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MCQ Trigonometric Identities ML Aggarwal Solutions ICSE Class-10

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 :–

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 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
Answer 1

(b) -1

cot2 θ – 1/sin2 θ

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= 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

Answer 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
Answer 3

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(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

Answer 4

(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

Answer 5

(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

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

Answer 6

(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 }

Answer 7

(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

ML aggarwal class-10 chapter 18 img 21

(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

Answer 8

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

Answer 9

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

Answer 10

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  :–


Return to  ML Aggarwal Solutions for ICSE Class-10

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