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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

(i)

(ii)

(iii)

(iv)

(5)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

### Solution 5

(i) 2 sinA - 1 = 0

(ii)

(i)

(ii)

(iii)

(iv)

(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)

4 cos2A - 3 = 0

### Solution 11

(i)

(ii) sin 3A - 1 = 0

(iii)

(iv)

(v)

(i)

(ii)

### 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)

To prove:

### 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)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

### 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)

(i)

(ii)

(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

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

### Solution 7

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,

(i)

(ii)

### 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