# Maths Solution Class 10 Chapter 8: Introduction to Trigonometry

### Exercise 8.1

1. In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :

1. sin A, cos A
2. sin C, cos C

Solution

By applying Pythagoras Theorem in ABC, we get

AC2 = AB2 + BC2

AC2 = 242 + 72

AC2 = 576 + 49

AC2 = 625

AC = 25 cm

I. sin A = Opposite Side/ Hypotenuse

= BC/AC = 7/25

cos A = Adjacent Side/ Hypotenus

= AB/AC = 24/25

II. sin C = Opposite Side/ Hypotenuse

= AB/AC = 24/25

cos C = Adjacent Side/ Hypotenus

= BC/AC = 7/25

2. In Fig. 8.13, find tan P - cot R.

Solution

By applying Pythagoras Theorem in PQR, we get

PR2 = PQ2 + QR2

QR2 = PR2 - PQ2

QR2 = 132 - 122

QR2 = 169 - 144

QR2 = 25

QR = 5 cm

tan P = Opposite side/ Adjacent Side

= QR/PQ = 5/12

cot R = Adjacent side/ Opposite Side

= QR/PQ = 5/12

tan P - cot R = 5/12 - 5/12 = 0

3. If sin A = 3/4, calculate cos A and tan A.

Solution

sin A = Opposite side/ Hypotenuse

3/4 = Opposite side/ Hypotenuse

Let Opposite side = 3x and Hypotenuse = 4x

By using the Pythagoras Theorem:

(4x)2 = (Side adjacent to A)2 + (3x)2

16x2 = (Side adjacent to A)2 + 9x2

7x2 = (Side adjacent to A)2

Side adjacent to A = x√7

cos A = Adjacent side/ Hypotenuse

= x√7/4x = √7/4

tan A = Opposite side/ Hypotenuse

= 3x/x√7 = 3/√7

Hence, cos A = √7/4 and tan A = 3/√7.

4. Given 15 cot A = 8, find sin A and sec A.

Solution

15 cot A = 8

cot A = 8/15

Adjacent Side/ Opposite side = 8/15

Let the side adjacent to A be 8x and let the side opposite to A be 15x.

By Pythagoras Theorem:

Hypotenuse2 = (8x)2 + (15x)2

Hypotenuse2 = 64x2 + 225x2

Hypotenuse2 = 289x2

Hypotenuse = 17x

Now,

sin A = Opposite Side/ Hypotenuse

= 15x/17x = 15/17

sec A = Hypotenuse/ Adjacent Side

= 17x/8x = 17/8

Hence, sin A = 15/17 and sec A = 17/8.

5. Given sec θ = 13/12, calculate all other trigonometric ratios.

Solution

sec θ = Hypotenuse/ Adjacent side

Let Hypotenuse = 13x and Adjacent side = 12x

By Pythagoras Theorem:

(13x)2 = (12x)2 + (Opposite side)2

169x2 = 144x2 + (Opposite side)2

25x2 = (Opposite side)2

5x = Opposite side

Now,

sin θ = Opposite side/ Hypotenuse

= 5x/13x = 5/13

cos θ = Adjacent side/ Hypotenuse

= 12x/13x = 12/13

tan θ = Opposite side/ Adjacent side

= 5x/12x = 5/12

cosec θ = Hypotenuse/ Opposite side

= 13x/5x = 13/5

cot θ = Adjacent side/ Opposite side = 12x/5x = 12/5

6. If ∠ A and ∠ B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B.

Solution

Let the triangle be ABC, with a line CD ? AB.

cos A = cos B

Let AD/BD = AC/BC = x

Therefore, AD = xBD and AC = xBC

By applying Pythagoras Theorem in CAD, we get

By applying Pythagoras Theorem in CBD, we get

BC2 = CD2 + BD2

CD2 = BC2 - BD2

AC2 - AD2 = BC2 - BD2

(xBC)2 - (xBD)2 = BC2 - BD2

x2(BC2 - BD2) = BC2 - BD2

x2 = 1

x = 1

Therefore, AC/BC = x = 1

AC = BC

This implies that ∠ A = ∠ B as angles opposite to equal sides of a triangle are also equal.

Hence, proved.

7. If cot θ = 7/8, evaluate :

1. (1 + sin θ)(1 - sin θ)/(1+cos θ)(1-cos θ)
2. cot2 θ

Solution

cot θ = 7/8

Adjacent side/ Opposite side = 7/8

Let adjacent side be 7x and opposite side be 8x.

By using Pythagoras Theorem:

Hypotenuse2 = Adjacent side2 + Opposite side2

Hypotenuse2 = (7x)2 + (8x)2

Hypotenuse2 = 49x2 + 64x2

Hypotenuse2 = 113x2

Hypotenuse = x√113

sin θ = Opposite side/ Hypotenuse

= 8x/x√113 = 8/√113

cos θ = Adjacent side/ Hypotenuse

= 7x/x√113 = 7/√113

I. (1 + sin θ)(1 - sin θ)/(1 + cos θ)(1 - cos θ)

= (1 + 8/√113)(1 - 8/√113)/(1 + 7/√113)(1 - 7/√113)

= 12 - (8/√113)2 /12 - (7/√113)2

= 49/64

II. cot2 θ = (7/8)2 = 49/64

8. If 3 cot A = 4, check whether (1 - tan2 A)/(1 + tan2 A) = cos2 A - sin2A or not.

Solution

3 cot A = 4

cot A = 4/3

Adjacent side/ Opposite side = 4/3

Let adjacent side = 4x and opposite side = 3x

By using Pythagoras Theorem:

Hypotenuse2 = Adjacent side2 + Opposite side2

Hypotenuse2 = (4x)2 + (3x)2

Hypotenuse2 = 16x2 + 9x2

Hypotenuse2 = 25x2

Hypotenuse = 5x

sin A = Opposite side/ Hypotenuse

= 3x/5x = 3/5

cos A = Adjacent side/ Hypotenuse

= 4x/5x = 4/5

tan A = Opposite side/ Adjacent side

= 3x/4x = 3/4

Now, LHS of the given relation:

Therefore, LHS = RHS.

Hence, proved.

9. In triangle ABC, right-angled at B, if tan A = 1/√3, find the value of:

1. sin A cos C + cos A sin C
2. cos A cos C - sin A sin C

Solution

tan A = 1/√3

Opposite side/ Adjacent side = 1/√3

Let BC = x and AB = x√3

By using Pythagoras Theorem:

AC2 = AB2 + BC2

AC2 = (x√3)2 + (x)2

AC2 = x2 + 3x2

AC2 = 4x2

AC = 2x

sin A = Opposite side/ Hypotenuse

= BC/AC = x/2x = 1/2

cos A = Adjacent side/ Hypotenuse

= AB/AC = x√3/2x = √3/2

sin C = AB/AC = x√3/2x = √3/2

cos C = BC/AC = x/2x = 1/2

I. sin A cos C + cos A sin C = (1/2)(1/2) + (√3/2)(√3/2)

= ¼ + ¾ = 4/4 = 1

II. cos A cos C - sin A sin C = (1/2)(√3/2) - (1/2)(√3/2)

= √3/4 - √3/4 = 0

10. In ∆ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Solution

PR + QR = 25

PR = 25 - QR

By applying Pythagoras Theorem in PQR, we get

PR2 = PQ2 + QR2

(25 - QR)2 = 52 + QR2

252 + QR2 - 2(25)(QR) = 25 + QR2

625 - 50QR = 25

600 = 50QR

QR = 12 cm

Now, PR = 25 - QR = 25 - 12 = 13 cm

sin P = QR/PR = 12/13

cos P = PQ/PR = 5/13

tan P = QR/PQ = 12/5

11. State whether the following are true or false. Justify your answer.

1. The value of tan A is always less than 1.
2. sec A = 12/5 for some value of angle A.
3. cos A is the abbreviation used for the cosecant of angle A.
4. cot A is the product of cot and A.
5. sin θ = 4/3 for some angle θ.

Solution

I. The given statement is false.

For a triangle ABC right angled at B:

tan A = Opposite side/ Adjacent side

It is possible to have side opposite to the angle greater than the side adjacent to it, which will make tan A greater than 1.

Example side lengths of ABC - 3 units, 4 units, and 5 units

II. The given statement is true.

Let there be a triangle ABC right angled at B.

sec A = 12/5

AC/AB = 12/5

Let AC = 12x and AB = 5x

By Pythagoras theorem:

AC2 = AB2 + BC2

(12x)2 = (5x)2 + BC2

144x2 = 25x2 + BC2

119x2 = BC2

Therefore it satisfies the Pythagoras Theorem.

Hence, sec A = 12/5 is possible.

III. The given statement is false.

cos A is the abbreviation used for cosine of angle A.

IV. The given statement is false.

cot A is not a product of cot and A. It means cotangent of angle A.

V. The given statement is false.

sin A = Opposite side/ Hypotenuse

For sin A to be 4/3, side of the triangle needs to be greater than the hypotenuse, which is impossible as hypotenuse is the longest side of a right angled triangle.

### Exercise 8.2

1. Evaluate the following :

1. sin 60° cos 30° + sin 30° cos 60°
2. 2 tan2 45° + cos2 30° - sin2 60°
3. cos 45°/(sec 30° + cosec 30°)
4. (sin 30° + tan 45° - cosec 60°)/(sec 30° + cos 60° + cot 45°)
5. (5 cos2 60° + 4 sec2 30° - tan2 45°)/(sin2 30° + cos2 30°)

Solution

I. sin 60° = √3/2

cos 30° = √3/2

sin 30° = 1/2

cos 60° = 1/2

sin 60° cos 30° + sin 30° cos 60° = (√3/2) × (√3/2) + (1/2) × (1/2)

= 3/4 + 1/4 = 4/4 = 1

II. tan 45° = 1

sin 60° = √3/2

cos 30° = √3/2

2 tan2 45° + cos2 30° - sin2 60° = 2 × (1)2 + (√3/2)2 - (√3/2)2

= 2 + 3/4 - 3/4 = 2

III. cos 45° = 1/√2

sec 30° = 2/√3

cosec 30° = 2

IV. sin 30° = 1/2

tan 45° = 1

cosec 60° = 2/√3

sec 30° = 2/√3

cos 60° = 1/2

cot 45° = 1

V. cos 60° = 1/2

sec 30° = 2/√3

tan 45° = 1

sin 30° = 1/2

cos 30° = √3/2

2. Choose the correct option and justify your choice :

1. 2 tan 30°/(1 + tan2 30°) =
(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
2. (1 - tan2 45°)/(1 + tan2 45°) =
(A) tan 90° (B) 1 (C) sin 45° (D) 0
3. sin 2A = 2 sin A is true when A =
(A) 0° (B) 30° (C) 45° (D) 60°
4. 2 tan 30°/(1 - tan2 30°) =
(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°

Solution

I. 2 tan 30°/(1 + tan2 30°) = 2(1/√3)/(1 + 1/√32)

= (2/√3)/(1 + 1/3) = (2/√3)/(4/3)

= 2√3/4 = √3/2 = sin 60°

Hence, (A) is the correct answer.

II. (1 - tan2 45°)/(1 + tan2 45°) = (1 - 12)/(1 + 12)

= (1 - 1)/(1 + 1) = 0/2 = 0

Hence, (D) is the correct answer.

III. Put A = 0°

sin 2A = sin 0° = 0

2 sin A = 2 sin 0° = 2 (0) = 0

Hence, (A) is the correct answer.

IV. 2 tan 30°/(1 - tan2 30°) = 2(1/√3)/(1 - 1/√32)

= (2/√3)/(1 - 1/3) = (2/√3)/(2/3) = 3/√3 = √3 = tan 60°

Hence, (C) is the correct answer.

3. If tan (A + B) = √3 and tan (A - B) = 1/√3; 0° < A + B ? 90°; A > B, find A and B.

Solution

tan (A + B) = √3

We know that tan 60° = √3. Therefore,

tan (A + B) = tan 60°

A + B = 60

tan (A - B) = 1/√3

We know that tan 30° =1/√3. Therefore,

tan (A - B) = tan 30°

A - B = 30

A + B + A - B = 60 + 30

2A = 90

A = 45

Now, to find B:

A - B = 30

45 - B = 30

B = 15

Hence, A = 45° and B = 15°

4. State whether the following are true or false. Justify your answer.

1. sin (A + B) = sin A + sin B.
2. The value of sin θ increases as θ increases.
3. The value of cos θ increases as θ increases.
4. sin θ = cos θ for all values of θ.
5. cot A is not defined for A = 0°.

Solution

I. If we put A = 30° and B = 60°, we will get:

LHS = sin (A + B) = sin (30° + 60°) = sin 90°

= 1

RHS = sin A + sin B = sin 30° + sin 60°

= 1/2 + √3/2 = (√3 + 1)/2

Therefore, LHS ? RHS.

Hence, the give statement is false.

II. sin 0° = 0

sin 30° = 1/2

sin 45° = 1/√2

sin 60° = √3/2

sin 90° = 1

Therefore, the value of sin θ increases with the value of θ.

Hence, the given statement is true.

III. cos 0° = 1

cos 30° = √3/2

cos 45° = 1/√2

cos 60° = 1/2

cos 90° = 0

Therefore, the value of cos θ decreases with the value of θ.

Hence, the given statement is false.

IV. sin 0° = 0 while cos 0° = 1

Hence, the given statement is false.

V. cot A = cot 0°

= cos 0°/sin 0° = 1/0 = Undefined

Hence, the given statement is true.

### Exercise 8.3

1. Evaluate :

1. sin 18°/cos 72°
2. tan 26°/cot 64°
3. cos 48° - sin 42°
4. cosec 31° - sec 59°

Solution

I. sin 18°/cos 72° = cos (90° - 18°)/cos 72°

= cos 72°/ cos 72° = 1

II. tan 26°/cot 64° = cot (90° - 24°)/cot 64°

= cot 64°/ cot 64° = 1

III. cos 48°/sin 42° = sin (90° - 48°)/sin 42°

= sin 42°/ sin 42° = 1

IV. cosec 31°/sec 59° = sec (90° - 31°)/sec 59°

= sec 59°/ sec 59° = 1

2. Show that :

1. tan 48° tan 23° tan 42° tan 67° = 1
2. cos 38° cos 52° - sin 38° sin 52° = 0

Solution

I. tan 48° tan 23° tan 42° tan 67° = tan (90° - 42°) tan (90° - 67°) tan 42° tan 67°

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°) = 1 × 1 = 1

Hence, proved.

II. cos 38° cos 52° - sin 38° sin 52° = cos (90° - 52°) cos (90°-38°) - sin 38° sin 52°

= sin 52° sin 38° - sin 38° sin 52°

= 0

Hence, proved.

3. If tan 2A = cot (A - 18°), where 2A is an acute angle, find the value of A.

Solution

We know that,

tan 2A = cot (90° - 2A) as 2A is an acute angle.

Therefore,

tan 2A = cot (A - 18°)

cot (90° - 2A) = cot (A - 18°)

90° - 2A = A - 18°

108° = 3A

A = 36°

Hence, A = 36°.

4. If tan A = cot B, prove that A + B = 90°.

Solution

tan A = cot B

tan A = tan (90° - B)

A = 90° - B

Therefore,

A + B = 90°

Hence, proved.

5. If sec 4A = cosec (A - 20°), where 4A is an acute angle, find the value of A.

Solution

We know that,

sec 4A = cosec (90° - 4A) as 4A is an acute angle.

Therefore,

sec 4A = cosec (A - 20°)

cosec (90° - 4A) = cosec (A - 20°)

90° - 4A = A - 20°

110° = 5A

A = 22°

Hence, A = 22°.

6. If A, B and C are interior angles of a triangle ABC, then show that

sin ((B + C)/2) = cos A/2

Solution

Since, A, B and C are interior angles of a triangle. Therefore,

A + B + C = 180°

B + C = 180° - A

Divide both sides by 2

(B + C)/2 = (180° - A)/2

(B + C)/2 = (90° - A/2)

Taking sin on both sides, we get

sin ((B + C)/2) = sin (90° - A/2)

We know that sin (90° - ?) = cos (?)

Therefore,

sin ((B + C)/2) = cos (A/2)

Hence, proved.

7. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Solution

sin 67° = cos (90° - 67°)

= cos 23°

cos 75° = sin (90° - 75°)

= sin 25°

Hence, sin 67° + cos 75° can be expressed as cos 23° + sin 25°

### Exercise 8.4

1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Solution

We know that,

cosec2 A- cot2 A = 1

cosec2 A = 1 + cot2 A

(1/sin2 A) = 1 + cot2 A as cosec is the inverse of sin function

sin2 A = 1/(1 + cot2 A)

Square root on both sides,

sin A = ±1/√(1 + cot2 A)

Hence, sin A in terms of cot A is ±1/√(1 + cot2 A).

Now,

sin2 A = 1/(1 + cot2 A)

1 - cos2 A = 1/(1 + cot2 A) as sin2 A + cos2 A = 1

cos2 A = 1 - 1/(1 + cot2 A)

cos2 A = (1 - 1 + cot2 A)/(1 + cot2 A)

1/sec2 A = (cot2 A)/(1 + cot2 A) as sec is the inverse of cos function

sec2 A = (1 + cot2 A)/cot2 A

Square root on both sides,

secA = ±√(1 + cot2 A)/cotA

Hence, sec A in terms of cot A is ±√(1 + cot2 A)/cot A.

We know that,

tan A = sin A/cos A

Reciprocal both sides,

1/tan A = cos A/sin A

1/tan A = cot A

tan A = 1/cot A

Hence, tan A in terms of cot A is 1/cot A.

2. Write all the other trigonometric ratios of ∠ A in terms of sec A.

Solution

We know that cos A is inverse of sec A. Therefore,

sec A = 1/cos A

cos A = 1/sec A

Hence, cos A in terms of sec A is 1/sec A.

We know that,

cos2 A + sin2 A = 1

sin2 A = 1 - cos2 A

sin2 A = 1 - (1/sec2 A)

sin2 A = (sec2 A - 1)/sec2 A

sin A = ±√(sec2 A - 1)/sec A

Hence, sin A in terms of sec A is ±√(sec2 A - 1)/sec A.

sin A = 1/cosec A

cosec A = 1/sin A

cosec A = ±1/√(sec2 A - 1)/sec A

Hence, cosec A in terms of sec A is ±1/√(sec2 A - 1)/sec A.

We know that,

sec2 A - tan2 A = 1

tan2 A = sec2 A - 1

tan A = ±√(sec2 A - 1)

Hence, tan A in terms of sec A is ±√(sec2 A - 1).

tan A = 1/cot A

cot A = 1/tan A

cot A = 1/±√(sec2 A - 1)

Hence, cot A in terms of sec A is ±1/√(sec2 A - 1).

3. Evaluate:

1. (sin2 63° + sin2 27°)/(cos2 17° + cos2 73°)
2. sin 25° cos 65° + cos 25° sin 65°

Solution

I. (sin2 63° + sin2 27°)/(cos2 17° + cos2 73°)

= (cos2 (90° - 63°) + sin2 27°)/(cos2 17° + sin2 (90° - 73°))

= (cos2 27° + sin2 27°)/(cos2 17° + sin2 17°)

We know that, cos2 A + sin2 A = 1. Therefore,

(cos2 27° + sin2 27°)/(cos2 17° + sin2 17°)

= 1/1 = 1

II. sin 25° cos 65° + cos 25° sin 65°

= sin 25° sin (90° - 65°) + cos 25° cos (90° - 65°)

= sin 25° sin 25° + cos 25° cos 25°

= sin2 25° + cos2 25°

We know that, cos2 A + sin2 A = 1. Therefore,

sin2 25° + cos2 25° = 1

4. Choose the correct option. Justify your choice.

1. 9 sec2 A - 9 tan2 A =
(A) 1 (B) 9 (C) 8 (D) 0
2. (1 + tan θ + sec θ) (1 + cot θ - cosec θ) =
(A) 0 (B) 1 (C) 2 (D) -1
3. (sec A + tan A) (1 - sin A) =
(A) sec A (B) sin A (C) cosec A (D) cos A
4. (1 + tan2 A)/(1 + cot2 A) =
(A) sec2 A (B) -1 (C) cot2 A (D) tan2 A

Solution

I. 9 sec2 A - 9 tan2 A = 9(sec2 A - tan2 A)

We know that sec2 θ - tan2 θ = 1. Therefore,

9(sec2 A - tan2 A) = 9 (1) = 9

Hence, (B) is the correct answer.

II. (1 + tan θ + sec θ) (1 + cot θ - cosec θ)

= (1 + sin θ/cos θ + 1/cos θ)(1 + cos θ/sin θ - 1/sin θ)

= (sin θ + cos θ + 1)/cos θ × (sin θ + cos θ - 1)/sin θ

= (cos θ + sin θ)2 - 12)/cos θ sin θ

= (cos2 θ + sin2 θ + 2cos θ sin θ - 1)/cos θ sin θ

We know that, cos2 A + sin2 A = 1. Therefore,

(cos2 θ + sin2 θ + 2cos θ sin θ - 1)/cos θ sin θ

= (1 + 2cos θ sin θ - 1)/cos θ sin θ

= 2cos θ sin θ/cos θ sin θ

= 2

Hence, (C) is the correct answer.

III. (sec A + tan A) (1 - sin A)

= (1/cos A + sin A/cos A) (1 - sin A)

= (1 + sin A)(1 - sin A)/cos A

= (12 - sin2 A)/cos A

= (1 - sin2 A)/cos A

We know that sin2 A - cos2 A = 1. Therefore,

(1 - sin2 A)/cos A = cos2 A/cos A

= cos A

Hence, (D) is the correct answer.

IV. (1 + tan2 A)/(1 + cot2 A)

= (1 + 1/cot2 A)/(1 + cot2 A)

= (cot2 A + 1)/cot2 A(1 + cot2 A)

= 1/cot2 A = tan2 A

Hence, (D) is the correct answer.

5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

1. (cosec θ - cot θ)2 = (1 - cos θ)/(1 + cos θ)
2. cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A
3. tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = 1 + sec θ cosec θ
[Hint : Write the expression in terms of sin θ and cos θ]
4. (1 + sec A)/sec A = sin2 A/(1 - cos A)
[Hint : Simplify LHS and RHS separately]
5. (cos A - sin A+1)/( cos A + sin A - 1) = cosec A + cot A, using the identity cosec2A = 1 + cot2A.
6. (sin θ - 2sin3 θ)/(2cos3 θ - cos θ) = tan θ
7. (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
8. (cosec A - sin A)(sec A - cos A) = 1/(tan A + cotA)
[Hint : Simplify LHS and RHS separately]
9. (x) (1 + tan2 A/1 + cot2 A) = (1 - tan A/1 - cot A)2 =tan2A

Solution

I. (cosec θ - cot θ)2 = (1 - cos θ)/(1 + cos θ)

= (cosec2 θ + cot2 θ - 2cosec θ cot θ)

= (1/sin2 θ + cos2 θ/sin2 θ - 2cos θ /(sin2 θ))

= (1 + cos2 θ - 2cos θ)/sin2 θ

= (12 + cos2 θ - 2(1)cos θ)/sin2 θ

= (1 - cos θ)2/(sin2 θ)

Since, sin2 A + cos2 A = 1. Therefore,

(1 - cos θ)2/(sin2 θ)

= (1 - cos θ)2/(1 - cos2 θ)

= (1 - cos θ)2/(1 - cos θ)(1 + cos θ)

= (1 - cos θ)/(1 + cos θ)

Hence, proved.

II. (cos A/(1 + sin A)) + ((1 + sin A)/cos A)

= (cos2 A + (1 + sin A)2)/(1 + sin A)cos A

= (cos2 A + sin2 A + 1 + 2sin A)/(1 + sin A)cos A

Since, sin2 a + cos2 a = 1. Therefore,

(cos2 A + sin2 A + 1 + 2sin A)/(1 + sin A)cos A

= (1 + 1 + 2sin A)/(1 + sin A)cos A

= 2(1 + sin A)/(1 + sin A)cos A

= 2/cos A

= 2 sec A

Hence, proved.

III. tan θ/(1 - cot θ) + cot θ/(1 - tan θ)

= sin θ/cos θ(1 - cos θ/sin θ) + cos θ/sin θ(1 - sin θ/cos θ)

= sin θ/cos θ((sin θ - cos θ)/sin θ) + cos θ/sin θ((cos θ - sin θ)/cos θ)

= sin2 θ/(sin θ - cos θ)cos θ + cos2 θ/(cos θ - sin θ)sin θ

= sin2 θ/(sin θ - cos θ)cos θ - cos2 θ/(sin θ - cos θ)sin θ

= 1/(sin θ - cos θ)[sin2 θ/cos θ - cos2 θ/sin θ]

= [sin2 θ/cos θ - cos2 θ/sin θ]/(sin θ - cos θ)

= [(sin3 θ - cos3 θ)/sin θ cos θ]/(sin θ - cos θ)

= [(sin θ - cos θ)(sin2 θ + cos2 θ + sin θ cos θ)/sin θ cos θ]/(sin θ - cos θ)

= (sin2 θ + cos2 θ + sin θ cos θ)/sin θ cos θ

Since, sin2 a + cos2 a = 1. Therefore,

(sin2 θ + cos2 θ + sin θ cos θ)/sin θ cos θ

= (1 + sin θ cos θ)/sin θ cos θ

= 1/sin θ cos θ + sin θ cos θ/sin θ cos θ

= cosec θ sec θ + 1

= 1 + sec θ cosec θ

Hence, proved.

IV. LHS = (1 + sec A)/sec A

= (1 + 1/cos A)/(1/cos A)

= ((cos A + 1)/cos A)/(1/cos A)

= cos A + 1

RHS = sin2 A/(1 - cos A)

= (1 - cos2 A)/(1 - cos A)

= (12 - cos2 A)/(1 - cos A)

= (1 - cos A)(1 + cos A)/(1 - cos A)

= cos A + 1

LHS = RHS

Hence, proved.

V. Divide the numerator and denominator by sin A

[(cos A - sin A + 1)/sin A]/[(cos A + sin A - 1)/sin A]

= (cot A - 1 + cosec A)/(cot A + 1 - cosec A)

Since, cosec2 A = 1 + cot2 A. Therefore, 1 = cosec2 A - cot2 A.

(cot A - 1 + cosec A)/(cot A + 1 - cosec A)

= (cot A - (cosec2 A - cot2 A) + cosec A)/(cot A + 1 - cosec A)

= (cot A + cosec A - (cosec A + cot A)(cosec A - cot A))/(cot A + 1 - cosec A)

= (cosec A + cot A)(1 + cot A - cosec A)/(cot A + 1 - cosec A)

= cosec A + cot A

Hence, proved.

VI.

Since, sec2 A - tan2 A = 1. Therefore,

= (sec A + tan A)/1

= sec A + tan A

VII. (sin θ - 2sin3 θ)/(2cos3 θ - cos θ)

= sin θ (1 - 2sin2 θ)/cos θ (2cos2 θ - 1)

Since, sin2 θ = 1 - cos2 θ. Therefore,

sin θ (1 - 2sin2 θ)/cos θ (2cos2 θ - 1)

= sin θ (1 - 2(1 - cos2 θ))/cos θ (2cos2 θ - 1)

= sin θ (1 - 2 + 2cos2 θ)/cos θ (2 cos2 θ - 1)

= sin θ (-1 + 2cos2 θ)/cos θ (2 cos2 θ - 1)

= sin θ/cos θ

= tan θ

Hence, proved.

VIII. (sin A + cosec A)2 + (cos A + sec A)2

= sin2 A + cosec2 A + 2sin A cosec A + cos2 A + sec2 A + 2cos A sec A

Since, cosec2 A = 1 + tan2 A and sec2 A = 1 + cot2 A. Therefore,

sin2 A + cosec2 A + 2sin A cosec A + cos2 A + sec2 A + 2cos A sec A

= sin2 A + 1 + tan2 A + 2sin A cosec A + cos2 A + 1 + cot2 A + 2cos A sec A

= 2 + tan2 A + cot2 A + sin2 A + cos2 A + 2sin A (1/sin A) + 2cos A (1/cos A)

Since, sin2 A + cos2 A = 1. Therefore,

= 2 + tan2 A + cot2 A + 1+ 2 + 2

= 7 + tan2 A + cot2 A

Hence, proved.

IX. LHS = (cosec A - sin A)(sec A - cos A)

= (1/sin A - sin A)(1/cos A - cos A)

= [(1 - sin2 A)/sin A][(1 - cos2 A)/cos A]

Since, sin2 a + cos2 a = 1. Therefore,

[(1 - sin2 A)/sin A][(1 - cos2 A)/cos A]

= [(cos2 A)/sin A][(sin2 A)/cos A]

= (cos A)(sin A)

= sin A cos A

RHS = 1/(tan A + cot A)

= 1/(sin A/cos A + cos A/sin A)

= 1/[(sin2 + cos2 A)/sin A cos A]

= 1/(1/sin A cos A)

= sin A cos A

Therefore, LHS = RHS

Hence, proved.

X. Part (i) (1 + tan2 A/1 + cot2 A)

= (1 + tan2 A)/(1 + 1/tan2 A)

= (1 + tan2 A)/[(tan2 A + 1)/tan2 A]

= tan2 A (1 + tan2 A)/(tan2 A + 1)

= tan2 A

Part (ii) (1 - tan A)2/(1 - cot A)2

= (1 - tan A)2/(1 - 1/tan A)2

= (1 - tan A)2/((tan A - 1)/tan A)2

= (1 - tan A)2/[(tan A - 1)2/tan2 A]

= tan2 A (1 - tan A)2/(-(1 - tan A))2

= tan2 A (1 - tan A)2/(1 - tan A)2

= tan2 A

Hence, proved.