# SELINA Solutions for Class 10 Maths Chapter 21 - Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables)

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Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

To prove:

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42

Solution 43

Solution 44

Solution 45

Solution 46

Solution 47

Solution 48

Solution 49

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

Solution 1

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

Solution 2

Solution 3

Solution 4

Solution 5

Given:

and

Solution 6

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)

Solution 1

(i)

(ii)

(iii)

Solution 2

Solution 3

(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

Solution 5

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

Solution 6

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

So, A + C = 90

Solution 7

(i)

Hence, x =

(ii)

Hence, x =

(iii)

Hence, x =

(iv)

Hence, x =

(v)

Hence, x =

(vi)

Hence, x =

(vii)

Hence,

Solution 8

(i)

(ii)

Solution 9

(i)

(ii)

Solution 10

Solution 11

Solution 12

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

Solution 1

(i) sin 21o = 0.3584

(ii) sin 34o 42'= 0.5693

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

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

(v) sin (10o 20' + 2045') = sin 30o65' = sin 31o5' = 0.5150 + 0.0012 = 0.5162

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 = 0.9042

Solution 3

(i) tan 37= 0.7536

(ii) tan 4218' = 0.9099

(iii) tan 17o  27' = tan (1724' + 3') = 0.3134 + 0.0010 = 0.3144

Solution 4

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

Hence, = 29o

(ii) From the tables, it is clear that sin 2230' = 0.3827

Hence, = 2230'

(iii) From the tables, it is clear that sin 4042' = 0.6521

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

From the tables, diff of 2' = 0.0004

Hence, = 40o  42' + 2' = 4044'

Solution 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’

Solution 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)

Solution 1

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

(xiii)

(xiv)

(xv)

(xvi)

(xvii)

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

(i) 2 sinA - 1 = 0

(ii)

Solution 7

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

Solution 8

(i)

(ii)

(iii)

(iv)

(v)

Solution 9

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

(i)

(ii)

(iii)

= cosec2A [sec(90 - B)]2

= cosec2A cosec2B

(iv)

Solution 10

Solution 11

4 cos2A - 3 = 0

Solution 12

(i)

(ii) sin 3A - 1 = 0

(iii)

(iv)

(v)

Solution 13

(i)

(ii)

Solution 14

Solution 15

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°

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