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

Here, we are providing the solutions for Chapter 18 Trigonometric Identities from ML Aggarwal Textbook for Class 10 ICSE Mathematics. Solutions of the eighteen chapter has been provided in detail. This will help the students in understanding the chapter more clearly. Class 10 Chapter 18 Trigonometric Identities of ML Aggarwal Solutions for ICSE is one of the most important chapter for the board exams which is based on finding all the trigonometric ratios from the given ratio, expressing trigonometric ratios in terms of other trigonometric ratios, finding all trigonometric ratios of the given triangle, evaluating the values using trigonometric table and solving questions.

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).

Answer

Given,

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).

Answer

Given,

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.

Answer

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.

Answer

Given, 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)

Answer

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,

(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):

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

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

Answer

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 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


7. (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°).

Answer

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[1×1×1] – 3

= 1 + 2 – 3

= 0


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

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

Answer

Given,

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

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

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

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

= 1/1

= 1

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

= 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


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)

Answer

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 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°) – (cos 28°cosec 62°)/(tan18° tan36° tan30° tan30° tan54° tan72°)

(ii) cosec2 (90 – θ) – tan2 θ)/(2(cos248° + cos242°) – (2tan230° sec252° sin238°)/(cosec270° - tan220°)

Answer

(i) 2(sec2 35° - cot2 55°) = (cos28° cosec62°)/(tan18° tan36° tan30° tan54° tan72°)

(ii) 


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)

Answer

(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.

(iv)


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.

Answer

Given,

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

L.H.S. = (sec A + tan A)(1 – sin A)

(ii) 


13. (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.


14. (i) 1/(1 + cos A) + 1/(1 – cos A) = 2cosec2A

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

Answer

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

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

(ii) 


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

Answer

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

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

(ii) 

(iii)


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) secA + 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)

(ii)

(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 = secA + 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

Answer

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

(ii)


18. (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)

(ii)

(iii)


19. (i) tan2θ – sin2θ = tan2 θ sin2 θ

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

Answer

(i) L.H.S = tan2 θ – sin2 θ


(ii) 

20. (i) cosec4θ – cosec2θ = 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 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.


21. (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

(ii)


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

(ii)= sin A/(1 + cos A)

Answer:

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

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

(ii)


23. (i)= tan A + sec A

(ii)= cosec A - cot A

Answer

(i)

(ii) 


24. (i)= 2 cosec A

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

Answer

(i)

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


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

Answer

(i)

(ii)


26. (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

(ii)


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

(ii) sin θ/(cot θ + cosec θ) = 2 + sin θ/(cot θ – cosec θ)

Answer

(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)

(ii) 


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

Answer

(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)

(ii) 


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

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

Answer

(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) + (sin3 A – cos3 A)/(sin A – cos A)


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


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)

Answer

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

(ii)

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


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

Answer

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

sin θ + cos θ = √2 cos θ

On dividing by sin θ, we have


32. If 7 sin2θ + 3 cos2θ = 4, 0° ≤ θ ≤ 90°, then find the value of θ.

Answer

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.

Answer

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.

Answer

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

= 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) – b2 (sec2θ – tan2θ)

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

= a2 – b2



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

Answer

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

= a(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. 

Answer

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

(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


2. Evaluate the following: 

(i) 2 × (cos2 20° cos2 70°)/(sin2 25° + sin2 65°) – tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°

(ii) (sin222° + sin268°)/(cos222° + cos268°) + sin263° + cos63° sin27°

Answer

(i) 2 × (cos2 20° cos2 70°)/(sin2 25° + sin2 65°) – tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°

(ii)

(iii)



3. If 4/3 (sec259o– cot2 31o) – 2/3 sin 90o + 3 tan2 56o tan2 34o = x/2, then find the value of x.

Answer

Given,

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



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

Answer

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

(ii)



5. (i) (cos θ – sin θ)(1 + tan θ)/2 cos2 θ – 1 = sec θ

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

Answer

(i) (cos θ – sin θ)(1 + tan θ)/2cos2 θ – 1 = sec θ

(ii)


6. (i) sin2θ + cos4θ = cos2 θ + sin4 θ 

(ii) cot θ/(cosec θ + 1) + (cosec θ + 1)/cot θ = 2 sec θ

Answer

Given,

(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)



7. (i) sec4A (1 – sin4A) – 2 tan2 A = 1 

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

Answer

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

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

= 1/cos4 A (1 + sin2 A)(1 – sin2 A) – 2 tan2 A

[∵ a2 – b2 = (a + b)(a – b)]

(ii)



8. (i) (sin3 θ + cos3 θ)/sin θ cos θ + sin θ cos θ = 1

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

Answer

(i) (sin3 θ + cos3 θ)/sinθ cosθ + sinθ cosθ = 1


(ii) 

9. (i) cos A/(1 – tan A) – sin2 A/(cos A – sin A)= sin A + cos A

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

(iii) tan2 θ/(tan2 θ – 1) – cosec2 θ/(sec2 θ – cosec2 θ) = 1/(sin2 θ – cos2 θ)

Answer

(i) cos A/(1 – tan A) – sin2 A/(cos A – sin A) = sin A + cos A

L.H.S. = cos A/(1 – tan A) – sin2 A/(cos A – sin A)

(ii)


(iii)



10. (sin A + cos A)/(sin A – cos A) + (sin A – cos A)/(sin A + cos A) = 2/(sin2A – cos2 A = 2/(1 – 2 cos2 A) = 2 sec2 A/(tan2 A – 1)

Answer

(sin A + cos A)/(sin A – cos A) + (sin A – cos A)/(sin A + cos A) = 2/(sin2A – cos2 A = 2/(1 – 2 cos2 A) = 2 sec2 A/(tan2 A – 1)

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


11. 2 (sin6θ + cos6θ) – 3 (sin4 θ + cos4 θ) + 1 = 0

Answer

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.

Answer

Given,

cot θ + cos θ = m …(i)

cot θ – cos θ = n …(ii)

Adding (i) and (ii), we get


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

Answer

Given, sec θ + tan θ = p

Prove that sin θ = (p2 – 1)/(p2 + 1)


14. If tan A = n tan B and sin A = m sin B, prove that cos2A = (m2– 1)/ (n2 – 1)

Answer

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

Answer

Given, sec A = x + 1/4x

We know that,


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).

Answer

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) = 0

⇒ (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 = 0or 60o


Multiple Choice Questions


Choose the correct answer from the given four options (1 to 12)

1. cot2 θ – 1/sin2 θ is equal to

(a) 1

(b) -1

(c) sin2 θ

(d) sec2 θ

Answer

(b) -1

cot2 θ – 1/sin2 θ

= cos2 θ/sin2 θ – 1/sin2 θ

= (cos2 θ – 1)/sin2 θ

= - sin2 θ/sin2 θ

= - 1


2. (sec2 θ – 1)(1 – cosec2 θ) is equal to

(a) -1

(b) 1

(c) 0

(d) 2

Answer

(a) -1

(sec2 θ – 1)(1 – cosec2 θ)

= (1/cos2 θ – 1)(1 – 1/sin2 θ)

= (1 – cos2 θ)/cos2 θ × (sin2 θ – 1)/sin2 θ

= (- sin2 θ cos2 θ)/(sin2 θ cos2 θ) = - 1

(∵ sin2 θ + cos2 θ = 1)


3. tan2 θ/(1 + tan2 θ) is equal to

(a) 2 sin2 θ

(b) 2 cos2 θ

(c) sin2 θ

(d) cos2 θ

Answer

(c)  sin2 θ

tan2 θ/(1 + tan2 θ)

tan2 θ/(1 + tan2 θ) = (sin2 θ/cos2 θ)/(1 + sin2 θ/cos2 θ)

= (sin2 θ/cos2 θ)/(cos2 θ + sin2 θ)/cos2 θ

= (sin2 θ/cos2 θ) × cos2 θ/(sin2 θ + cos2 θ)

(∵ sin2 θ + cos2 θ = 1)

= sin2 θ/1

= sin2 θ 


4. (cos θ + sin θ)2 + (cos θ – sin θ)2 is equal to

(a) -2

(b) 0

(c) 1

(d) 2

Answer

(d) 2

(cos θ + sin θ)2 + (cos θ – sin θ)2

= cos2 θ + sin2 θ + 2 sin θ cos θ + cos2 θ + sin2 θ – 2 sin θ cos θ

= 2(sin2 θ + cos2 θ)

= 2 × 1

= 2

(∵ sin2 θ + cos2 θ = 1)


5. (sec A + tan A)(1 – sin A) is equal to

(a) sec A

(b) sin A

(c) cosec A

(d) cos A

Answer

(d) cos A

(sec A + tan A)(1 – sin A)

= (1/cos A + sin A/cos A) (1 – sin A)

= (1 + sin A)/cos A × 1 – sin A

= {(1 + sin A)(1 – sin A)}/cos A

= (1 – sin2 A)/cos A

= cos2 A/cos A

= cos A


6. (1 + tan2 A)/(1 + cot2 A) is equal to

(a) sec2 A

(b) -1

(c) cot2 A

(d) tan2 A

Answer

(d) tan2 A

(1 + tan2 A)/(1 + cot2 A)

(1 + tan2 A)/(1 + cot2 A) = (1 + sin2 A/cos2 A)/(1 + cos2 A/sin2 A)

= {(cos2 A + sin2 A)/cos2 A}/(sin2 A + cos2 A)/sin2 A}

= (1/cos2 A)/(1/sin2 A)

= 1/cos2 A × sin2 A/1

= sin2 A/cos2 A

= tan2 A


7. If sec θ – tan θ = k, then the value of sec θ + tan θ is

(a) 1 – 1/k

(b) 1 – k

(c) 1 + k

(d) 1/k

Answer

(d) 1/k

sec θ – tan θ = k

1/cos θ – sin θ/cos θ = k

(1 – sin θ)/cos θ = k

Squaring both sides, we get

(1 – sin θ)/cos θ = k

Squaring both sides, we get

{(1 – sin θ)/cos θ}2 = (k)2

⇒ (1 – sin θ)2/cos2 θ = k2

= (1 – sin θ)2/(1 – sin2 θ)

= k2

⇒ (1 – sin θ)2/(1 + sin θ)(1 – sin θ) = k2

= (1 – sin θ)/(1 + sin θ) 

= k2

⇒ (1 + sin θ)/(1 – sin θ) = 1/k2

(1 + sin θ)/cos θ = 1/k

= 1/cos θ + sin θ/cos θ

= 1/k

⇒ sec θ + tan θ = 1/k


8. Which of the following is true for all values of θ (0° < θ < 90°) :

(a) cos2 θ – sin2 θ = 1

(b) cosec2 θ – sec2 θ = 1

(c) sec2 θ – tan2 θ = 1

(d) cot2 θ – tan2 θ = 1

Answer

(c) sec2 θ – tan2 θ = 1

∴ sec2 θ – tan2 θ = 1 is true for all values of θ as it is an identity.

(0° < θ < 90°)


9. If θ is an acute angle of a right triangle, then the value of sin θ cos (90° - θ) + cos θ sin (90° – θ) is

(a) 0

(b) 2 sin θ cos θ

(c) 1

(d) 2 sin2 θ

Answer

(c) 1

sin θ cos (90° - θ) + cos θ sin (90° - θ)

= sin θ sin θ + cos θ cos θ

{∵ sin(90° - θ) = cos θ, cos (90° - θ) = sin θ}

= sin2 θ + cos2 θ

= 1


10. The value of cos 65° sin 25° + sin 65° cos 25° is

(a) 0

(b) 1

(c) 2

(d) 4

Answer

(b) 1

cos 65° sin 25° + sin 65°cos 25°

= cos (90° - 25°) sin 25° + sin (90° - 25°) cos 25°

= sin 25°. sin 25° + cos 25°. cos 25°

∴ sin2 25° + cos2 25°

(∵ sin2 θ + cos2 θ = 1)

= 1


11. The value of 3 tan2 26° - 3 cosec2 64° is

(a) 0

(b) 3

(c) -3

(d) -1

Answer

(c) -3

3 tan2 26° - 3 cosec2 64°

= 3 tan2 26° - 3 cosec (90° - 26°)

= 3 tan2 26° - 3 sec2 26°

= 3(tan2 26° - sec2 26°) {∵ sec2 θ – tan2 θ = 1}

= 3 × (-1)

= -3


12. The value of sin {(90° - θ)sin θ}/tan θ – 1 is

(a) -cot θ

(b) -sin2 θ

(c) -cos2 θ

(d) -cosec2 θ

Answer

(b) -sin2 θ

{(sin(90° - θ)sin θ}/tan θ – 1

= (cosθ sinθ)/(sinθ)/cos θ – 1

= (sin θ cos θ × cos θ)/(sin θ) – 1

= cos2 θ – 1

= -(1 – cos2 θ)

= - sin2 θ


The solutions provided for Chapter 18 Trigonometric Identities of ML Aggarwal Textbook. This solutions of ML Aggarwal Textbook of Chapter 18 Trigonometric Identities contains answers to all the exercises given in the chapter. These solutions are very important if you are a student of ICSE boards studying in Class 10. While preparing the solutions, we kept this in our mind that these should based on the latest syllabus given by ICSE Board.

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