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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)

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 + tan2 A) × cos2 A – 1

= sec2A × cos2 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) x2 + y2

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

Solution 1(g)

Correct option: (i) tan 30o

Solution 1(h)

Correct option: (iii) 1

Solution 1(i)

Correct option: (i) 45o

sin 2x = 2 sin 45o cos 45o

Then, sin 2x = sin 90o

2x = 90o x = 45o

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)

= cosec2A [sec(90 - B)]2

= cosec2A cosec2B

 

 

 

 

 

 

 

 

(iv)

 

 

 

 

 

 

 

 

 

 

 

 

Solution 9

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


 

  

  

 

 

 

 

 

 

 

 

 

Solution 10

4 cos2A - 3 = 0

 

 

 

 

 

 

 

 

 

Solution 11

(i)

 

 

(ii) sin 3A - 1 = 0

 

 

 

(iii)

 

 

 

(iv)

 

 

(v)

 

 

 

Solution 12

(i)

 

 

 

(ii)

 

 

 

 

 

Solution 13

Solution 14

Solution 15

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° 

 

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) 2sin2 Ɵ – 1

sin4 Ɵ – cos4 Ɵ = (sin2 Ɵ)2 – (cos2 Ɵ)2

= (sin2 Ɵ – cos2 Ɵ)(sin2 Ɵ + cos2 Ɵ)

= (sin2 Ɵ – cos2 Ɵ)(1)

= sin2 Ɵ – (1 – sin2 Ɵ)

= sin2 Ɵ – 1 + sin2 Ɵ

= 2sin2 Ɵ – 1

Solution 1(b)

Correct option: (ii) 2sec2 Ɵ

(1 + tan Ɵ)2 + (1 – tan Ɵ)2

= (1 + tan2 Ɵ + 2tan Ɵ) + (1 + tan2 Ɵ – 2tan Ɵ)

= (sec2 Ɵ + 2tan Ɵ) + (sec2 Ɵ – 2tan Ɵ)

= 2sec2 Ɵ

Solution 1(c)

Correct option: (iv) cosec4 Ɵ – cosec2 Ɵ

cot4 Ɵ + cot2 Ɵ = cot2 Ɵ(cot2 Ɵ + 1)

= cot2 Ɵ.cosec2 Ɵ

= (cosec2 Ɵ – 1)cosec2 Ɵ

= cosec4 Ɵ – cosec2 Ɵ

Solution 1(d)

Correct option: (ii) sec2 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 = (m2 + n2) cos2B

Hence, (m2 + n2) cos2B = n2.

Solution 1(a)

Correct option: (iv) sec2 A.cosec2 A

(1 + cot2 A) + (1 + tan2 A)

= cosec2 A + sec2 A

Solution 1(b)

Correct option: (iii) b2 – a2 = 1

b2 – a2 = sec2Ɵ – tan2 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

= (sec2 Ɵ + cos2 Ɵ – 2sec Ɵ cos Ɵ) – (sec2 Ɵ + cos2 Ɵ + 2sec Ɵ cos Ɵ)

= sec2 Ɵ + cos2 Ɵ – 2sec Ɵ cos Ɵ – sec2 Ɵ – cos2 Ɵ – 2sec Ɵ cos Ɵ

= –4sec Ɵ cos Ɵ

= –4

Solution 1(e)

Correct option: (iv) cosec2 A . cosec2 B

(cot A – cot B)2 + (1 + cot A cot B)2

= cot2 A + cot2 B – 2cot A cot B + 1 + cot2 A cot2 B + 2cot A cot B

= (cot2 A + 1) + cot2 B(1 + cot2 A)

= (cot2 A + 1)(cot2 B + 1)

= cosec2 A . cosec2 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

sin2 A + sin2(90o – A)

= sin2 A + cos2 A

= 1

Solution 1(b)

Correct option: (i)

In a triangle ABC,

A + B + C = 180o

A + C = 180o – B

Now, sec

Solution 1(c)

Correct option: (iii) 3

Solution 1(d)

Correct option: (iv) 1

sin 67o . cos 23o + cos 67o . sin 23o

= sin (90o – 23o) . cos 23o + cos (90o – 23o) . sin 23o

= cos 23o . cos 23o + sin 23o . sin 23o

= cos2 23o + sin2 23o

= 1

Solution 1(e)

Correct option: (ii) sin2 Ɵ

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

Solution 2

(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

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= 0.7536

(ii) tan 4218' = 0.9099

Solution 5

(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'

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 + x2

1 + sin2 27o20’

= 1 + sin2(90o – 62o40’)

= 1 + sin2(89o60’ – 62o40’)

= 1 + cos2 62o40’

= 1 + x2

Solution 1(c)

Correct option: (iii)

Then,

Solution 1(d)

Correct option: (iii) 0.4446

cos 63o36’ = 0.4446

cos(90o – 26o24’) = 0.4446

cos(89o60’ – 26o24’) = 0.4446

sin 26o24’ = 0.4446

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