ICSE Class 10 Trigonometrical Identities Solution By Clarify Knowledge

ICSE Class 10 Trigonometrical Identities Solution New Pattern 2022

Chapter 21 - Trigonometrical Identities Ex. 21(A)

Question 1

Prove:

Solution 1

Question 2

Prove:

Solution 2

Question 3

Prove:

Solution 3

Question 4

Prove:

Solution 4

Question 5

Prove:

Solution 5

Question 6

Prove:

Solution 6

Question 7

Prove:

Solution 7

Question 8

Prove:

Solution 8

Question 9

Prove:

Solution 9

Question 10

Prove:

Solution 10

Question 11

Prove:

Solution 11

Question 12

Prove:

Solution 12

Question 13

Prove:

Solution 13

Question 14

Prove:

Solution 14

Question 15

Prove:

Solution 15

Question 16

Prove:

Solution 16

Question 17

Prove:

Solution 17

Question 18

Prove:

Solution 18

Question 19

Prove:

Solution 19

Question 20

Prove:

Solution 20

Question 21

Prove:

Solution 21

Question 22

Prove:

Solution 22

Question 23

Prove:

Solution 23

Question 24

Prove:

Solution 24

Question 25

Prove:

Solution 25

Question 26

Prove:

Solution 26

Question 27

Prove:

Solution 27

Question 28

Prove:

Solution 28

Question 29

Prove:

Solution 29

Question 30

Prove:

Solution 30

Question 31

Prove:

Solution 31

Question 32

Prove:

Solution 32

Question 33

Prove:

Solution 33

Question 34

Solution 34

To prove: 

Question 35

Prove:

Solution 35

Question 36

Prove:

Solution 36

Question 37

Prove:

Solution 37

Question 38

Prove:Solution 38

Question 39

Prove:

Solution 39

Question 40

Prove:

Solution 40

Question 41

Prove:

Solution 41

Question 42

Prove:

Solution 42

Question 43

Prove:

Solution 43

Question 44

Prove:

Solution 44

Question 45

Prove:

Solution 45

Question 46

Prove:

Solution 46

Question 47

Prove:

Solution 47

Question 48

Prove:

Solution 48

Question 49

Prove:

Solution 49

Chapter 21 - Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) Exercise Ex. 21(B)

Question 1

Prove that:

(i) 

(ii) 

(iii) 

(iv) 

(v) 

(vi) 

(vii) 

(viii) 

(ix) Solution 1

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

Question 2

If = m and = n, then prove that x² + y² = m² + n².Solution 2

Question 3

If m= and n= , prove thatSolution 3

Question 4

If , prove that Solution 4

Question 5

If and , prove that Solution 5

Given:

and

Question 6

If , prove that Solution 6

Question 7

If  and , show that (m² + n²) cos²B = n².Solution 7

LHS = (m² + n²) cos²B

Hence, (m² + n²) cos²B = n².

Chapter 21 - Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) Exercise Ex. 21(C)

Question 1

Without using trigonometric tables, show that:

(i) 

(ii) 

(iii) Solution 1

(i)

(ii)

(iii)

Question 2

Express each of the following in terms of angles between 0°and 45°:

(i) sin 59°+ tan 63°

(ii) cosec 68°+ cot 72°

(iii)cos 74°+ sec 67°Solution 2

Question 3

Show that:

(i) 

(ii) Solution 3

(i) 

(ii) 

Question 4

For triangle ABC, show that:

(i) 

(ii) Solution 4

(i) We know that for a triangle ABC

  A + B + C = 180°

(ii) We know that for a triangle ABC

 A + B + C = 180°

B + C = 180° - A

Question 5

Evaluate:

(i) 

(ii) 

(iii) 

(iv) 

(v) 

(vi) 

(vii) 

(viii) 

(ix) Solution 5

(i)

(ii) 

(iii) 

(iv) 

(v) 

(vi) 

(vii) 

(viii) 

(ix) 

Question 6

A triangle ABC is right angled at B; find the value of Solution 6

Since, ABC is a right angled triangle, right angled at B.

So, A + C = 90

Question 7

Find (in each case, given below) the value of x if:

(i) 

(ii) 

(iii) 

(iv) 

(v) 

(vi) 

(vii) Solution 7

(i) 

Hence, x = 

(ii) 

Hence, x = 

(iii) 

Hence, x = 

(iv) 

Hence, x = 

(v) 

Hence, x = 

(vi) 

Hence, x = 

(vii) 

Hence, Question 8

In each case, given below, find the value of angle A, where 

(i) 

(ii) Solution 8

(i) 

(ii) 

Question 9

Prove that:

(i) 

(ii) Solution 9

(i)

(ii)

Question 10

Evaluate :

Solution 10

Question 11

Evaluate

sin² 34° + sin² 56° + 2tan 18° tan72° - cot² 30°Solution 11

Question 12

Without using trigonometrical tables, evaluate:

Solution 12

Chapter 21 - Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) Exercise Ex. 21(D)

Question 1

Use tables to find sine of:

(i) 21°

(ii) 34° 42'

(iii) 47° 32'

(iv) 62° 57'

(v) 10° 20' + 20° 45'Solution 1

(i) sin 21^o = 0.3584

(ii) sin 34^o 42'= 0.5693

(iii) sin 47^o 32'= sin (47^o 30' + 2') =0.7373 + 0.0004 = 0.7377

(iv) sin 62^o 57' = sin (62^o 54' + 3') = 0.8902 + 0.0004 = 0.8906

(v) sin (10^o 20' + 20^o 45') = sin 30^o65' = sin 31^o5' = 0.5150 + 0.0012 = 0.5162Question 2

Use tables to find cosine of:

(i) 2° 4’

(ii) 8° 12’

(iii) 26° 32’

(iv) 65° 41’

(v) 9° 23’ + 15° 54’Solution 2

(i) cos 2° 4’ = 0.9994 - 0.0001 = 0.9993

(ii) cos 8° 12’ = cos 0.9898

(iii) cos 26° 32’ = cos (26° 30’ + 2’) = 0.8949 - 0.0003 = 0.8946

(iv) cos 65° 41’ = cos (65° 36’ + 5’) = 0.4131 -0.0013 = 0.4118

(v) cos (9° 23’ + 15° 54’) = cos 24° 77’ = cos 25° 17’ = cos (25° 12’ + 5’) = 0.9048 - 0.0006 = 0.9042Question 3

Use trigonometrical tables to find tangent of:

(i) 37°

(ii) 42° 18'

(iii) 17° 27'Solution 3

(i) tan 37^o = 0.7536

(ii) tan 42^o 18' = 0.9099

(iii) tan 17^o  27' = tan (17^o 24' + 3') = 0.3134 + 0.0010 = 0.3144Question 4

Use tables to find the acute angle , if the value of sin  is:

(i) 0.4848

(ii) 0.3827

(iii) 0.6525Solution 4

(i) From the tables, it is clear that sin 29^o = 0.4848

Hence, = 29^o

(ii) From the tables, it is clear that sin 22^o 30' = 0.3827

Hence,  = 22^o 30'

(iii) From the tables, it is clear that sin 40^o 42' = 0.6521

sin  - sin 40^o 42' = 0.6525 -; 0.6521 = 0.0004

From the tables, diff of 2' = 0.0004

Hence,  = 40^o  42' + 2' = 40^o 44'Question 5

Use tables to find the acute angle , if the value of cos  is:

(i) 0.9848

(ii) 0.9574

(iii) 0.6885Solution 5

(i) From the tables, it is clear that cos 10° = 0.9848

Hence,  = 10°

(ii) From the tables, it is clear that cos 16° 48’ = 0.9573

cos  - cos 16° 48’ = 0.9574 - 0.9573 = 0.0001

From the tables, diff of 1’ = 0.0001

Hence,  = 16° 48’ - 1’ = 16° 47’

(iii) From the tables, it is clear that cos 46° 30’ = 0.6884

cos q - cos 46° 30’ = 0.6885 - 0.6884 = 0.0001

From the tables, diff of 1’ = 0.0002

Hence,  = 46° 30’ - 1’ = 46° 29’Question 6

Use tables to find the acute angle , if the value of tan q is:

(i) 0.2419

(ii) 0.4741

(iii) 0.7391Solution 6

(i) From the tables, it is clear that tan 13° 36’ = 0.2419

Hence,  = 13° 36’

(ii) From the tables, it is clear that tan 25° 18’ = 0.4727

tan  - tan 25° 18’ = 0.4741 - 0.4727 = 0.0014

From the tables, diff of 4’ = 0.0014

Hence,  = 25° 18’ + 4’ = 25° 22’

(iii) From the tables, it is clear that tan 36° 24’ = 0.7373

tan  - tan 36° 24’ = 0.7391 - 0.7373 = 0.0018

From the tables, diff of 4’ = 0.0018

Hence,  = 36° 24’ + 4’ = 36° 28’

Chapter 21 - Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) Exercise Ex. 21(E)

Question 1

Prove the following identities:

(i) 

(ii) 

(iii) 

(iv) 

(v) 

(vi) 

(vii) 

(viii) 

(ix) 

(x) 

(xi) 

(xii) 

(xiii) 

(xiv) 

(xv) 

(xvi) 

(xvii) Solution 1

(i) 

(ii)

(iii) 

(iv) 

(v) 

(vi) 

(vii) 

(viii) 

(ix) 

(x) 

(xi)

(xii) 

(xiii) 

(xiv) 

(xv) 

(xvi) 

(xvii) 

Question 2

If and , then prove that:

q(p² - 1) = 2pSolution 2

Question 3

If , show that:

Solution 3

Question 4

If , show that:

Solution 4

Question 5

If tan A = n tan B and sin A = m sin B, prove that:

Solution 5

Question 6

(i) If 2 sinA - 1 = 0, show that:

sin 3A = 3 sinA - 4 sin³A

(ii) If 4 cos²A - 3 = 0, show that:

cos 3A = 4 cos³A - 3 cosASolution 6

(i) 2 sinA - 1 = 0

(ii)

Question 7

Evaluate:

(i) 

(ii) 

(iii) 

(iv) 

(v) 

(vi) 

(vii) 

(viii) Solution 7

(i) 

(ii) 

(iii) 

(iv) 

(v) 

(vi) 

(vii) 

(viii) 

Question 8

Prove that:

(i) 

(ii) 

(iii) 

(iv) 

(v) Solution 8

(i) 

(ii) 

(iii) 

(iv) 

(v) 

Question 9

If A and B are complementary angles, prove that:

(i) 

(ii) 

(iii) cosec²A + cosec²B = cosec²A cosec²B

(iv) Solution 9

Since, A and B are complementary angles, A + B = 90°

(i)

(ii)

(iii)

= cosec²A [sec(90 - B)]²

= cosec²A cosec²B

(iv) 

Question 10

Prove that:

Solution 10

Question 11

If 4cos²A - 3 = 0 and 0°A 90°, then prove that:

(i) sin3A= 3 sinA - 4 sin³A

(ii) cos3A= 4 cos³A - 3 cosASolution 11

4 cos²A - 3 = 0

Question 12

Find A, if 0°A 90° and:

(i) 

(ii) sin 3A - 1 = 0

(iii) 

(iv) 

(v) Solution 12

(i) 

(ii) sin 3A - 1 = 0

(iii) 

(iv) 

(v) 

Question 13

If 0° < A < 90°; find A, if:

(i) 

(ii) Solution 13

(i) 

(ii) 

Question 14

Prove that:

(cosec A - sin A) (sec A - cos A) sec²A = tan ASolution 14

Question 15

Prove the identity (sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ.Solution 15

Question 16

Evaluate without using trigonometric tables,

sin² 28° + sin² 62° + tan² 38° - cot² 52° + sec² 30° Solution 16

sin² 28° + sin² 62° + tan² 38° - cot² 52° + sec² 30° 

= sin² 28° + [sin (90 - 28)°]² + tan² 38° - [cot(90 - 38)°]² + sec² 30° 

= sin² 28°  + cos² 28° + tan² 38° - tan² 38° + sec² 30°