# 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 TEST YOURSELF

### Solution 1(a)

Correct option: (i) cot γ < cot β

The value of cot decreases from 0 to infinity.

Therefore, for three acute angles ⍺, β and γ, where ⍺ < β < γ,

cot γ < cot β.

### Solution 1(b)

Correct option: (iv)

### Solution 1(c)

Correct option: (iii) 0

(1 + tan^{2} A) × cos^{2} A – 1

= sec^{2}A × cos^{2} A – 1

= 1 – 1

= 0

### Solution 1(d)

Correct option: (ii) sec A

### Solution 1(e)

Correct option: (ii) B and C

The value of sin increases from 0 to 1 and the value of tan increases from 0 to infinity.

Therefore, for two acute angles ⍺ and β, if ⍺ > β, then

tan ⍺ > tan β and sin ⍺ > sin β.

### Solution 1(f)

Correct option: (iii) x^{2} + y^{2}

M = x cos A + y sin A and N = x sin A – y cos A

### Solution 1(g)

Correct option: (i) tan 30^{o}

### Solution 1(h)

Correct option: (iii) 1

### Solution 1(i)

Correct option: (i) 45^{o}

sin 2x = 2 sin 45^{o} cos 45^{o}

Then, sin 2x = sin 90^{o}

2x = 90^{o}⇒ x = 45^{o}

### Solution 2

(i)

(ii)

(iii)

(iv)

(5)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

### Solution 3

### Solution 4

### Solution 5

(i) 2 sinA - 1 = 0

(ii)

### Solution 6

(i)

(ii)

(iii)

(iv)

### Solution 7

(i)

(ii)

(iii)

(iv)

### Solution 8

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 9

### Solution 10

4 cos^{2}A - 3 = 0

### Solution 11

(i)

(ii) sin 3A - 1 = 0

(iii)

(iv)

(v)

### Solution 12

(i)

(ii)

### Solution 13

### Solution 14

### Solution 15

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°

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

### 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

To prove:

### Solution 24

### Solution 25

### Solution 26

### Solution 27

### Solution 28

### Solution 29

### Solution 30

### Solution 31

### Solution 32

### Solution 33

### Solution 1(a)

Correct option: (iii) 2sin^{2} Ɵ – 1

sin^{4} Ɵ – cos^{4 }Ɵ = (sin^{2} Ɵ)^{2} – (cos^{2 }Ɵ)^{2}

= (sin^{2} Ɵ – cos^{2 }Ɵ)(sin^{2} Ɵ + cos^{2 }Ɵ)

= (sin^{2} Ɵ – cos^{2 }Ɵ)(1)

= sin^{2} Ɵ – (1 – sin^{2 }Ɵ)

= sin^{2} Ɵ – 1 + sin^{2 }Ɵ

= 2sin^{2} Ɵ – 1

### Solution 1(b)

Correct option: (ii) 2sec^{2} Ɵ

(1 + tan Ɵ)^{2} + (1 – tan Ɵ)^{2}

= (1 + tan^{2} Ɵ + 2tan Ɵ) + (1 + tan^{2} Ɵ – 2tan Ɵ)

= (sec^{2} Ɵ + 2tan Ɵ) + (sec^{2} Ɵ – 2tan Ɵ)

= 2sec^{2} Ɵ

### Solution 1(c)

Correct option: (iv) cosec^{4} Ɵ – cosec^{2} Ɵ

cot^{4} Ɵ + cot^{2} Ɵ = cot^{2} Ɵ(cot^{2} Ɵ + 1)

= cot^{2} Ɵ.cosec^{2} Ɵ

= (cosec^{2} Ɵ – 1)cosec^{2} Ɵ

= cosec^{4} Ɵ – cosec^{2} Ɵ

### Solution 1(d)

Correct option: (ii) sec^{2} A(1 + sin A)

### Solution 1(e)

Correct option: (ii)

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

### Solution 2

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

### Solution 3

### Solution 4

### Solution 5

### Solution 6

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

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

### Solution 1(a)

Correct option: (iv) sec^{2} A.cosec^{2} A

(1 + cot^{2 }A) + (1 + tan^{2 }A)

= cosec^{2 }A + sec^{2 }A

### Solution 1(b)

Correct option: (iii) b^{2} – a^{2} = 1

b^{2} – a^{2} = sec^{2}Ɵ – tan^{2} Ɵ = 1

### Solution 1(c)

Correct option: (ii) sin Ɵ . cos Ɵ

### Solution 1(d)

Correct option: (iv) –4

(sec Ɵ – cos Ɵ)^{2} – (sec Ɵ + cos Ɵ)^{2}

= (sec^{2} Ɵ + cos^{2} Ɵ – 2sec Ɵ cos Ɵ) – (sec^{2} Ɵ + cos^{2} Ɵ + 2sec Ɵ cos Ɵ)

= sec^{2} Ɵ + cos^{2} Ɵ – 2sec Ɵ cos Ɵ – sec^{2} Ɵ – cos^{2} Ɵ – 2sec Ɵ cos Ɵ

= –4sec Ɵ cos Ɵ

= –4

### Solution 1(e)

Correct option: (iv) cosec^{2} A . cosec^{2 }B

(cot A – cot B)^{2} + (1 + cot A cot B)^{2}

= cot^{2} A + cot^{2} B – 2cot A cot B + 1 + cot^{2} A cot^{2} B + 2cot A cot B

= (cot^{2} A + 1) + cot^{2 }B(1 + cot^{2} A)

= (cot^{2} A + 1)(cot^{2} B + 1)

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

### Solution 2(i)

L.H.S. = (sec A – tan A)^{2}(1 + sin A)

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

### Solution 2

(i)

(ii)

### Solution 3

### Solution 4

(i)

(ii)

### Solution 5

(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 6

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

### Solution 7

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

So, A + C = 90

### Solution 8

(i)

Hence, x =

(ii)

Hence, x =

(iii)

Hence, x =

(iv)

Hence, x =

(v)

Hence,

### Solution 9

(i)

(ii)

### Solution 10

### Solution 11

### Solution 12

### Solution 13

### Solution 1(a)

Correct option: (ii) 1

sin^{2} A + sin^{2}(90^{o} – A)

= sin^{2} A + cos^{2} A

= 1

### Solution 1(b)

Correct option: (i)

In a triangle ABC,

A + B + C = 180^{o}

A + C = 180^{o} – B

Now, sec

### Solution 1(c)

Correct option: (iii) 3

### Solution 1(d)

Correct option: (iv) 1

sin 67^{o} . cos 23^{o} + cos 67^{o} . sin 23^{o}

= sin (90^{o} – 23^{o}) . cos 23^{o} + cos (90^{o} – 23^{o}) . sin 23^{o}

= cos 23^{o} . cos 23^{o} + sin 23^{o} . sin 23^{o}

= cos^{2} 23^{o} + sin^{2} 23^{o}

= 1

### Solution 1(e)

Correct option: (ii) sin^{2} Ɵ

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

### Solution 2

(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

### Solution 3

(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

### Solution 3

(i) tan 37^{o }= 0.7536

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

### Solution 5

(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'

### Solution 6

(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’

### Solution 7

(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’

### Solution 1(a)

Correct option: (iv) 2cos Ɵ = 1

### Solution 1(b)

Correct option: (i) 1 + x^{2}

1 + sin^{2} 27^{o}20’

= 1 + sin^{2}(90^{o} – 62^{o}40’)

= 1 + sin^{2}(89^{o}60’ – 62^{o}40’)

= 1 + cos^{2} 62^{o}40’

= 1 + x^{2}

### Solution 1(c)

Correct option: (iii)

Then,

### Solution 1(d)

Correct option: (iii) 0.4446

cos 63^{o}36’ = 0.4446

cos(90^{o} – 26^{o}24’) = 0.4446

cos(89^{o}60’ – 26^{o}24’) = 0.4446

sin 26^{o}24’ = 0.4446