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Chapter-Test Trigonometric Identities ML Aggarwal Solutions ICSE Class-10

Chapter-Test 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  (Chapter-Test)
Writer ML Aggarwal
Book Name Understanding
Topics Solution of Exe-18, MCQ and Chapter Test Questions
Academic Session 2021-2022

Chapter-Test 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.


Chapter- Test ML Aggarwal Solutions Trigonometric Identities

(Page 460)

Question 1

(i) If θ is an acute angle and cosec θ = √5 find the value of cot θ – cos θ.
(ii) If θ is an acute angle and tan θ = \\ \frac { 8 }{ 15 } , find the value of sec θ + cosec θ.

Answer 1

(i) θ is an acute angle.
cosec θ = √5

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sin θ = 1/√5

And, cos θ = √(1 – sin2 θ)

cos θ = √(1 – (1/√5)2)

= √(1 – (1/5))

= √(4/5)

cos θ = 2/√5

Now, cot θ – cos θ = (cos θ/sin θ) – cos θ

ML aggarwal class-10 chapter 18 img 22

(ii) Given, θ is an acute angle and tan θ = 8/15

In fig. we have

tan θ = BC/AB = 8/15

So, BC = 8 and AB = 15

By Pythagoras theorem, we have

AC = √(AB2 + BC2) = √(52 + 82) = √(25 + 64) = √289

⇒ AC = 17

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Now,

sec θ = AC/AB = 17/15

cosec θ = AC/BC = 17/8

So,

sec θ + cosec θ = 17/15 + 17/8

= (136 + 255)/ 120

= 391/120

= 3.31/120

Question 2

Evaluate the following:

(i) 2 × (cos20° cos70°)/(sin2 25° + sin2 65°) – tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°

(ii) (sin222° + sin268°)/(cos222° + cos268°) + sin263° + cos63° sin27°


Answer 2

(i) 2\times \left( \frac { { cos }^{ 2 }{ 20 }^{ O }+{ cos }^{ 2 }{ 70 }^{ O } }{ { sin }^{ 2 }{ 25 }^{ O }+{ sin }^{ 2 }{ 65 }^{ O } } \right)  – tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°

(ii) \frac { { sin }^{ 2 }{ 22 }^{ O }+{ sin }^{ 2 }{ 68 }^{ O } }{ { cos }^{ 2 }{ 22 }^{ O }+{ cos }^{ 2 }{ 68 }^{ O } }  + sin2 63° + cos 63° sin 27°

Question 3

If \\ \frac { 4 }{ 3 }  (sec2 59° – cot2 31°) – \\ \frac { 2 }{ 2 }  sin 90° + 3tan2 56° tan2 34° = \\ \frac { x }{ 2 } , then find the value of x.

Answer 3

Given
\\ \frac { 4 }{ 3 }  (sec2 59° – cot2 31°) – \\ \frac { 2 }{ 2 }  sin 90° + 3tan2 56° tan2 34° = \\ \frac { x }{ 2 }
ML aggarwal class-10 chapter 18 img 23
Class-10 Trigonometric Identities ML Aggarwal img 26


(Page 461)

Prove the following (4 to 11) identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

Question 4


(i) \frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA } =2secA
(ii) \frac { cosA }{ cosecA+1 } +\frac { cosA }{ cosecA-1 } =2tanA


Answer 4


(i) \frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA } =2secA
L.H.S = \frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA }
ML aggarwal class-10 chapter 18 img 24
Class-10 Trigonometric Identities ML Aggarwal img 27

Question 5  (Chapter-Test Trigonometric Identities ML Aggarwal )


(i) sin2 θ + cos4 θ = cos2 θ + sin4 θ
(ii) \frac { cot\theta }{ cosec\theta +1 } +\frac { cosec\theta +1 }{ cot\theta } =2sec\theta


Answer 5

(i) L.H.S. = sin2 θ + cos4 θ = cos2 θ + sin4 θ

L.H.S. = sin2 θ + cos4 θ

= (1 – cos2 θ) + cos4 θ

= cos4 θ – cos2 θ + 1

= cos2 θ (cos2 θ – 1) + 1

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= cos2 θ (- sin2 θ) + 1

= 1 – sin2 θ cos2 θ

Now,

R.H.S. = cos2 θ + sin4 θ

= (1 – sin2 θ) + sin4 θ

= sin4 θ – sin2 θ + 1

= sin2 θ (sin2 θ – 1) + 1

= sin2 θ (-cos2 θ) + 1

= 1 – sin2 θ cos2 θ

Hence, L.H.S. = R.H.S.

(ii)

ML aggarwal class-10 chapter 18 img 25

Question 6

(i) sec4 A (1 – sin4 A) – 2 tan2 A = 1
(ii) \frac { 1 }{ sinA+cosA+1 } +\frac { 1 }{ sinA+cosA-1 } =secA+cosecA

Answer 6

(i) sec4 A (1 – sin4 A) – 2 tan2 A = 1
L.H.S = sec4 A (1 – sin4 A) – 2 tan2 A
ML aggarwal class-10 chapter 18 img 26
Class-10 Trigonometric Identities ML Aggarwal img 29

Question 7

(i) \frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta =1
(ii) (sec A – tan A)2 (1 + sin A) = 1 – sin A.

Answer 7

(i) \frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta =1
L.H.S = \frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta
ML aggarwal class-10 chapter 18 img 27
Class-10 Trigonometric Identities ML Aggarwal img 30

Question 8

(i) \frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA } =sinA+cosA
(ii) (sec A – cosec A) (1 + tan A + cot A) = tan A sec A – cot A cosec A
(iii) \frac { { tan }^{ 2 }\theta }{ { tan }^{ 2 }\theta -1 } -\frac { { cosec }^{ 2 }\theta }{ { sec }^{ 2 }\theta -{ cosec }^{ 2 }\theta } =\frac { 1 }{ { sin }^{ 2 }\theta -{ cos }^{ 2 }\theta }

Answer 8

(i) \frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA } =sinA+cosA
L.H.S = \frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA }
ML aggarwal class-10 chapter 18 img 29
Class-10 Trigonometric Identities ML Aggarwal img 31
ML aggarwal class-10 chapter 18 img 28
Class-10 Trigonometric Identities ML Aggarwal img 32

Question 9

\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } =\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } =\frac { 2 }{ 1-2{ cos }^{ 2 }A } =\frac { { 2sec }^{ 2 }A }{ { tan }^{ 2 }A-1 }
Answer 9
\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } =\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } =\frac { 2 }{ 1-2{ cos }^{ 2 }A } =\frac { { 2sec }^{ 2 }A }{ { tan }^{ 2 }A-1 }
L.H.S = \frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA }
ML aggarwal class-10 chapter 18 img 30
Class-10 Trigonometric Identities ML Aggarwal img 33

Question 10

2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = θ

Answer 10

2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = θ
L.H.S = 2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1

= 2 [(sin2 θ)3 + (cos2 θ)3] – 3 (sin4 θ + cos4 θ) + 1

= 2 [(sin2 θ + cos2 θ) (sin4 θ + cos4 θ – sin2 θ cos2 θ)] – 3 (sin4 θ + cos4 θ) + 1

= 2 (sin4 θ + cos4 θ – sin2 θ cos2 θ) – 3 (sin4 θ + cos4 θ) + 1

= 2 sin4 θ + 2 cos4 θ – 2 sin2 θ cos2 θ – 3 sin4 θ – 3 cos4 θ + 1

= 1 – sin4 θ – cos4 θ – 2 sin2 θ cos2 θ

= 1 – (sin2 θ + cos2 θ)2 = 1 – 1 = 0

= R. H. S

Question 11

If cot θ + cos θ = m, cot θ – cos θ = n, then prove that (m2 – n2)2 = 16 mn.

Answer 11

cot θ + cos θ = m…..(i)
cot θ – cos θ = n……(ii)
Adding (i)&(ii) we get
ML aggarwal class-10 chapter 18 img 31

Question 12  (Chapter-Test Trigonometric Identities ML Aggarwal )

When 0° < θ < 90°, solve the following equations:
(i) 2 cos2 θ + sin θ – 2 = 0
(ii) 3 cos θ = 2 sin2 θ
(iii) sec2 θ – 2 tan θ = 0
(iv) tan2 θ = 3 (sec θ – 1).

Answer 12

0° < θ < 90°
(i) 2 cos2 θ + sin θ – 2 = 0

⇒ 2 (1 – sin2 θ) + sin θ – 2 = 0

⇒ 2 – 2 sin2 θ + sin θ – 2 = 0

⇒ -2 sin2 θ + sin θ = 0

⇒ sin θ (1 – 2 sin θ) = 0

So, either sin θ = 0 or 1 – 2 sin θ = 0

If sin θ = 0

⇒ θ = 0o

And, if 1 – 2 sin θ = 0

sin θ = ½

⇒ θ = 30o

Thus, θ = 0o or 30o

(ii) 3 cos θ = 2 sin2 θ

⇒ 3 cos θ = 2 (1 – cos2 θ)

⇒ 3 cos θ = 2 – 2 cos2 θ

⇒ 2 cos2 θ + 3 cos θ – 2 = 0

⇒ 2 cos2 θ + 4 cos θ – cos θ – 2 = 0

⇒ 2 cos θ (cos θ + 2) – 1(cos θ + 2) = 0

⇒ (2 cos θ – 1) (cos θ + 2) = 0

So, either 2 cos θ – 1 = 0 or cos θ + 2 = 0

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