MCQ Trigonometric Identities ML Aggarwal Solutions ICSE Class-10 Chapter-18 . We Provide Step by Step Answer of Exercise-18 of Trigonometric Identities with  MCQs and Chapter-Test Questions  / Problems related  for ICSE Class-10 APC Understanding Mathematics  .  Visit official Website CISCE  for detail information about ICSE Board Class-10.

 Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 10th Chapter-18 Trigonometric Identities  (MCQ) Writer ML Aggarwal Book Name Understanding Topics Solution of Exe-18, MCQ and Chapter Test Questions Academic Session 2021-2022

## MCQ Trigonometric Identities ML Aggarwal Solutions ICSE Class-10 Chapter-18

–: Select Exercise :–

Exercise 18 ,

MCQS ,

Chapter Test

### How to Solve Trigonometric Identities Problems/Questions / Exercise of ICSE Class-10 Mathematics

Before viewing Answer of Chapter-18 Trigonometric Identities of ML Aggarwal Solutions. Read the Chapter Carefully with formula and then solve all example of Exe-17 given in  your text book . For more practice on Trigonometric Identities  related problems /Questions / Exercise try to solve Trigonometric Identities  exercise of other famous publications also such as Goyal Brothers Prakshan (RS Aggarwal ICSE)  / Concise Selina Publications ICSE  Mathematics. Get the Formula of Trigonometric Identities    for ICSE Class 10 Maths  to understand the topic more clearly in effective way.

### MCQ Solutions of ML Aggarwal Trigonometric Identities Chapter 18

(Page 458)

Choose the correct answer from the given four options (1 to 12) :

#### Question 1

${ cot }^{ 2 }\theta -\frac { 1 }{ { sin }^{ 2 }\theta }$ is equal to
(a) 1
(b) -1
(c) sin2 θ
(d) sec2 θ

(b) -1

cot2 θ – 1/sin2 θ

= cos2 θ/sin2 θ – 1/sin2 θ

= (cos2 θ – 1)/sin2 θ

= – sin2 θ/sin2 θ

= – 1

#### Question 2

(sec2 θ – 1) (1 – cosec2 θ) is equal to
(a) – 1
(b) 1
(c) 0
(d) 2

(a) -1

(secθ – 1)(1 – cosec2 θ)

= (1/cos2 θ – 1)(1 – 1/sin2 θ)

= (1 – cosθ)/cos2 θ × (sin2 θ – 1)/sinθ

= (- sin2 θ cos2 θ)/(sin2 θ cos2 θ) = – 1

(Page 459)

#### Question 3

$\frac { { tan }^{ 2 }\theta }{ 1+{ tan }^{ 2 }\theta }$ is equal to
(a) 2 sin2 θ
(b) 2 cos2 θ
(c) sin2 θ
(d) cos2 θ

(c)  sin2 θ

tan2 θ/(1 + tan2 θ)

tan2 θ/(1 + tan2 θ) = (sin2 θ/cos2 θ)/(1 + sin2 θ/cos2 θ)

= (sin2 θ/cos2 θ)/(cos2 θ + sin2 θ)/cos2 θ

= (sin2 θ/cos2 θ) × cos2 θ/(sin2 θ + cos2 θ)

(∵ sin2 θ + cos2 θ = 1)

= sin2 θ/1

= sin2 θ

#### Question 4

(cos θ + sin θ)2 + (cos θ – sin θ)2 is equal to
(a) – 2
(b) 0
(c) 1
(d) 2

(cos θ + sin θ)2 + (cos θ – sin θ)2
= cos2 θ + sin2 θ + 2 sin θ cos θ + cos2 θ + sin2 θ – 2 sin θ cos θ
so = 2(sin2 θ + cos2 θ)
Hence = 2 × 1 = 2 (d)
(∵ sin2 θ + cos2 θ = 1)

#### Question 5

(sec A + tan A) (1 – sin A) is equal to
(a) sec A
(b) sin A
(c) cosec A
(d) cos A

(sec A + tan A) (1 – sin A)

= (1/cos A + sin A/cos A) (1 – sin A)

= (1 + sin A)/cos A × 1 – sin A

= {(1 + sin A)(1 – sin A)}/cos A

= (1 – sin2 A)/cos A

= cos2 A/cos A

= cos A

#### Question 6

$\frac { { 1+tan }^{ 2 }A }{ { 1+cot }^{ 2 }A }$ is equal to
(a) sec2 A
(b) – 1
(c) cot2 A
(d) tan2 A

(d) tan2 A

(1 + tan2 A)/(1 + cot2 A)

(1 + tan2 A)/(1 + cot2 A) = (1 + sin2 A/cosA)/(1 + cos2 A/sinA)

= {(cos2 A + sin2 A)/cos2 A}/(sin2 A + cos2 A)/sin2 A}

= (1/cos2 A)/(1/sin2 A)

= 1/cos2 A × sin2 A/1

= sin2 A/cosA

= tan2 A

#### Question 7

If sec θ – tan θ = k, then the value of sec θ + tan θ is
(a) $1-\frac { 1 }{ k }$
(b) 1 – k
(c) 1 + k
(d) $\\ \frac { 1 }{ k }$

(d) 1/k

sec θ – tan θ = k

1/cos θ – sin θ/cos θ = k

(1 – sin θ)/cos θ = k

Squaring both sides, we get

(1 – sin θ)/cos θ = k

Squaring both sides, we get

{(1 – sin θ)/cos θ}2 = (k)2

⇒ (1 – sin θ)2/cos2 θ = k2

= (1 – sin θ)2/(1 – sin2 θ)

= k2

⇒ (1 – sin θ)2/(1 + sin θ)(1 – sin θ) = k2

= (1 – sin θ)/(1 + sin θ)

= k2

⇒ (1 + sin θ)/(1 – sin θ) = 1/k2

(1 + sin θ)/cos θ = 1/k

= 1/cos θ + sin θ/cos θ

= 1/k

⇒ sec θ + tan θ = 1/k

#### Question 8

If θ is an acute angle of a right triangle, then the value of sin θ cos (90° – θ) + cos θ sin (90° – θ) is
(a) 0
(b) 2 sin θ cos θ
(c) 1
(d) 2 sin2 θ

sin θ cos (90° – θ) + cos θ sin (90° – θ)
= sin θ sin θ + cos θ cos θ
{ ∵ sin(90° – θ) = cosθ, cos (90° – θ) = sin θ }
= sin2 θ + cos2 θ = 1 (c)

#### Question 9

The value of cos 65° sin 25° + sin 65° cos 25° is
(a) 0
(b) 1
(b) 2
(d) 4

cos 65° sin 25° + sin 65° cos 25°
so = cos (90° – 25°) sin 25° + sin (90° – 25°) cos 25°
and = sin 25° . sin 25° + cos 25° . cos 25°
therefore = sin2 25° + cos2 25°
( ∵ sin2 θ + cos2 θ = 1)
hence = 1 (b)

#### Question 10

The value of 3 tan2 26° – 3 cosec2 64° is
(a) 0
(b) 3
(c) – 3
(d) – 1

3 tan2 26° – 3 cosec2 64°
= 3 tan2 26° – 3 cosec (90° – 26°)
= 3 tan2 26° – 3 sec2 26°

= 3 tan2 26° – 3 sec2 26°

= 3(tan2 26° – sec2 26°) {∵ sec2 θ – tan2 θ = 1}

= 3 × (-1)

= -3

–: End of Trigonometric Identities ML Aggarwal (MCQ) Solutions  :–

Thanks

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