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

To prove:

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

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

Given:

and

### Solution 7

LHS = (m2 + n2) cos2B

Hence, (m2 + n2) cos2B = n2.

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

(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

(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

(i)

Hence, x =

(ii)

Hence, x =

(iii)

Hence, x =

(iv)

Hence, x =

(v)

Hence, x =

(vi)

Hence, x =

(vii)

Hence,

(i)

(ii)

(i)

(ii)

(i)

(ii)

(iii)

## 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’

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

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

(xiii)

(xiv)

(xv)

(xvi)

(xvii)

### Solution 6

(i) 2 sinA - 1 = 0

(ii)

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(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)

4 cos2A - 3 = 0

### Solution 12

(i)

(ii) sin 3A - 1 = 0

(iii)

(iv)

(v)

(i)

(ii)

### 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°