# 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. Itâ€™s highly suggested that students use these solutions to solve problems as itâ€™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 BYJUâ€™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.

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

Exercise 18

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

Solution:

Given,

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

So, in âˆ†ABC we have âˆ B = 90o

And,

AC = 5 and BC = 3

By Pythagoras theorem,

AB = âˆš(AC2 â€“ BC2)

= âˆš(52 – 32) = âˆš(25 – 9) = âˆš16

= 4

Now,

cos A = AB/AC = 4/5

tan A = BC/AB = 3/4

cot A = 1/tan Î¸ = 4/3

sec A = 1/cos Î¸ = 5/4

cosec A = 1/sin Î¸ = 5/3

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

Solution:

Given,

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

So, in âˆ† ABC we have âˆ B = 90o

And,

AC = 17 and AB = 8

By Pythagoras theorem,

BC = âˆš(AC2 â€“ AB2)

= âˆš(172 – 82) = âˆš(289 – 64) = âˆš225

= 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 = âˆš(1 â€“ sin2 A)

tan A = sin A/cos A = sin A/ âˆš(1 â€“ sin2 A)

sec A = 1/cos A = 1/ (âˆš1 â€“ sin2 A)

4. If tan A =Â 1/âˆš3, find all other trigonometric ratios of angle A.

Solution:

Given, tan A = 1/âˆš3

In right âˆ† ABC,

tan A = BC/AB = 1/âˆš3

So,

BC = 1 and AB = âˆš3

By Pythagoras theorem,

AC = âˆš(AB2 + BC2) = âˆš[(âˆš3)2Â + (1)2]

= âˆš(3 + 1) = âˆš4 = 2

Hence,

sin A = BC/AC = Â½

cos A = AB/AC = âˆš3/2

cot A = 1/tan A = âˆš3

sec A = 1/cos A = 2/âˆš3

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

5. If 12 cosec Î¸ = 13, find the value of (2 sin Î¸ â€“ 3 cos Î¸)/ (4 sin Î¸ â€“ 9 cos Î¸)

Solution:

Given,

12 cosec Î¸ = 13

â‡’ cosec Î¸ = 13/12

In right âˆ† ABC,

âˆ A = Î¸

So, cosec Î¸ = AC/BC = 13/12

AC = 13 and BC = 12

By Pythagoras theorem,

AB = âˆš(AC2 â€“ BC2)

= âˆš[(13)2Â – (12)2]

= âˆš(169 – 144)

= âˆš25

= 5

Now,

sin Î¸ = BC/AC = 12/13

cos Î¸ = AB/AC = 5/13

Hence,

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 â€“ 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) â€“ sin 28o sec 62o + cosec2 30o

(ii) (sin 29o/ cosec 61o) + 2 cot 8Â° cot 17Â° cot 45Â° cot 73Â° cot 82Â° â€“ 3(sinÂ² 38Â° + sinÂ² 52Â°).

Solution:

Given,

(i) (sin 65o/ cos 25o) + (cos 32o/sin 58o) â€“ sin 28o sec 62o + cosec2 30o

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

= (sin 65o/sin 65o)Â + (cos 32o/cos 32o) â€“ [sin 28o x cosec 28o] + 4

= 1 + 1 â€“ 1 + 4

= 5

(ii) (sin 29o/ cosec 61o) + 2 cot 8Â° cot 17Â° cot 45Â° cot 73Â° cot 82Â° â€“ 3(sinÂ² 38Â° + sinÂ² 52Â°).

= (sin 29o/ cosec (90o – 29o)) + [2 cot 8Â° cot 17Â° cot 45Â° cot (90Â° – 17o) cot (90o – 8o)] â€“ 3(sinÂ² 38Â° + sinÂ² (90Â° – 38o))

= (sin 29o/ sin 29o) + [2 cot 8Â° cot 17Â° cot 45Â° tan 17o tan 8o] â€“ 3(sinÂ² 38Â° + cosÂ² 38Â°)

= 1 + 2[(cot 8Â° tan 8o) (cot 17Â° tan 17o) cot 45Â°] â€“ 3(1)

= 1 + 2[1 x 1 x 1] â€“ 3

= 1 + 2 â€“ 3

= 0

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

(ii) sin2 34o + sin2 56o + 2 tan18o tan 72o â€“ cot2 30o

Solution:

Given,

(i) (sin 35o cos 55o + cos 35o sin 55 o)/ (cosec2 10o â€“ tan2 80 o)

(ii) sin2 34o + sin2 56o + 2 tan18o tan 72o â€“ cot2 30o

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

= [sin2 34o + cos2 34o] + 2 tan18o cot 18o â€“ cot2 30o

= 1 + 2 x 1 – (âˆš3)2

= 1 + 2 â€“ 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 Î¸ sec (90o â€“ Î¸) â€“ cot Î¸ tan (90o – Î¸)

Solution:

Given,

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

(ii) (cos2 25o + cos2 65o) + cosec Î¸ sec (90o â€“ Î¸) â€“ cot Î¸ tan (90o – Î¸)

= cos2 25o + cos2 (90o â€“ 25o) + cosec Î¸ sec (90o â€“ Î¸) â€“ cot Î¸. cot Î¸

= (cos2 25o + sin2 25o) + (cosec2 Î¸ â€“ cot2 Î¸)

= 1 + 1 = 2

10. (i) 2(secÂ² 35Â° â€“ cotÂ² 55Â°) â€“Â

(ii)

Solution:

Given,

11. Prove that following:Â

(i) cos Î¸ sin (90Â° â€“ Î¸) + sin Î¸ cos (90Â° â€“ Î¸) = 1

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

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

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

Solution:

(i) L.H.S. = cos Î¸ sin (90Â° â€“ Î¸) + sin Î¸ cos (90Â° â€“ Î¸)

= cos Î¸ x cos Î¸ + sin Î¸ x sin Î¸

= cos2 Î¸ + sin2 Î¸

= 1 = R.H.S.

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

= tan Î¸/ cot Î¸ + cos Î¸/ cos Î¸

= tan Î¸/ (1/tan Î¸) + 1

= tan2 Î¸ + 1 = sec2 Î¸ = R.H.S.

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

= (sin Î¸ cos Î¸)/ tan Î¸ + sin2 Î¸

= (sin Î¸ cos Î¸)/ (sin Î¸/ cos Î¸) + sin2 Î¸

= cos2 Î¸ + sin2 Î¸

= 1 = R.H.S.

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 â€“ sin A) = cos AÂ

(ii) (1 + tan2Â A) (1 â€“ sin A) (1 + sin A) = 1.

Solution:

Given,

(i) L.H.S = (sec A + tan A) (1 â€“ sin A)

(ii) L.H.S. = (1 + tan2Â A) (1 â€“ sin A) (1 + sin A)

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

(ii) (1 â€“ 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 â€“ cos A) (1 + sec A)

14. (i) 1/ (1 + cos A) + 1/ (1 â€“ cos A) = 2 cosec2 A

(ii) 1/(sec A + tan A) + 1/(sec A â€“ tan A) = 2 sec A

Solution:

(i) L.H.S. = 1/ (1 + cos A) + 1/ (1 â€“ cos A)

(ii)

15. (i) sin A/ (1 + cos A) = (1 â€“ cos A)/ sin A

(ii) (1 â€“ tan2 A)/ (cot2 A – 1) = tan2 A

(iii) sin A/ (1 + cos A) = cosec A â€“ cot A

Solution:

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

On multiplying and dividing by (1 â€“ cos A), we have

16. (i) (sec A â€“ 1)/(sec A + 1) = (1 â€“ cos A)/(1 + cos A)

(ii) tan2 Î¸/ (sec Î¸ â€“ 1)2 = (1 + cos Î¸)/ (1 â€“ cos Î¸)

(iii) (1 + tan A)2 + (1 â€“ tan A)2 = 2 sec2 A

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

Solution:

(iii) L.H.S. = (1 + tan A)2 + (1 â€“ tan A)2

= 1 + 2 tan A + tan2 A + 1 â€“ 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)

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

(ii) cot A â€“ tan A = (2cos2Â A – 1)/ (sin A â€“ cos A)

(iii) (cot A â€“ 1)/ (2 â€“ sec2 A) = cot A/ (1 + tan A)

Solution:

(i) L.H.S. =

19. (i) tan2 Î¸ â€“ sin2 Î¸ = tan2 Î¸ sin2 Î¸

(ii) cos Î¸/ (1 â€“ tan Î¸) â€“ sin2 Î¸/ (cos Î¸ – sin Î¸) = cos Î¸ + sin Î¸

Solution:

(i) L.H.S = tan2 Î¸ â€“ sin2 Î¸

20. (i) cosec4Â Î¸ â€“ cosec2Â Î¸ = cot4Â Î¸ + cot2Â Î¸Â

(ii) 2 sec2Â Î¸ â€“ sec4Â Î¸ â€“ 2 cosec2Â Î¸ + cosec4Â Î¸ = cot4Â Î¸ â€“ tan4Â Î¸.

Solution:

(i) L.H.S. = cosec4Â Î¸ â€“ cosec2Â Î¸

= cosec2Â Î¸ (cosec2Â Î¸ – 1)

= cosec2Â Î¸ cot2Â Î¸ [cosec2Â Î¸ â€“ 1 = cot2Â Î¸]

= (cot2Â Î¸ + 1) cot2Â Î¸

= cot4Â Î¸ + cot2Â Î¸

= R.H.S.

(ii) L.H.S. = 2 sec2Â Î¸ â€“ sec4Â Î¸ â€“ 2 cosec2Â Î¸ + cosec4Â Î¸

= 2 (tan2Â Î¸ + 1) â€“ (tan2Â Î¸ + 1)2 â€“ 2 (1 + cot2Â Î¸) + (1 + cot2Â Î¸)2

= 2 tan2 Î¸ + 2 â€“ (tan4 Î¸ + 2 tan2 Î¸ + 1) â€“ 2 â€“ 2 cot2 Î¸ + (1 + 2 cot2 Î¸ + cot4 Î¸)

= 2 tan2 Î¸ + 2 â€“ tan4 Î¸ â€“ 2 tan2 Î¸ â€“ 1 â€“ 2 â€“ 2 cot2 Î¸ + 1 + 2 cot2 Î¸ + cot4 Î¸

= cot4 Î¸ â€“ tan4 Î¸ = R.H.S.

21. (i) = cot Î¸

(ii) (tan3 Î¸ â€“ 1)/ (tan Î¸ – 1) = sec2 Î¸ + tan Î¸

Solution:

22. (i) (1 + cosec A)/ cosec A = cos2 A/ (1 â€“ sin A)

(ii)

Solution:

23. (i) = tan A + sec A

(ii) = cosec A â€“ cot A

Solution:

24. (i) = 2 cosec A

(ii) cos A cot A/ (1 â€“ sin A) = 1 + cosec A

Solution:

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

(ii) sec4 A â€“ tan4 A = 1 + 2 tan2 A

Solution:

(ii) sec4 A â€“ tan4 A = 1 + 2 tan2 A

L.H.S. = sec4 A â€“ tan4 A

= (sec2 A â€“ tan2 A) (sec2 A + tan2 A)

= (1 + tan4 A â€“ 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) cosec6Â A â€“ cot6Â A = 3 cot2Â A cosec2Â A + 1Â

(ii) sec6Â A â€“ tan6Â A = 1 + 3 tan2Â A + 3 tan4Â A

Solution:

(i) cosec6Â A â€“ cot6Â A = 3 cot2Â A cosec2Â A + 1

27.

Solution:

(i)

= (cos Î¸ – 1) (2 â€“ 1 â€“ cos Î¸)/ (cos Î¸ – 1)

= 1 â€“ cos Î¸

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

28. (i) (sin Î¸ + cos Î¸) (sec Î¸ + cosec Î¸) = 2 + sec Î¸ cosec Î¸Â

(ii) (cosec A â€“ sin A) (sec A â€“ cos A) sec2A = tan A

Solution:

(i) (sin Î¸ + cos Î¸) (sec Î¸ + cosec Î¸) = 2 + sec Î¸ cosec Î¸

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

= R.H.S.

(ii)

29.

Solution:

30. (i) 1/ (sec A + tan A) â€“ 1/cos A = 1/cos A â€“ 1/(sec A â€“ 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 â€“ sin A) = (sec A + 1)/ (sec A â€“ 1)

Solution:

31. If sin Î¸ + cos Î¸ = âˆš2 sin (90Â° â€“ Î¸), show that cot Î¸ = âˆš2 + 1

Solution:

Given, sin Î¸ + cos Î¸ = âˆš2 sin (90Â° â€“ Î¸)

sin Î¸ + cos Î¸ = âˆš2 cos Î¸

On dividing by sin Î¸, we have

1 + cot Î¸ = âˆš2 cot Î¸

1 = âˆš2 cot Î¸ – cot Î¸

(âˆš2 – 1) cot Î¸ = 1

cot Î¸ = 1/ (âˆš2 â€“ 1)

Hence, cot Î¸ = âˆš2 + 1

32. If 7 sin2Â Î¸ + 3 cos2Â Î¸ = 4, 0Â° â‰¤ Î¸ â‰¤ 90Â°, then find the value of Î¸.

Solution:

Given,

7 sin2Â Î¸ + 3 cos2Â Î¸ = 4, 0Â° â‰¤ Î¸ â‰¤ 90Â°

3 sin2Â Î¸ + 3 cos2Â Î¸ + 4 sin2Â Î¸ = 4

3 (sin2Â Î¸ + 3 cos2Â Î¸) + 4 sin2Â Î¸ = 4

3 (1) + 4 sin2Â Î¸ = 4

4 sin2Â Î¸ = 4 â€“ 3

sin2Â Î¸ = Â¼

Taking square-root on both sides, we get

sinÂ Î¸ = Â½

Thus, Î¸ = 30o

33. If sec Î¸ + tan Î¸ = m and sec Î¸ â€“ tan Î¸ = n, prove that mn = 1.

Solution:

Given,

sec Î¸ + tan Î¸ = m

sec Î¸ â€“ tan Î¸ = n

Now,

mn = (sec Î¸ + tan Î¸) (sec Î¸ â€“ tan Î¸)

= sec2 Î¸ â€“ tan2 Î¸ = 1

Thus, mn = 1

34. If x â€“ a sec Î¸ + b tan Î¸ and y = a tan Î¸ + b sec Î¸, prove that x2Â â€“ y2Â = a2Â â€“ b2.

Solution:

Given,

x =
a sec Î¸ + b tan Î¸,

y = a tan Î¸ + b sec Î¸

Now,

x2Â â€“ y2Â = (a sec Î¸ + b tan Î¸)2 â€“ (a tan Î¸ + b sec Î¸)2

– Hence proved.

35. If x = h + a cos Î¸ and y = k + a sin Î¸, prove that (x â€“ h)2Â + (y â€“ k)2Â = a2.

Solution:

Given,

x = h + a cos Î¸

y = k + a sin Î¸

Now,

x â€“ h = a cos Î¸

y â€“ k = a sin Î¸

On squaring and adding we get

(x â€“ h)2 + (y â€“ k)2 = a2 cos2 Î¸ + a2 sin2 Î¸

= a2 (sin2 Î¸ + cos2 Î¸)

= a2 (1) [Since, sin2 Î¸ + cos2 Î¸ = 1]

– Hence proved

Chapter Test

1. (i) If Î¸ is an acute angle and cosec Î¸ = âˆš5, find the value of cot Î¸ â€“ cos Î¸.Â

(ii) If Î¸ is an acute angle and tan Î¸ =Â 8/15, find the value of sec Î¸ + cosec Î¸.Â

Solution:

Given, Î¸ is an acute angle and cosec Î¸ = âˆš5

So,

sin Î¸ = 1/âˆš5

And, cos Î¸ = âˆš(1 â€“ sin2 Î¸)

cos Î¸ = âˆš(1 â€“ (1/âˆš5)2)

= âˆš(1 â€“ (1/5))

= âˆš(4/5)

cos Î¸ = 2/âˆš5

Now,

cot Î¸ â€“ cos Î¸ = (cos Î¸/sin Î¸) â€“ cos Î¸

(ii) Given, Î¸ is an acute angle and tan Î¸ =Â 8/15

In fig. we have

tan Î¸ = BC/AB = 8/15

So, BC = 8 and AB = 15

By Pythagoras theorem, we have

AC = âˆš(AB2 + BC2) = âˆš(52 + 82) = âˆš(25 + 64) = âˆš289

â‡’ AC = 17

Now,

sec Î¸ = AC/AB = 17/15

cosec Î¸ = AC/BC = 17/8

So,

sec Î¸ + cosec Î¸ = 17/15 + 17/8

= (136 + 255)/ 120

= 391/120

=

2. Evaluate the following:Â

(i) 2 x () â€“ tan 45o + tan 13o tan 23o tan 30o tan 67o tan 77o

(ii) + sin2 63o + cos 63o sin 27o

Solution:

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

Solution:

Given,

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

4. (i) cos A/ (1 â€“ 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:

5. (i)

(ii) (cosec Î¸ â€“ sin Î¸) (sec Î¸ â€“ cos Î¸) (tan Î¸ + cot Î¸) = 1.

Solution:

6. (i) sin2Â Î¸ + cos4Â Î¸ = cos2Â Î¸ + sin4Â Î¸Â

(ii)

Solution:

Given,

(i) sin2Â Î¸ + cos4Â Î¸ = cos2Â Î¸ + sin4Â Î¸

L.H.S. = sin2Â Î¸ + cos4Â Î¸

= (1 – cos2Â Î¸) + cos4Â Î¸

= cos4Â Î¸ â€“ cos2Â Î¸ + 1

= cos2Â Î¸ (cos2Â Î¸ – 1) + 1

= cos2Â Î¸ (- sin2Â Î¸) + 1

= 1 – sin2Â Î¸ cos2Â Î¸

Now,

R.H.S. = cos2Â Î¸ + sin4Â Î¸

= (1 – sin2Â Î¸) + sin4Â Î¸

= sin4Â Î¸ – sin2Â Î¸ + 1

= sin2Â Î¸ (sin2Â Î¸ – 1) + 1

= sin2Â Î¸ (-cos2Â Î¸) + 1

= 1 – sin2Â Î¸ cos2Â Î¸

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

7. (i) sec4Â A (1 â€“ sin4Â A) â€“ 2 tan2Â A = 1Â

(ii)Â  = sec A + cosec A

Solution:

(i) sec4Â A (1 â€“ sin4Â A) â€“ 2 tan2Â A = 1

L.H.S. = sec4Â A (1 â€“ sin4Â A) â€“ 2 tan2Â A

8. (i) + sin Î¸ cos Î¸ = 1

(ii) (sec A â€“ tan A)2Â (1 + sin A) = 1 â€“ sin A.

Solution:

9. (i) = sin A + cos A

(ii) (sec A â€“ cosec A) (1 + tan A + cot A) = tan A sec A â€“ cot A cosec AÂ

(iii)

Solution:

10.

Solution:

11. 2 (sin6Â Î¸ + cos6Â Î¸) â€“ 3 (sin4Â Î¸ + cos4Â Î¸) + 1 = 0

Solution:

Given,

2 (sin6Â Î¸ + cos6Â Î¸) â€“ 3 (sin4Â Î¸ + cos4Â Î¸) + 1 = 0

L.H.S. = 2 (sin6Â Î¸ + cos6Â Î¸) â€“ 3 (sin4Â Î¸ + cos4Â Î¸) + 1

= 2 [(sin2Â Î¸)3 + (cos2Â Î¸)3] â€“ 3 (sin4Â Î¸ + cos4Â Î¸) + 1

= 2 [(sin2Â Î¸ + cos2Â Î¸) (sin4Â Î¸ + cos4Â Î¸ – sin2Â Î¸ cos2Â Î¸)] â€“ 3 (sin4Â Î¸ + cos4Â Î¸) + 1

= 2 (sin4Â Î¸ + cos4Â Î¸ – sin2Â Î¸ cos2Â Î¸) – 3 (sin4Â Î¸ + cos4Â Î¸) + 1

= 2 sin4Â Î¸ + 2 cos4Â Î¸ â€“ 2 sin2Â Î¸ cos2Â Î¸ – 3 sin4Â Î¸ â€“ 3 cos4Â Î¸ + 1

= 1 â€“ sin4Â Î¸ â€“ cos4Â Î¸ â€“ 2 sin2Â Î¸ cos2Â Î¸

= 1 â€“ [sin4Â Î¸ + cos4Â Î¸ + 2 sin2Â Î¸ cos2Â Î¸]

= 1 â€“ 1

= 0 = R.H.S.

12. If cot Î¸ + cos Î¸ = m, cot Î¸ â€“ cos Î¸ = n, then prove that (m2Â â€“ n2)2Â = 16.

Solution:

Given,

cot Î¸ + cos Î¸ = m â€¦ (i)

cot Î¸ â€“ cos Î¸ = n â€¦ (ii)

Adding (i) and (ii), we get

13. If sec Î¸ + tan Î¸ = p, prove that sin Î¸ = (p2 â€“ 1)/ (p2 + 1)

Solution:

Given, sec Î¸ + tan Î¸ = p

14. If tan A = n tan B and sin A = m sin B, prove that cos2Â A = (m2 â€“ 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

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,

16. When 0Â° < Î¸ < 90Â°, solve the following equations:Â

(i) 2 cos2Â Î¸ + sin Î¸ â€“ 2 = 0Â

(ii) 3 cos Î¸ = 2 sin2Â Î¸Â

(iii) sec2Â Î¸ â€“ 2 tan Î¸ = 0Â

(iv) tan2Â Î¸ = 3 (sec Î¸ â€“ 1).

Solution:

Given, 0Â° < Î¸ < 90Â°

(i) 2 cos2Â Î¸ + sin Î¸ â€“ 2 = 0

2 (1 – sin2Â Î¸) + sinÂ Î¸ â€“ 2 = 0

2 â€“ 2 sin2Â Î¸ + sinÂ Î¸ â€“ 2 = 0

â€“ 2 sin2Â Î¸ + sinÂ Î¸ = 0

sin Î¸ (1 â€“ 2 sin Î¸) = 0

So, either sin Î¸ = 0 or 1 â€“ 2 sin Î¸ = 0

If sin Î¸ = 0

â‡’ Î¸ = 0o

And, if 1 â€“ 2 sin Î¸ = 0

sin Î¸ = Â½

â‡’ Î¸ = 30o

Thus, Î¸ = 0o or 30o

(ii) 3 cos Î¸ = 2 sin2Â Î¸

3 cos Î¸ = 2 (1 – cos2Â Î¸)

3 cos Î¸ = 2 â€“ 2 cos2 Î¸

2 cos2 Î¸ + 3 cos Î¸ â€“ 2 = 0

2 cos2 Î¸ + 4 cos Î¸ â€“ cos Î¸ â€“ 2 = 0

2 cos Î¸ (cos Î¸ + 2) â€“ 1(cos Î¸ + 2)

(2 cos Î¸ â€“ 1) (cos Î¸ + 2) = 0

So, either 2 cos Î¸ â€“ 1 = 0 or cos Î¸ + 2 = 0

If 2 cos Î¸ â€“ 1 = 0

cos Î¸ = Â½

â‡’ Î¸ = 60o

And, for cos Î¸ + 2 = 0

â‡’ cos Î¸ = -2 which is not possible being out of range.

Thus, Î¸ = 60o

(iii) sec2Â Î¸ â€“ 2 tan Î¸ = 0

(1 + tan2Â Î¸) â€“ 2 tan Î¸ = 0

tan2Â Î¸ â€“ 2 tanÂ Î¸ + 1 = 0

(tanÂ Î¸ – 1)2 = 0

tanÂ Î¸ â€“ 1 = 0

â‡’ tan Î¸ = 1

Thus, Î¸ = 45o

(iv) tan2Â Î¸ = 3 (sec Î¸ â€“ 1)

(sec2 Î¸ â€“ 1) = 3 sec Î¸ â€“ 3

sec2 Î¸ â€“ 1 â€“ 3 sec Î¸ + 3 = 0

sec2 Î¸ â€“ 3 sec Î¸ + 2 = 0

sec2 Î¸ â€“ 2 sec Î¸ â€“ sec Î¸ + 2 = 0

sec Î¸ (sec Î¸ â€“ 2) â€“ 1 (sec Î¸ = 2) = 0

(sec Î¸ – 1) (sec Î¸ – 2) = 0

So, either sec Î¸ â€“ 1 = 0 or sec Î¸ â€“ 2 = 0

If sec Î¸ â€“ 1 = 0

sec Î¸ = 1

â‡’ Î¸ = 0o

And, if sec Î¸ â€“ 2 = 0

sec Î¸ = 2

â‡’ Î¸ = 60o

Thus, Î¸ = 0o or 60o