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

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

### 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

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### 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

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### Solution 40

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### Solution 46

### Solution 47

### Solution 48

## 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 = (m^{2}
+ n^{2}) cos^{2}B

Hence, (m^{2} + n^{2})
cos^{2}B = n^{2}.

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

### 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 11

### Solution 1

(i)

(ii)

(iii)

### Solution 10

### Solution 12

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

### 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^{o}65' = sin 31^{o}5' = 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^{o }= 0.7536

(ii) tan 42^{o }18' = 0.9099

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

### Solution 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'

### 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’

## 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)

= cosec^{2}A [sec(90 - B)]^{2}

= cosec^{2}A cosec^{2}B

(iv)

### Solution 10

### Solution 11

4 cos^{2}A - 3 = 0

### Solution 12

(i)

(ii) sin 3A - 1 = 0

(iii)

(iv)

(v)

### Solution 13

(i)

(ii)

### Solution 14

### Solution 15

### Solution 16

sin^{2}
28° + sin^{2}
62° + tan^{2}
38° - cot^{2}
52° + sec^{2} 30°^{ }

= sin^{2}
28° + [sin (90 -
28)°]^{2} +
tan^{2} 38° - [cot(90 - 38)°]^{2} +
sec^{2} 30°

= sin^{2}
28°^{ } + cos^{2} 28° + tan^{2}
38°^{ }- tan^{2}
38° + sec^{2} 30°