

**Explanation:** Since

(A) 3

(B) 2

(C) 1

(D) 0

**Answer: (B)**

**Explanation: **Using trigonometric properties, we have:

**12.** If cos9α = sin α and 9α < 90°, then the value of tan 5α is

(A) √3

(B) 1/√3

(C) 0

(D) 1

**Answer: (D)**

**Explanation: **Since

cos9α = sinα

⇒ sin (90° - 9α) = sinα

⇒ (90° - 9α) = α

⇒ α = 9°

Therefore,

tan 5α = tan 5 (9°)

= tan45°

= 1

**13. **If a pole 6m high casts a shadow 2√3 m long on the ground, then the sun’s elevation is

(A) 60°

(B) 45°

(C) 30°

(D)90°

**Answer: (A)**

**Explanation: **Given condition can be represented as follows:

**14.** If cos (A + B) = 0, then sin (A – B) is reduced to:

(A) cos A

(B) cos 2B

(C) sin A

(D) sin 2B

**Answer: (B)**

**Explanation: **Since

cos (A + B) = 0

⇒ cos (A + B) = cos90°

⇒ (A + B) = 90°

⇒ A = 90° - B

This implies

sin (A – B) = sin (90° - B - B)

⇒ sin (A – B) = sin (90° - 2B)

sin (A – B) = cos 2B

(A) 2/3

(B) 1/3

(C) 1/2

(D) 3/4

**Answer:(C)**

**Explanation: **This can be solved as,

**All the above MCQ questions can also be downloaded in PDF from the following link: **

**Also check: CBSE Class 10 Maths MCQs - All Chapters**

**Important Articles for the Preparations of Class 10 Maths Exam 2021:**

We have prepared some articles which will help the class 10 students in preparations of their Maths exam by summarising all important resources at one place. Students can check following links to explore the important articles to prepare in effective manner and perform well in the exam:

**CBSE Class 10 Maths Exam Pattern 2021 with Blueprint & Marking Scheme**

**CBSE Class 10 Maths Important Questions and Answers for Board Exam 2021**

**CBSE Class 10 Maths Solved Previous Year Question Papers**

**CBSE Class 10th Maths Chapter-wise Important Formulas, Theorems & Properties**