ML Aggarwal Solutions for Class 10 Maths Chapter 18 Trigonometric Identities

ML Aggarwal Solutions for Class 10 Maths Chapter 18 Trigonometric Identities is the best resource for students who are looking for a reference material. It’s highly suggested that students use these solutions to solve problems as it’s in accordance with the latest syllabus of ICSE board. Students aiming to score well in the examinations can access the ML Aggarwal Solutions for Class 10 Maths Chapter 18 Trigonometric Identities PDF which is available in the link given below.

Chapter 18 explains the proof of various trigonometric identities and solution of trigonometric equations which are crucial from the exam perspective. The solutions are created by subject experts at BYJU’S in order to help students prepare confidently for the board exams and other competitive exams at various levels. Referring to the PDF comprising the ML Aggarwal Solutions will build a strong foundation over concepts covered in this chapter.

PDF Solution for ML Aggarwal Class 10 Maths Chapter 18 :-Click Here to Download PDF

 

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Access ML Aggarwal Solutions for Class 10 Maths Chapter 18 Trigonometric Identities

Exercise 18

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

Solution:

ML Aggarwal Solutions for Class 10 Chapter 18 - 1Given,

sin A = 3/5 and A is an acute angle

So, in ∆ABC we have ∠B = 90o

And,

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

 

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

Solution:

ML Aggarwal Solutions for Class 10 Chapter 18 - 2Given,

sec A = 17/8 and A is an acute angle

So, in ∆ ABC we have ∠B = 90o

And,

AC = 17 and 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

3. Express the ratios cos A, tan A and sec A in terms of sin A.

Solution:

We know that,

sin2 A + cos2 A = 1

So,

cos A = √(1 – sin2 A)

tan A = sin A/cos A = sin A/ √(1 – sin2 A)

sec A = 1/cos A = 1/ (√1 – sin2 A)

4. If tan A = 1/√3, find all other trigonometric ratios of angle A.

Solution:

ML Aggarwal Solutions for Class 10 Chapter 18 - 3Given, tan A = 1/√3

In right ∆ ABC,

tan A = BC/AB = 1/√3

So,

BC = 1 and AB = √3

By Pythagoras theorem,

AC = √(AB2 + BC2) = √[(√3)2 + (1)2]

= √(3 + 1) = √4 = 2

Hence,

sin A = BC/AC = ½

cos A = AB/AC = √3/2

cot A = 1/tan A = √3

sec A = 1/cos A = 2/√3

cosec A = 1/sin A = 2/1 = 2

5. If 12 cosec θ = 13, find the value of (2 sin θ – 3 cos θ)/ (4 sin θ – 9 cos θ)

Solution:

Given,

12 cosec θ = 13

⇒ cosec θ = 13/12

In right ∆ ABC,

∠A = θ

So, cosec θ = AC/BC = 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,

ML Aggarwal Solutions for Class 10 Chapter 18 - 4

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

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

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

Solution:

Given,

(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 90o – 22o)/ cot 22o] + cos2 (90o – 44o) + cos2 46o

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

= 1 + 1 + 1

= 3

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

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

Solution:

Given,

(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 x 1 x 1] – 3

= 1 + 2 – 3

= 0

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

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

Solution:

Given,

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

ML Aggarwal Solutions for Class 10 Chapter 18 - 5

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

= sin2 34o + sin2 (90o – 34o) + 2 tan18o tan (90o – 18o) – cot2 30o

= [sin2 34o + cos2 34o] + 2 tan18o cot 18o – cot2 30o

= 1 + 2 x 1 – (√3)2

= 1 + 2 – 3

= 0

9. (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)

Solution:

Given,

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

ML Aggarwal Solutions for Class 10 Chapter 18 - 6

(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

10. (i) 2(sec² 35° – cot² 55°) – ML Aggarwal Solutions for Class 10 Chapter 18 - 7

(ii) ML Aggarwal Solutions for Class 10 Chapter 18 - 8

Solution:

Given,

ML Aggarwal Solutions for Class 10 Chapter 18 - 9

11. Prove that following: 

(i) cos θ sin (90° – θ) + sin θ cos (90° – θ) = 1

(ii) tan θ/ tan (90o – θ) + sin (90o – θ)/ cos θ = sec2 θ

(iii) (cos (90o – θ) cos θ)/ tan θ + cos2 (90o – θ) = 1

(iv) sin (90o – θ) cos (90o – θ) = tan θ/ (1 + tan2 θ)

Solution:

(i) L.H.S. = cos θ sin (90° – θ) + sin θ cos (90° – θ)

= cos θ x cos θ + sin θ x sin θ

= cos2 θ + sin2 θ

= 1 = R.H.S.

(ii) L.H.S = tan θ/ tan (90o – θ) + sin (90o – θ)/ cos θ

= tan θ/ cot θ + cos θ/ cos θ

= tan θ/ (1/tan θ) + 1

= tan2 θ + 1 = sec2 θ = R.H.S.

(iii) L.H.S. = (cos (90o – θ) cos θ)/ tan θ + cos2 (90o – θ)

= (sin θ cos θ)/ tan θ + sin2 θ

= (sin θ cos θ)/ (sin θ/ cos θ) + sin2 θ

= cos2 θ + sin2 θ

= 1 = R.H.S.

ML Aggarwal Solutions for Class 10 Chapter 18 - 10

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

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

(ii) (1 + tan2 A) (1 – sin A) (1 + sin A) = 1.

Solution:

Given,

(i) L.H.S = (sec A + tan A) (1 – sin A)

ML Aggarwal Solutions for Class 10 Chapter 18 - 11

(ii) L.H.S. = (1 + tan2 A) (1 – sin A) (1 + sin A)

ML Aggarwal Solutions for Class 10 Chapter 18 - 12

13. (i) tan A + cot A = sec A cosec A 

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

Solution:

(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)

ML Aggarwal Solutions for Class 10 Chapter 18 - 13

14. (i) 1/ (1 + cos A) + 1/ (1 – cos A) = 2 cosec2 A

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

Solution:

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

ML Aggarwal Solutions for Class 10 Chapter 18 - 14

(ii)

ML Aggarwal Solutions for Class 10 Chapter 18 - 15

15. (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

Solution:

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

On multiplying and dividing by (1 – cos A), we have

ML Aggarwal Solutions for Class 10 Chapter 18 - 16

16. (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

Solution:

ML Aggarwal Solutions for Class 10 Chapter 18 - 17

(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

17. (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

Solution:

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

ML Aggarwal Solutions for Class 10 Chapter 18 - 18

18. (i) cosec A/ (cosec A – 1) + cosec A/ (cosec A + 1) = 2 sec2 A

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

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

Solution:

(i) L.H.S. =
ML Aggarwal Solutions for Class 10 Chapter 18 - 19

ML Aggarwal Solutions for Class 10 Chapter 18 - 20

ML Aggarwal Solutions for Class 10 Chapter 18 - 21

19. (i) tan2 θ – sin2 θ = tan2 θ sin2 θ

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

Solution:

(i) L.H.S = tan2 θ – sin2 θ

ML Aggarwal Solutions for Class 10 Chapter 18 - 22

ML Aggarwal Solutions for Class 10 Chapter 18 - 23

20. (i) cosec4 θ – cosec2 θ = cot4 θ + cot2 θ 

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

Solution:

(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

ML Aggarwal Solutions for Class 10 Chapter 18 - 24

= 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.

21. (i) ML Aggarwal Solutions for Class 10 Chapter 18 - 25 = cot θ

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

Solution:

ML Aggarwal Solutions for Class 10 Chapter 18 - 26

ML Aggarwal Solutions for Class 10 Chapter 18 - 27

22. (i) (1 + cosec A)/ cosec A = cos2 A/ (1 – sin A)

(ii) ML Aggarwal Solutions for Class 10 Chapter 18 - 28

Solution:

ML Aggarwal Solutions for Class 10 Chapter 18 - 29

ML Aggarwal Solutions for Class 10 Chapter 18 - 30

23. (i) ML Aggarwal Solutions for Class 10 Chapter 18 - 31 = tan A + sec A

(ii) ML Aggarwal Solutions for Class 10 Chapter 18 - 32 = cosec A – cot A

Solution:

ML Aggarwal Solutions for Class 10 Chapter 18 - 33

ML Aggarwal Solutions for Class 10 Chapter 18 - 34

24. (i) ML Aggarwal Solutions for Class 10 Chapter 18 - 35 = 2 cosec A

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

Solution:

ML Aggarwal Solutions for Class 10 Chapter 18 - 36

ML Aggarwal Solutions for Class 10 Chapter 18 - 37

25. (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

Solution:

ML Aggarwal Solutions for Class 10 Chapter 18 - 38

(ii) sec4 A – tan4 A = 1 + 2 tan2 A

L.H.S. = sec4 A – tan4 A

= (sec2 A – tan2 A) (sec2 A + tan2 A)

= (1 + tan4 A – tan4 A) (1 + tan4 A + tan4 A) [As sec2 A = tan4 A + 1]

= 1 (1 + 2 tan2 A)

= 1 + 2 tan2 A = R.H.S.

26. (i) cosec6 A – cot6 A = 3 cot2 A cosec2 A + 1 

(ii) sec6 A – tan6 A = 1 + 3 tan2 A + 3 tan4 A

Solution:

(i) cosec6 A – cot6 A = 3 cot2 A cosec2 A + 1

ML Aggarwal Solutions for Class 10 Chapter 18 - 39

27.

ML Aggarwal Solutions for Class 10 Chapter 18 - 40

Solution:

(i)

ML Aggarwal Solutions for Class 10 Chapter 18 - 41

ML Aggarwal Solutions for Class 10 Chapter 18 - 42

= (cos θ – 1) (2 – 1 – cos θ)/ (cos θ – 1)

= 1 – cos θ

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

28. (i) (sin θ + cos θ) (sec θ + cosec θ) = 2 + sec θ cosec θ 

(ii) (cosec A – sin A) (sec A – cos A) sec2A = tan A

Solution:

(i) (sin θ + cos θ) (sec θ + cosec θ) = 2 + sec θ cosec θ

L.H.S. = (sin θ + cos θ) (sec θ + cosec θ)

ML Aggarwal Solutions for Class 10 Chapter 18 - 43

= R.H.S.

(ii)

ML Aggarwal Solutions for Class 10 Chapter 18 - 44

29.

ML Aggarwal Solutions for Class 10 Chapter 18 - 45

Solution:

ML Aggarwal Solutions for Class 10 Chapter 18 - 46

ML Aggarwal Solutions for Class 10 Chapter 18 - 47

30. (i) 1/ (sec A + tan A) – 1/cos A = 1/cos A – 1/(sec A – tan A)

(ii) (sin A + sec A)2 + (cos A + cosec A)2 = (1 + sec A cosec A)2

(iii) (tan A + sin A)/ (tan A – sin A) = (sec A + 1)/ (sec A – 1)

Solution:

ML Aggarwal Solutions for Class 10 Chapter 18 - 48

ML Aggarwal Solutions for Class 10 Chapter 18 - 49

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

Solution:

Given, sin θ + cos θ = √2 sin (90° – θ)

sin θ + cos θ = √2 cos θ

On dividing by sin θ, we have

1 + cot θ = √2 cot θ

1 = √2 cot θ – cot θ

(√2 – 1) cot θ = 1

cot θ = 1/ (√2 – 1)

ML Aggarwal Solutions for Class 10 Chapter 18 - 50

Hence, cot θ = √2 + 1

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

Solution:

Given,

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

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

Solution:

Given,

sec θ + tan θ = m

sec θ – tan θ = n

Now,

mn = (sec θ + tan θ) (sec θ – tan θ)

= sec2 θ – tan2 θ = 1

Thus, mn = 1

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

Solution:

Given,

x =
a sec θ + b tan θ,

y = a tan θ + b sec θ

Now,

x2 – y2 = (a sec θ + b tan θ)2 – (a tan θ + b sec θ)2

ML Aggarwal Solutions for Class 10 Chapter 18 - 51

– Hence proved.

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

Solution:

Given,

x = h + a cos θ

y = k + a sin θ

Now,

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

= a2 (sin2 θ + cos2 θ)

= a2 (1) [Since, sin2 θ + cos2 θ = 1]

– Hence proved

Chapter Test

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 = 8/15, find the value of sec θ + cosec θ. 

Solution:

Given, θ is an acute angle and cosec θ = √5

So,

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 Solutions for Class 10 Chapter 18 - 52

ML Aggarwal Solutions for Class 10 Chapter 18 - 53(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

=
ML Aggarwal Solutions for Class 10 Chapter 18 - 54

2. Evaluate the following: 

(i) 2 x (ML Aggarwal Solutions for Class 10 Chapter 18 - 55) – tan 45o + tan 13o tan 23o tan 30o tan 67o tan 77o

(ii) ML Aggarwal Solutions for Class 10 Chapter 18 - 56 + sin2 63o + cos 63o sin 27o

Solution:

ML Aggarwal Solutions for Class 10 Chapter 18 - 57

3. If 4/3 (sec2 59o – cot2 31o) – 2/3 sin 90o + 3 tan2 56o tan2 34o = x/2, then find the value of x.

Solution:

Given,

4/3 (sec2 59o – cot2 31o) – 2/3 sin 90o + 3 tan2 56o tan2 34o = x/2

ML Aggarwal Solutions for Class 10 Chapter 18 - 58

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

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

Solution:

ML Aggarwal Solutions for Class 10 Chapter 18 - 59

ML Aggarwal Solutions for Class 10 Chapter 18 - 60

5. (i) ML Aggarwal Solutions for Class 10 Chapter 18 - 61

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

Solution:

ML Aggarwal Solutions for Class 10 Chapter 18 - 62

ML Aggarwal Solutions for Class 10 Chapter 18 - 63

6. (i) sin2 θ + cos4 θ = cos2 θ + sin4 θ 

(ii) ML Aggarwal Solutions for Class 10 Chapter 18 - 64

Solution:

Given,

(i) 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.

ML Aggarwal Solutions for Class 10 Chapter 18 - 65

ML Aggarwal Solutions for Class 10 Chapter 18 - 66

7. (i) sec4 A (1 – sin4 A) – 2 tan2 A = 1 

(ii) ML Aggarwal Solutions for Class 10 Chapter 18 - 67 = sec A + cosec A

Solution:

(i) sec4 A (1 – sin4 A) – 2 tan2 A = 1

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

ML Aggarwal Solutions for Class 10 Chapter 18 - 68

ML Aggarwal Solutions for Class 10 Chapter 18 - 69

8. (i) ML Aggarwal Solutions for Class 10 Chapter 18 - 70 + sin θ cos θ = 1

(ii) (sec A – tan A)2 (1 + sin A) = 1 – sin A.

Solution:

ML Aggarwal Solutions for Class 10 Chapter 18 - 71

ML Aggarwal Solutions for Class 10 Chapter 18 - 72

9. (i) ML Aggarwal Solutions for Class 10 Chapter 18 - 73 = sin A + cos A

(ii) (sec A – cosec A) (1 + tan A + cot A) = tan A sec A – cot A cosec A 

(iii) ML Aggarwal Solutions for Class 10 Chapter 18 - 74

Solution:

ML Aggarwal Solutions for Class 10 Chapter 18 - 75

ML Aggarwal Solutions for Class 10 Chapter 18 - 76

10. ML Aggarwal Solutions for Class 10 Chapter 18 - 77

Solution:

ML Aggarwal Solutions for Class 10 Chapter 18 - 78

11. 2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = 0

Solution:

Given,

2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = 0

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 – [sin4 θ + cos4 θ + 2 sin2 θ cos2 θ]

= 1 – 1

= 0 = R.H.S.

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

Solution:

Given,

cot θ + cos θ = m … (i)

cot θ – cos θ = n … (ii)

Adding (i) and (ii), we get

ML Aggarwal Solutions for Class 10 Chapter 18 - 79

ML Aggarwal Solutions for Class 10 Chapter 18 - 80

13. If sec θ + tan θ = p, prove that sin θ = (p2 – 1)/ (p2 + 1)

Solution:

Given, sec θ + tan θ = p

ML Aggarwal Solutions for Class 10 Chapter 18 - 81

14. If tan A = n tan B and sin A = m sin B, prove that cos2 A = (m2 – 1)/ (n2 – 1)

Solution:

Given,

tan A = n tan B and sin A = m sin B

n = tan A/ tan B

m = sin A/ sin B

ML Aggarwal Solutions for Class 10 Chapter 18 - 82

15. If sec A = x + 1/4x, then prove that sec A + tan A = 2x or 1/2x

Solution:

Given, sec A = x + 1/4x

We know that,

ML Aggarwal Solutions for Class 10 Chapter 18 - 83

ML Aggarwal Solutions for Class 10 Chapter 18 - 84

16. 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).

Solution:

Given, 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)

(2 cos θ – 1) (cos θ + 2) = 0

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

If 2 cos θ – 1 = 0

cos θ = ½

⇒ θ = 60o

And, for cos θ + 2 = 0

⇒ cos θ = -2 which is not possible being out of range.

Thus, θ = 60o

(iii) sec2 θ – 2 tan θ = 0

(1 + tan2 θ) – 2 tan θ = 0

tan2 θ – 2 tan θ + 1 = 0

(tan θ – 1)2 = 0

tan θ – 1 = 0

⇒ tan θ = 1

Thus, θ = 45o

(iv) tan2 θ = 3 (sec θ – 1)

(sec2 θ – 1) = 3 sec θ – 3

sec2 θ – 1 – 3 sec θ + 3 = 0

sec2 θ – 3 sec θ + 2 = 0

sec2 θ – 2 sec θ – sec θ + 2 = 0

sec θ (sec θ – 2) – 1 (sec θ = 2) = 0

(sec θ – 1) (sec θ – 2) = 0

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

If sec θ – 1 = 0

sec θ = 1

⇒ θ = 0o

And, if sec θ – 2 = 0

sec θ = 2

⇒ θ = 60o

Thus, θ = 0o or 60o

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