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 CISCE for detail information about ICSE Board Class-10.

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

Board	ICSE
Subject	Maths
Class	10th
Chapter-18	Trigonometric Identities
Writer / Book	Understanding
Topics	Solutions of Exe-18
Academic Session	2024-2025

ML Aggarwal Trigonometric Identities Exe-18 Solutions

ICSE Class-10 Maths Ch-18

Question 1. If A is an acute angle and sin A = 3/5 find all other trigonometric ratios of angle A (using trigonometric identities).

Answer :

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).

Answer :

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)

Answer :

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

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

Question 4.

(i) cos²26^o+ cos 64^o sin 26^o + (tan 36^o/cot 54^o)

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

Answer :

Given that:

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

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

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

= 1 + 1 = 2

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

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

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

= 1 + 1 + 1

= 3

Question 5.

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

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

Answer :

given that

Given,

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

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

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

= 1 + 1 – 1 + 4

= 5

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

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

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

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

= 1 + 2[1×1×1] – 3

= 1 + 2 – 3

= 0

Question 6.

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

(ii) sin² 34^o + sin² 56^o + 2 tan18^o tan 72^o – cot² 30^o

Answer :

Given that:

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

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

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

= (sin² 35^o + cos² 35^o)/(cosec² 10^o – cot² 10^o)

= 1/1

= 1

(ii) sin² 34^o + sin² 56^o + 2 tan18^o tan 72^o – cot² 30^o

= sin² 34^o + sin² (90^o – 34^o) + 2 tan18^o tan (90^o – 18^o) – cot² 30^o

= [sin² 34^o + cos² 34^o] + 2 tan18^o cot 18^o – cot² 30^o

= 1 + 2 x 1 – (√3)²

= 1 + 2 – 3

= 0

Question 7.

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

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

Answer :

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

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

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

= (cos² 25^o + sin² 25^o) + (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.

Answer :

(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.

Answer :

(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)

= sin² A/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.

Answer :

(i) LHS = cot² A – cos² 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

Answer :

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

Answer :

(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

Answer :

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

Answer :

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

Question 15.

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

Answer :

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

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.

Answer :

(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 cot² θ + cot⁴ θ)

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

Answer :

(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)

Answer :

(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

Answer :

(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)

Answer :

(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

Answer :

(i) (1 + tan A)/(sin A) + (1 + cot A)/cos A = 2(sec A + cosec A)
L.H.S = (1 + tan A)/(sin A) + (1 + cot A)/cos 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

Answer :

(i) cosec⁶ A – cot⁶ A = 3 cot² A cosec² A + 1
L.H.S = cosec⁶ A – cot⁶ A

Question 23.

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

Answer :

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