# ML Aggarwal Trigonometric Identities Exe-18 Solutions ICSE Class-10 Maths Ch-18

ML Aggarwal Trigonometric Identities Exe-18 Solutions ICSE Class-10 Maths Ch-18. We Provide Step by Step Answer of Exercise-18 Trigonometric Identities Questions for ICSE Class-10 APC Understanding Mathematics. Visit official Website for detail information about ICSE Board Class-10.

## ML Aggarwal Trigonometric Identities Exe-18 Solutions ICSE Class-10 Maths Ch-18

 Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 10th Chapter-18 Trigonometric Identities Writer ML Aggarwal Book Name Understanding Topics Solution of Exe-18 Questions Edition 2022-2023

### ML Aggarwal Trigonometric Identities Exe-18 Solutions

ICSE Class-10 Maths Ch-18

Page-456

sin A = 3/5
In ∆ABC, ∠B = 90°
AC = 5 and BC = 3

By Pythagoras theorem,

AB = √(AC2 – BC2)

= √(52 – 32) = √(25 – 9)

= √16

= 4

Now,

cos A = AB/AC = 4/5

tan A = BC/AB = 3/4

cot A = 1/tan θ = 4/3

sec A = 1/cos θ = 5/4

cosec A = 1/sin θ = 5/3

#### Question 2. If A is an acute angle and sec A = 17/8, find all other trigonometric ratios of angle A (using trigonometric identities).

sec A = 17/8 (A is an acute angle)
In right ∆ABC
sec A = AC/AB = 17/8
AC = 17, AB = 8

By Pythagoras theorem,

BC = √(AC2 – AB2)

= √(172 – 82) = √(289 – 64) = √225

= 15

Now,

sin A = BC/AC = 15/17

cos A = 1/sec A = 8/17

tan A = BC/AB = 15/8

cot A = 1/tan A= 8/15

cosec A = 1/sin A = 17/15

#### Question 3. If 12 cosec θ = 13, find the value of (2sinθ-3cosθ)/(4sinθ-9cosθ)

12 cosec θ = 13
⇒ cosec θ = 13/12
In right ∆ABC,
∠A = θ
cosec θ = AC/AB = 13/12

AC = 13 and BC = 12

By Pythagoras theorem,

AB = √(AC2 – BC2)

= √[(13)2 – (12)2]

= √(169 – 144)

= √25

= 5

Now,

sin θ = BC/AC = 12/13

cos θ = AB/AC = 5/13

Hence,

(2sin θ – 3cos θ)/(4sin θ – 9 cos θ)

= (2 × 12/13 – 3 × 5/13)/(4 × 12/13 – 9 × 5/13)

= (24/13 – 15/13)/(48/13 – 45/13)

= 9/13 ÷ 3/13

= 9/13 × 13/3

= 3

### ML Aggarwal Trigonometric Identities Exe-18 Solutions

ICSE Class-10 Maths Ch-11

Page-457

Without using trigonometric tables, evaluate the following (6 to 10) :

#### Question 4.

(i) cos226o+ cos 64o sin 26o + (tan 36o/cot 54o)

(ii) (sec 17o/ cosec 73o) + (tan 68o/ cot 22o) + cos2 44o + cos2 46o

Given that:

##### (i) cos2 26o + cos 64o sin 26o + (tan 36o/cot 54o)

= cos2 26o + cos (90o – 16o) sin 26o + [tan 36o/cot (90o – 54o)]

= [cos2 26o + sin2 26o] + (tan 36o/tan 36o)

= 1 + 1 = 2

##### (ii) (sec 17o/cosec 73o) + (tan 68o/cot 22o) + cos2 44o + cos2 46o

= [sec 17o/cosec (90o – 73o)] + [(tan 90– 22o)/cot 22o] + cos2 (90o – 44o) + cos2 46o

= [sec 17o/sec 17o] + [cot 22o/cot 22o] + [sin2 46o + cos2 46o]

= 1 + 1 + 1

= 3

#### Question 5.

(i) (sin 65o/cos 25o) + (cos 32o/sin 58o) – sin 28osec 62o+ cosec2 30o

(ii) (sin 29o/ cosec 61o) + 2 cot 8° cot 17° cot 45° cot 73° cot 82° – 3(sin² 38° + sin² 52°).

given that

##### (i) (sin 65o/cos 25o) + (cos 32o/sin 58o) – sin 28o sec 62o + cosec2 30o

= (sin 65o/cos (90o – 65o)) + (cos 32o/sin (90o – 32o)) – sin 28o sec (90o – 28o) + 22

= (sin 65o/sin 65o)+ (cos 32o/cos 32o) – [sin 28o x cosec 28o] + 4

= 1 + 1 – 1 + 4

= 5

##### (ii) (sin 29o/cosec 61o) + 2 cot 8° cot 17° cot 45° cot 73° cot 82° – 3(sin² 38° + sin² 52°).

= (sin 29o/cosec (90o – 29o)) + [2 cot 8° cot 17° cot 45° cot (90° – 17o) cot (90o – 8o)] – 3(sin² 38° + sin² (90° – 38o))

= (sin 29o/sin 29o) + [2 cot 8° cot 17° cot 45° tan 17o tan 8o] – 3(sin² 38° + cos² 38°)

= 1 + 2[(cot 8° tan 8o) (cot 17° tan 17o) cot 45°] – 3(1)

= 1 + 2[1×1×1] – 3

= 1 + 2 – 3

= 0

#### Question 6.

(i) (sin 35ocos 55o+ cos 35o sin 55 o)/ (cosec10o – tan2 80 o)

(ii) sin34o + sin56+ 2 tan18o tan 72o – cot30o

Given that:

##### (i) (sin 35o cos 55o + cos 35o sin 55o)/ (cosec2 10o – tan2 80o)

= sin 35o cos (90o – 35o) + cos 35o sin (90o – 35o)/(cosec2 10o – tan2(90o – 10o)

= sin 35sin 35o + cos 35cos 35o)/(cosec2 10o – cot2 10o)

= (sin2 35o + cos2 35o)/(cosec10o – cot2 10o)

= 1/1

= 1

##### (ii) sin2 34o + sin2 56o + 2 tan18o tan 72o – cot2 30o

= sin34o + sin(90o – 34o) + 2 tan18o tan (90o – 18o) – cot30o

= [sin34o + cos2 34o] + 2 tan18o cot 18o – cot30o

= 1 + 2 x 1 – (√3)2

= 1 + 2 – 3

= 0

#### Question 7.

(i) (tan 25o/cosec 65o)2+ (cot 25o/sec 65o)2+ 2 tan 18o tan 45o tan 75o

(ii) (cos2 25o + cos2 65o) + cosec θ sec (90o – θ) – cot θ tan (90o – θ)

##### (ii) (cos2 25o + cos2 65o) + cosec θ sec (90o – θ) – cot θ tan (90o – θ)

= cos2 25o + cos2 (90o – 25o) + cosec θ sec (90o – θ) – cot θ. cot θ

= (cos2 25o + sin2 25o) + (cosec2 θ – cot2 θ)

= 1 + 1 = 2

Prove that following (12 to 30) identities, where the angles involved are acute angles for which the trigonometric ratios as defined:

#### Question 8.

(i) (sec A + tan A) (1 – sin A) = cos A
(ii) (1 + tan2 A) (1 – sin A) (1 + sin A) = 1.

##### (i) (sec A + tan A) (1 – sin A) = cos A

L.H.S. = (sec A + tan A) (1 – sin A)

#### Question 9.

(i) tan A + cot A = sec A cosec A
(ii) (1 – cos A)(1 + sec A) = tan A sin A.

##### (i) L.H.S. = tan A + cot A

= sin A/cos A + cos A/sin A

= (sin2 A + cos2 A)/(sin A cos A)

= 1/(sin A cos A)

= sec A cosec A

= R.H.S

##### (ii) L.H.S. = (1 – cos A) (1 + sec A)

= (1 – cos A)(1 + 1/cos A)

= (1 – cos A)(cos A + 1)/cos A

= (1 – cos2 A)/(cos A)

= (sin2A)/(cos A)

= sinA/cos A

= sin A × sin A/cos A {1 – cos2 A = sin2 A}

= tan A sin A

= R.H.S.

#### Question 10.

(i) cot² A – cos² A = cot² A cos² A

(ii) 1 + tan²A/ 1 + sec A = sec A

(iii) (1 + sec A)/ sec A = sin² A/ 1- cos A

(iv) sin A/(1 – cos A) = cosec A + cot A.

##### (ii) 1 + tan²A/ 1 + sec A = sec A

We know that tan² 2A + 1 = sec² 2A

tan² 2A  = sec² 2A – 1

1 + sec A – 1

= sec A.

L.H.S = R.H.S.

##### (iii) (1 + sec A)/ sec A = sin² A/ 1- cos A

LHS = ( 1 + sec A )/sec A

= ( 1 + 1/cos A ) / ( 1/cosA )

= [ ( Cos A + 1 ) / cos A ] / ( 1/ cos A )

= Cos A + 1

= ( 1 + cos A ) ( 1 – cos A ) / ( 1 – cos A )

= ( 1 – cos² A ) / ( 1 – cos A )

= Sin² A / ( 1 – cos A )

= RHS

##### (iv) sin A/(1 – cos A) = cosec A + cot A.

If A = = θ

then,

We know that sin² θ = cos² θ = 1

= cosec θ + cot θ.

= cosec A + cot A.

#### Question 11.

(i) sin A/ (1 + cos A) = (1 – cos A)/ sin A

(ii) (1 – tan2 A)/ (cot2 A – 1) = tan2 A

(iii) sin A/ (1 + cos A) = cosec A – cot A

##### (i) sin A/ (1 + cos A) = (1 – cos A)/ sin A L.H.S = sin A/ (1 + cos A)

(multiplying and dividing by (1 – cosA))

#### Question 12.

(i) (sec A – 1)/(sec A + 1) = (1 – cos A)/(1 + cos A)

(ii) tan2 θ/ (sec θ – 1)2 = (1 + cos θ)/ (1 – cos θ)

(iii) (1 + tan A)2 + (1 – tan A)2 = 2 sec2 A

(iv) secA + cosec2 A = sec2 A. cosec2 A

##### (i) (sec A – 1)/(sec A + 1) = (1 – cos A)/(1 + cos A)

L.H.S. = (sec A – 1)/(sec A + 1)

##### (iii) L.H.S. = (1 + tan A)2 + (1 – tan A)2

= 1 + 2 tan A + tan2 A + 1 – 2 tan A + tan2 A

= 2 + 2 tan2 A

= 2(1 + tan2 A) [As 1 + tan2 A = sec2 A]

= 2 sec2 A

= R.H.S.

##### (iv) L.H.S = sec2 A + cosec2 A

= 1/cos2 A + 1/sin2 A

= (sin2 A + cos2 A)/(sin2 A cos2 A)

= 1/ (sin2 A cos2 A)

= sec2 A cosec2 A = R.H.S

#### Question 13.

(i) (1 + sin A)/cos A + cos A/(1 + sin A) = 2 sec A

(ii) tan A/(sec A – 1) + tan A/(sec A + 1) = 2cosec A

#### Question 14.

(i) cosec A/(cosec A – 1) + cosec A/(cosec A + 1) = 2 sec2A

(ii) cot A – tan A = (2cos2 A – 1)/ (sin A – cos A)

(iii) (cot A – 1)/ (2 – sec2 A) = cot A/ (1 + tan A)

#### Question 15.

(i) tan2θ – sin2θ = tan2 θ sin2 θ

(ii) cos θ/ (1 – tan θ) – sin2 θ/ (cos θ – sin θ) = cos θ + sin θ

(i) tan2θ – sin2θ = tan2 θ sin2 θ
L.H.S = tan2θ – sin2θ

### ML Aggarwal Trigonometric Identities Exe-18 Solutions

ICSE Class-10 Maths Ch-11

Page-458

#### Question 16.

(i) cosec4 θ – cosec2 θ = cot4 θ + cot2 θ
(ii) 2 sec2 θ – sec4 θ – 2 cosec2 θ + cosec4 θ = cot4 θ – tan4 θ.

##### (i) L.H.S. = cosec4 θ – cosec2 θ

= cosec2 θ (cosec2 θ – 1)

= cosec2 θ cot2 θ [cosec2 θ – 1 = cot2 θ]

= (cot2 θ + 1) cot2 θ

= cot4 θ + cot2 θ

= R.H.S.

##### (ii) L.H.S. = 2 sec2 θ – sec4 θ – 2 cosec2 θ + cosec4 θ

= 2 (tan2 θ + 1) – (tan2 θ + 1)2 – 2 (1 + cot2 θ) + (1 + cot2 θ)2

{∵ sec2 θ = tan2 θ + 1 cosec2 θ = 1 + cot2 θ}

= 2 tan2 θ + 2 – (tan4 θ + 2 tan2 θ + 1) – 2 – 2 cot2 θ + (1 + 2 cotθ + cot4 θ)

= 2 tan2 θ + 2 – tan4 θ – 2 tan2 θ – 1 – 2 – 2 cot2 θ + 1 + 2 cotθ + cot4 θ

= cot4 θ – tan4 θ = R.H.S.

#### Question 17.

(i) (1 + cos θ – sin2 θ)/(sin θ(1 + cos θ) = cot θ

(ii) (tan3 θ – 1)/(tan θ – 1) = sec2 θ + tan θ

##### (i) (1 + cos θ – sin2 θ)/(sin θ(1 + cos θ) = cot θ

L.H.S = (1 + cos θ – sin2 θ)/(sin θ(1 + cos θ)

#### Question 18.

(i) (1 + cosec A)/cosec A = cos2 A/(1 – sin A)
(ii) √(1-cosA)/(1+cosA)=sinA/(1+cosA)

(i)  (1 + cosec A)/cosec A = cos2 A/(1 – sin A)
L.H.S = (1 + cosec A)/cosec A

#### Question 19.

(i) √(1+sinA)/(1-sinA)=tanA+secA
(ii) √(1-cosA)/(1+cosA)=cosecA-cotA

(i) √(1+sinA)/(1-sinA)=tanA+secA

#### Question 20.

(i) √(secA-1)/(secA+1) +√(secA+1)/(secA-1)=2cosecA
(ii) cotA.cotA/(1-sinA)=(1+cosecA)

##### (i) √(secA-1)/(secA+1) +√(secA+1)/(secA-1)=2cosecA

L.H.S = √(secA-1)/(secA+1) +√(secA+1)/(secA-1)

#### Question 21.

(i) (1 + tan A)/(sin A) + (1 + cot A)/cos A = 2(sec A + cosec A)

(ii) sec4 A – tan4 A = 1 + 2 tan2 A

#### Question 22.

(i) cosec6 A – cot6 A = 3 cot2 A cosec2 A + 1
(ii) sec6 A – tan6 A = 1 + 3 tan2 A + 3 tan4 A

#### Question 23.

(i)( cotθ+cosecθ-1)/(cotθ-cosecθ+1)=(1+cosθ)/sinθ
(ii) sinθ/(cotθ+cosecθ)=2+[sinθ/(cotθ-cosecθ)]

(i) (cot θ + cosec θ – 1)/(cot θ – cosec θ + 1) = (1 + cos θ)/sin θ

L.H.S. = (cot θ – cosec θ – 1)/(cot θ – cosec θ + 1)

#### Question 24.

(i) (sinθ + cos θ)(secθ + cosecθ) = 2 + sec θ cosec θ
(ii) (cosec A – sin A) (sec A – cos A) sec2A = tan A

##### (i) (sinθ + cosθ)(secθ + cosecθ) = 2 + secθ cosecθ

L.H.S = (sinθ + cosθ)(secθ + cosecθ)

#### Question 25.

(i) (sin3 A + cos3 A)/(sin A + cos A) + (sinA – cos3 A)/(sin A – cos A) = 2
(ii) tan²A/(1+tan²A)+cot²A/(1+cot²A)=1

##### (i) (sin3 A + cos3 A)/(sin A + cos A) + (sin3 A – cos3 A)/(sin A – cos A) = 2

L.H.S. = (sin3 A + cos3 A)/(sin A + cos A) + (sinA – cos3 A)/(sin A – cos A)

#### Question 26.

(i) 1/(secA+tanA)-1/cosA=1/cosA-1/(secA-tanA)
(ii)( sinA+secA)² + (cosA +cosecA)²=(1+secAcosecA)²
(iii) (tanA+sinA)/(tanA-sinA)=(secA+1)/(secA-1)

(i) 1/(secA+tanA)-1/cosA=1/cosA-1/(secA-tanA)
L.H.S = 1/(secA+tanA)-1/cosA

(ML Aggarwal Trigonometric Identities Exe-18 Solutions)

#### Question 27. If sin θ + cos θ = √2 sin (90° – θ), show that cot θ = √2 + 1

sin θ + cos θ = √2 sin (90° – θ)
sin θ + cos θ = √2 cos θ
dividing by sin θ

#### Question 28. If 7 sin2 θ + 3 cos2 θ = 4, 0° ≤ θ ≤ 90°, then find the value of θ.

7 sin2 θ + 3 cos2 θ = 4, 0° ≤ θ ≤ 90°
3 sin2 θ + 3 cos2 θ + 4 sin2 θ = 4

⇒ 3 (sin2 θ + 3 cos2 θ) + 4 sin2 θ = 4

⇒ 3 (1) + 4 sin2 θ = 4

⇒ 4 sin2 θ = 4 – 3

⇒ sin2 θ = ¼

Taking square-root on both sides, we get

sin θ = ½

Thus, θ = 30o

#### Question 29. If sec θ + tan θ = m and sec θ – tan θ = n, prove that mn = 1.

sec θ + tan θ = m and sec θ – tan θ = n
mn = (sec θ + tan θ) (sec θ – tan θ) = sec2 θ – tan2 θ = 1
(∴ sec2 θ – tan2 θ = 1)
Hence proved.

#### Question 30. If x – a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2 – y2 = a2 – b2.

x – a sec θ + b tan θ and y = a tan θ + b sec θ
To prove that x2 – y2 = a2 – b2.

= a2 sec2θ + b2 tan2θ + 2ab secθ tanθ – (a2 tan2θ + b2sec2θ + 2ab secθ tanθ)

= a2 sec2θ + b2tan2θ + 2ab secθ tanθ – a2tan2θ – b2sec2θ – 2ab secθ tanθ

= a2 (sec2θ – tan2θ) – b(sec2θ – tan2θ)

= a2 × 1 – b2 × 1  (sec2θ – tan2θ = 1)

= a2 – b2

#### Question 31. If x = h + a cos θ and y = k + a sin θ, prove that (x – h)2 + (y – k)2 = a2.

x = h + a cos θ and y = k + a sin θ
To prove that (x – h)2 + (y – k)2 = a2.

x – h = a cos θ

y – k = a sin θ

On squaring and adding we get

(x – h)2 + (y – k)2 = a2 cos2 θ + a2 sin2 θ

= a(sin2 θ + cos2 θ)

= a2 (1) [Since, sin2 θ + cos2 θ = 1]

Hence, proved

{ sin² θ + cos² θ = 1}

—  : End of ML Aggarwal Trigonometric Identities Exe-18 Solutions ICSE Class-10 Maths Ch-18 : –

Thanks

### 4 thoughts on “ML Aggarwal Trigonometric Identities Exe-18 Solutions ICSE Class-10 Maths Ch-18”

1. This helped me a lot.
Tomorrow is my exam and now it’s 10.45 pm .
It helped me to do revision.
Thanks to the website owner.
🙂🙂

• thanks for positive response

2. In the Chapter : Trigonometric Identities, Excercise 18- Sum 23(i): The question displayed is wrong. The question displayed is-

(cotθ – cosecθ -1)÷(cotθ – cosecθ+1)=(1+cosθ)÷sinθ

The correct question is-(Both according to the book and the answer solved):

(cotθ + cosecθ -1)÷(cotθ – cosecθ+1)=(1+cosθ)÷sinθ

That is ,
In the L.H.S numerator, cosecθ is to be added instead of being subtracted. Meaning ,
(cotθ + cosecθ -1) instead of (cotθ – cosecθ -1).

There is no problem with the steps of the answer as it is solved according to the correct form of question.

Kindly solve this issue if possible.
Regards.