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

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Exercise 18 ,

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### 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.

(i) θ is an acute angle.
cosec θ = √5

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 (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

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°

(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.

Given $\\ \frac { 4 }{ 3 }$ (sec2 59° – cot2 31°) – $\\ \frac { 2 }{ 2 }$ sin 90° + 3tan2 56° tan2 34° = $\\ \frac { x }{ 2 }$  (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) $\frac { cosA }{ cosecA+1 } +\frac { cosA }{ cosecA-1 } =2tanA$

(i) $\frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA } =2secA$
L.H.S = $\frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA }$  #### Question 5  (Chapter-Test Trigonometric Identities ML Aggarwal )

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

(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

= 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) #### Question 6

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

(i) sec4 A (1 – sin4 A) – 2 tan2 A = 1
L.H.S = sec4 A (1 – sin4 A) – 2 tan2 A  #### 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.

(i) $\frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta =1$
L.H.S = $\frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta$  #### 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) $\frac { { tan }^{ 2 }\theta }{ { tan }^{ 2 }\theta -1 } -\frac { { cosec }^{ 2 }\theta }{ { sec }^{ 2 }\theta -{ cosec }^{ 2 }\theta } =\frac { 1 }{ { sin }^{ 2 }\theta -{ cos }^{ 2 }\theta }$

(i) $\frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA } =sinA+cosA$
L.H.S = $\frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA }$    #### Question 9 $\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } =\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } =\frac { 2 }{ 1-2{ cos }^{ 2 }A } =\frac { { 2sec }^{ 2 }A }{ { tan }^{ 2 }A-1 }$ $\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } =\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } =\frac { 2 }{ 1-2{ cos }^{ 2 }A } =\frac { { 2sec }^{ 2 }A }{ { tan }^{ 2 }A-1 }$
L.H.S = $\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA }$  #### Question 10

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

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.

cot θ + cos θ = m…..(i)
cot θ – cos θ = n……(ii) #### 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).

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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