Chapter 17 Trigonometric Ratios ML Aggarwal ICSE Solutions for Class 9 Maths

ML Aggarwal Solutions for Chapter 17 Trigonometric Ratios Class 9 Maths ICSE

Here, we are providing the solutions for Chapter 17 Trigonometric Ratios from ML Aggarwal Textbook for Class 9 ICSE Mathematics. Solutions of the seventeen chapter has been provided in detail. This will help the students in understanding the chapter more clearly. Class 9 Chapter 17 Trigonometric Ratios ML Aggarwal Solutions for ICSE is one of the most important chapter for the board exams which is based on finding the values of trigonometric ratios, learning trigonometric identities and application of trigonometric ratios in the finding the sides and angles of a triangles. 

Exercise 17

1. (a) From the figure (1) given below, find the values of:

(i) sin θ

(ii) cos θ

(iii) tan θ

(iv) cot θ

(v) sec θ

(vi) cosec θ

(b) From the figure (2) given below, find the values of:

(i) sin 

(ii) cos A

(iii) sin² A + cos² A

(iv) sec² A – tan² A.

Solution

(a) From right angled triangle OMP,

By Pythagoras theorem, we get

OP² = OM² +MP²

⇒ MP² = OP² + OM²

⇒ MP² = (15)² – (12)²

⇒ MP² = 225 – 144

⇒ MP² = 81

⇒ MP² = 9²

⇒ MP = 9

(i) sin θ = MP/OP

= 9/15

= 3/5

(ii) cos θ = OM/OP

= 12/15

= 4/5

(iii) tan θ = MP/OP

= 9/12

= ¾

(iv) cot θ = OM/MP

= 12/9

= 4/3

(v) sec θ = OP/OM

= 15/12

= 5/4

(vi) cosec θ = OP/MP

= 15/9

= /3

(b) From right angled triangle ABC,

By Pythagoras theorem, we get

AB² = AC² + BC²

⇒ AB² = (12)² + (5)²

⇒ AB² = 144 + 25

⇒ AB² = 169

⇒ AB² = 13²

⇒ AB = 13

(i) sin A = BC/AB

= 5/13

(ii) cos A = AC/AB

= 12/13

(iii) sin² A + cos² A = (BC/AB)² + (AC/AB)²

= (5/13)² + (12/13)²

= (25/169) + (144/169)

= (25 + 144)/ 169

= 169/169

= 1

sin² A + cos² A = 1

(iv) sec² A – tan² A = (AB/AC)² – (BC/AC)²

= (13/12)² – (5/12)²

= (169/144) – (25/144)

= (169 – 25)/144

= 144/144

= 1

sec² A – tan² A = 1

2. (a) From the figure (1) given below, find the values of:
(i) sin B

(i) cos C

(iii) sin B + sin C

(iv) sin B cos C + sin C cos B.

(b) From the figure (2) given below, find the values of:

(i) tan x

(ii) cos y

(iii) cosec² y – cot² y

(iv) 5/sin x + 3/sin y – 3 cot y.

Solution

(a) From right angled triangle ABC,

By Pythagoras theorem, we get

BC² = AC² + AB²

⇒ AC² = BC² – AB²

⇒ AC² = (10)² – (6)²

⇒ AC² = 100 – 36

⇒ AC² = 64

⇒ AC² = 8²

⇒ AC = 8

(i) sin B = perpendicular/ hypotenuse

= AC/BC

= 8/10

= 4/5

(ii) cos C = Base/hypotenuse

= AC/BC

= 8/10

= 4/5

(iii) sin B = Perpendicular/hypotenuse

= AC/BC

= 8/10

= 4/5

sin C = perpendicular/hypotenuse

= AB/BC

= 6/10

= 3/5

Now,

sin B + sin C = (4/5) + (3/5)

= (4 + 3)/5

= 7/5

(iv) sin B = 4/5

cos C = 4/5

sin C = perpendicular/ hypotenuse

= AB/BC

= 6/10

= 3/5

cos B = Base/Hypotenuse

= AB/BC

= 6/10

= 3/5

sin B cos C + sin C cos B

= (4/5)×(4/5) + (3/5)×(3/5)

= (26/25) + (9/25)

= (16+9)/25

= 25/25

= 1

(b) From Figure

AC = 13, CD = 5, BC =21,

BD = BC – CD

= 21 – 5

= 16

From right angled ∆ACD,

By Pythagoras theorem we get

AC = AD² + CD²

⇒ AD² = AC² – CD²

⇒ AD² = (13)² – (5)²

⇒ AD² = 169 – 25

⇒ AD² = 144)

⇒ AD² = (12)²

⇒ AD = 12

From right angled ∆ABD,

By Pythagoras angled ∆ABD

By Pythagoras theorem we get

AB² = AD² + BD²

⇒ AB² = 400

⇒ AB² = (20)²

⇒ AB = 20

(i) tan x = perpendicular/Base (in right angled ∆ACD)

=CD/AD

= 5/12

(ii) cos y = Base/Hypotenuse (in right angled ∆ABD)

= BD/AB

= (20)/12 – (5/3)

cot y = Base/Perpendicular (in right ∆ABD)

=BD/AB

= 16/20 = 4/5

(iii) cos y = Hypotenuse/ perpendicular (in right angled ∆ABD)

BD/AB

= 20/12

= 5/3

cot y = Base/Perpendicular (in right ∆ABD)

AB/AD

= 16/12

= 4/3

cosec² y – cot² y = (5/3)² – (4/3)²

= (25/9) – (16/9)

= (25-16)/9

= 9/9

= 1

Hence, cosec² y – cot² y = 1

(iv) sin x = Perpendicular/Hypotenuse (in right angled ∆ACD)

= AD/AB

= 12/20

= 3/5

cot y = Base/Perpendicular (in right angled ∆ABD)

= BD/AD

= 16/12

= 4/3

(5/sin x) + (3/sin y) – 3cot y

= [5/(5/13)] + 3/(3/5) – (3 × 4/3)

= (5× 13/5) + (3× 5/3) – (3× 4/3)

= (1× 13/1) + (1× 5/1) – (1× 4/1)

= 13 + 5 – 4

= 18 – 4

= 14

Hence 5/sin x + 3/sin y – 3cot y = 14

3. (a) From the figure (1) given below, find the value of sec θ.

(b) From the figure (2) given below, find the values of:

(i) sin x

(ii) cot x

(iii) cot² x- cosec² x

(iv) sec y

(v) tan² y – 1/cos² y.

Solution

(a) From the figure, Sec θ = AB/BD

But in ∆ADC, ∠D = 90^o

AC² =AD² + DC² (Pythagoras Theorem)

⇒ (13)² =AD² + 25

⇒ AD² = 169 -25

= 144

= (12)²

⇒ AD = 12

AB² = AD² + BD² (in right ∆ABD)

= (12)² + (16)²

= 144 + 256

= 400

= (20)²

⇒ AB = 20

Now, Sec θ = AB/BD

= 20/16

= 5/4

(b) let given ∆ABC

BD = 3, AC = 12, AD = 4

In right angled ∆ABD

By Pythagoras theorem

AB² =AD² + BD²

⇒ AB² = (4)² + (3)²

⇒ AB² = 16 + 9

⇒ AB² = 25

⇒ AB² = (5)²

⇒ AB = 5

In right angled triangle ACD

By Pythagoras theorem,

AC² = AD² + CD²

⇒ CD² = AC² – AD²

⇒ CD² = (12)² – (4)²

⇒ CD² = 144 – 16

⇒ CD² = 128

⇒ CD = √128

⇒ CD = √64 × 2 CD

= 8√2

(i) sin x = perpendicular/Hypotenuse

= AD/AB

= 4/5

(ii) cot x = Base/Perpendicular

= BD/AD

= ¾

(iii) cot x = Base/Perpendicular

BD/AD

= 3/4

(iv) cosec x = Hypotenuse/Perpendicular

AB/BD

= 5/4

cot² x – cosec² x

= (3/4)² – (5/4)²

= 9/16 – 25/16

(9 -25)/16

= -16/16

= -1

Perpendicular = Hypotenuse/Base (in right angled ∆ACD)

= AD/CD

= 12/(8 √2)

= 3/(2 √2)

cot y = Base/Hypotenuse

= AD/CD

= 4/8 √ 2

= 1/2 √2

cot y = Base/Hypotenuse (in right angled ∆ACD)

= CD/AC

= 8√2/12

= 2√/3

Now tan² y = 1/cos² y

= (1/2√2)² – 1/(2√2/3)²

= ¼ × – ¼ × 2

= (1/8) – (9/8)

= (1-9)/8

= -8/8

= -1

tan² y – 1/cos² y = –1

4. (a) From the figure (1) given below, find the values of:

(i) 2 sin y – cos y

(ii) 2 sin x – cos x

(iii) 1 – sin x + cos y

(iv) 2 cos x – 3 sin y + 4 tan x

(b) In the figure (2) given below, ∆ABC is right-angled at B. If AB = y units, BC = 3 units and CA = 5 units, find

(i) sin x°

(ii) y.

Solution

(a) In a right angled ∆BCD,

Using Pythagoras theorem

BC² = BD² + CD²

Substituting the values

BC² = 9² + 12²

By further calculation

BC² = 81 + 144 = 225

⇒ BC² = 15²

⇒ BC = 15

In a right angled ∆ABC,

Using Pythagoras theorem

AC² = AB² + BC²

We can write it as

AB² = AC² – BC²

Substituting the values

AB² = 25² – 15²

By further calculation

AB² = 625 – 225 = 400

So we get

AB² = 20²

AB = 20

(i) We know that

In right angled ∆BCD

sin y = perpendicular/ hypotenuse

⇒ sin y = BD/ BC

Substituting the values

sin y = 9/15 = 3/5

In right angled ∆BCD

cos y = base/hypotenuse

⇒ cos y = CD/BC

Substituting the values

cos y = 12/15 = 4/5

Here

2sin y – cos y = 2× 3/5 – 4/5

We can write it as

= 6/5 – 4/5

= 2/5

Therefore, 2 sin y – cos y = 2/5

(ii) In right angled ∆ABC

sin x = perpendicular/ hypotenuse

sin x = BC/AC

Substituting the values

sin x = 15/25 = 3/5

In right angled ∆ABC

cos x = base/hypotenuse

⇒ cos x = AB/AC

Substituting the values

cos x = 20/25 = 4/5

Here

2 sin x – cos x = 2× 3/5 – 4/5

We can write it as

= 6/5 – 4/5

= 2/5

Therefore, 2 sin x – cos x = 2/5.

(iii) In right angled ∆ABC

sin x = perpendicular/hypotenuse

⇒ sin x = BC/AC

Substituting the values

sin x = 12/25 = 3/5

In right angled ∆BCD

cos y = base/hypotenuse

⇒ cos y = CD/BC

Substituting the values

cos y = 12/15 = 4/5

Here

1 – sin x + cos y

= 1 – 3/5 + 4/5

By further calculation

= (5 – 3 + 4)/ 5

So we get

= (9 – 3)/ 5

= 6/5

Therefore, 1 – sin x + cos y = 6/5.

(iv) In right angled ∆BCD

cos x = base/hypotenuse

⇒ cos x = AB/AC

Substituting the values

cos x = 20/25 = 4/5

In right angled ∆BCD

sin y = perpendicular/hypotenuse

⇒ sin y = BD/BC

Substituting the values

sin y = 9/15 = 3/5

In right angled ∆ABC

tan x = perpendicular/base

⇒ tan x = BC/AB

Substituting the values

tan x = 15/20 = ¾

Here,

2 cos x – 3 sin y + 4 tan x = (2× 4/5) – (3× 3/5) + (4× ¾)

By further calculation

= 8/5 – 9/5 3/1

Taking LCM

= (8 – 9 + 15)/5

So we get

= (23 – 9)/ 5

= 14/5

(b) It is given that

AB = y units, BC = 3 units, CA = 5 units

(i) In right angled ∆ABC

sin x = perpendicular/hypotenuse

⇒ sin x = BC/AC

Substituting the values

sin x = 3/5

(ii) In right angled ∆ABC

Using Pythagoras theorem

AC² = BC² + AB²

We can write it as

AB² = AC² – BC²

Substituting the values

AB² = 5² – 3²

By further calculation

AB² = 25 – 9 = 16

So we get

AB² = 4²

⇒ AB = 4

y = 4 units

Therefore, y = 4 units.

5. In a right-angled triangle, it is given that angle A is an acute angle and that
tan A=5/12. Find the values of:
(i) cos A
(ii) cosec A- cot A.

Solution

Here, ABC is right angled triangle

∠A is an acute angle and ∠C = 90^o

tan A = 5/12

⇒ BC/AC =5/12

Let BC = 5x and AC = 12x

From right angled ∆ABC

By Pythagoras theorem, we get

AB² = (5x)² + (12x)²

⇒ AB² = 25x² + 144x²

⇒ AB² = 169x²

(i) cos A = Base/ Hypotenuse

= AC / AB

= 12x/13x

=12/13

(ii) cosec A = Hypotenuse/perpendicular

= AC / BC

= 13x /5x

= 13/5

cosec A – cot A = 13/5 – 12/5

= (13-12)/5

= 1/5

6. (a) In ∆ABC, ∠A = 90°. If AB = 7 cm and BC – AC = 1 cm, find:

(i) sin C

(ii) tan B

(b) In ∆PQR, ∠Q = 90°. If PQ = 40 cm and PR + QR = 50 cm, find:

(i) sin P

(ii) cos P

(iii) tan R.

Solution

(a) In right ∆ABC

∠A = 90°

AB = 7 cm

BC – AC = 1 cm

⇒ BC = 1 + AC

We know that

BC² = AB² + AC²

Substituting the value of BC

(1 + AC)² = AB² + AC²

⇒ 1 + AC² + 2AC = 7² + AC²

By further calculation

1 + AC² + 2AC = 49 AC²

⇒ 2AC = 49 – 1 – 48

So we get

AC = 48/2 = 24 cm

Here,

BC = 1 + AC

Substituting the value

BC = 1 + 24 = 25 cm

(i) sin C = AB/BC = 7/25

(ii) tan B = AC/AB = 24/7

(b) In right ∆PQR

∠Q = 90°

PQ = 40 cm

⇒ PQ + QR = 50 cm

We can write it as

PQ = 50 – QR

Using Pythagoras theorem

PR² = PQ² + QR²

⇒ (50 – QR)² = (40)² + QR²

By further calculation

2500 + QR² – 100QR = 1600 + QR²

So we get

2500 – 1600 = 100QR

⇒ 100QR = 900

By division

QR = 900/100 = 9

We get

PR = 50 – 9 = 41

(i) sin P = QR/PR = 9/41

(ii) cos P = PQ/PR = 40/41

(iii) tan R = PQ/QR = 40/9

7. In triangle ABC, AB = 15 cm, AC = 15 cm and BC = 18 cm. Find

(i) cos ∠ABC

(ii) sin ∠ACB.

Solution

Here ABC is a triangle in which

AB = 15 cm, AC = 15 cm and BC = 18 cm

Draw AD perpendicular to BC , D is mid-point of BC.

Then, BD – DC = 9 cm

in right angled triangle ABD,

By Pythagoras theorem, we get

AB² = AD² + BD²

⇒ AD² = AB² – BD²

⇒ AD² = (15)² – (9)²

⇒ AD² = 225 – 81

⇒ AD² = 144

⇒ AD – 12 cm

(i) cos ∠ABC = Base/Hypotenuse

(In right angled ∆ABD, ∠ABC = ∠ABD)

= BD / AB

= 9/15

= 3/5

(ii) sin ∠ACB = sin ∠ACD

= perpendicular/ Hypotenuse

= AD/AC

= 12/15

= 4/5

8. (a) In the figure (1) given below, ∆ABC is isosceles with AB = AC = 5 cm and BC = 6 cm. Find

(i) sin C

(ii) tan B

(iii) tan C – cot B.

(b) In the figure (2) given below, ∆ABC is right-angled at B. Given that ∠ACB = θ, side AB = 2 units and side BC = 1 unit, find the value of sin² θ + tan² θ.

(c) In the figure (3) given below, AD is perpendicular to BC, BD = 15 cm, sin B = 4/5 and tan C = 1.

(i) Calculate the lengths of AD, AB, DC and AC.

(ii) Show that tan² B – 1/cos² B = – 1.

Solution

(a) It is given that

∆ABC is isosceles with AB = AC = 5 cm and BC = 6 cm

Construct AD perpendicular to BC

D is the mid point of BC

So BD = CD

Here

BD = CD = 6/2 = 3 cm

In right angled ∆ABD

Using Pythagoras theorem

AB² = AD² + BD²

We can write it as

AD² = AB² – BD²

Substituting the values

AD² = 5² – 3²

By further calculation

AD² = 25 – 9 = 16

So we get

AD² = 4²

AD = 4 cm

(i) In right angled ∆ACD

sin C = perpendicular/hypotenuse

sin C = AD/AC = 4/5

(ii) In right angled ∆ABD

tan B = perpendicular/base

tan B = AD/BD = 4/3

(iii) In right angled ∆ACD

tan C = perpendicular/base

⇒ tan C = AD/CD = 4/3

In right angled ∆ABD

cot B = base/perpendicular

⇒ cot B = BD/AD = ¾

Here

tan C – cot B = 4/3 – ¾

Taking LCM,

tan C – cot B = (16 – 9)/12 = 7/12

(b) It is given that

∆ABC is right-angled at B

AB = 2 units and BC = 1 unit

In right angled ∆ABC

Using Pythagoras theorem

AC² = AB² + BC²

Substituting the values

AC² = 2² + 1²

⇒ AC² = 4 + 1 = 5

So we get

AC² = 5

⇒ AC = √5 units

In right angled ∆ABC

sin θ = perpendicular/hypotenuse

⇒ sin θ = AB/AC = 2/√5

In right angled ∆ABC

tan θ = perpendicular/base

⇒ tan θ = AB/BC = 2/1

We know that

sin² θ + tan² θ = (2/√5)² + (2/1)²

By further calculation

= 4/5 + 4/1

Taking LCM

= (4 + 20)/5

= 24/5

= 4 4/5

(c) (i) In ∆ABC

AD is perpendicular to BC

BD = 15 cm

sin B = 4/5

tan C = 1

In ∆ABD

sin B = perpendicular/hypotenuse

⇒ sin B = AD/AB = 4/5

Consider AD = 4x and AB = 5x

Using Pythagoras theorem

In right angled ∆ABD,

AB² = AD² + BD²

We can write it as

BD² = AB² – AD²

Substituting the values

(15)² = (5x)² – (4x)²

⇒ 225 = 25x² – 16x²

By further calculation

225 = 9x²

⇒ x² = 225/9 = 25

So we get

x = √25 = 5

Here

AD = 4 × 5 = 20

AB = 5 × 5 = 25

In right angled ∆ACD

tan C = perpendicular/base

So we get

tan C = AD/CD = 1/1

Consider AD = X then CD = x

In right angled ∆ADC

Using Pythagoras theorem

AC² = AD² + CD²

Substituting the values

AC² = x² + x² …(1)

So the equation becomes

AC² = 20² + 20²

⇒ AC² = 400 + 400 = 800

So we get

AC = √800 = 20√2

Length of AD = 20 cm

Length of AB = 25 cm

Length of DC = 20 cm

Length of AC = 20√2 cm

(ii) In right angled ∆ABD

tan B = perpendicular/base

So we get

tan B = AD/BD

Substituting the values

tan B = 20/15 = 4/3

In right angled ∆ABD,

cos B = base/hypotenuse

So we get,

cos B = BD/AB

Substituting the values

cos B = 15/25 = 3/5

Here,

LHS = tan² B – 1/cos² B

Substituting the values

= (4/3)² – 1/(3/5)²

By further calculation

= (4)²/(3)² – (5)²/(3)²

= 16/9 – 25/9

So we get

= (16 – 25)/9

= -9/9

= – 1

= RHS

Hence, proved.

9. If sin θ =3/5 and θ is acute angle, find

(i) cos θ

(ii) tan θ.

Solution

Let ∆ ABC be a right angled at B

Let ∠ACB = θ

Given that, sin θ = 3/5

AB/AC = 3/5

Let AB = 3x

then AC = 5x

In right angled ∆ ABC,

By Pythagoras theorem,

We get,

(5x)² = (3x)² + BC²

⇒ BC² = (5x)² – (3x)²

⇒ BC² = (2x)²

⇒ BC = 4x

(i) cos θ = Base/ Hypotenuse

= BC / AC

= 4x /5x

= 4/5

(ii) tan θ = perpendicular/Base

= AB/BC

= 3x/4x

= ¾

10. Given that tan θ = 5/12 and θ is an acute angle, find sin θ and cos θ.

Solution

Consider ∆ ABC be right angled at B and ∠ACB = θ

It is given that

tan θ = 5/12

AB/BC = 5/12

Consider AB = 5x and BC = 12x

In right angled ∆ ABC

Using Pythagoras theorem

AC² = AB² + BC²

Substituting the values

AC² = (5x)² + (12x)²

By further calculation

AC² = 25x² + 144x² = 169x²

So we get

AC² = (13x)²

⇒ AC = 13x

In right angled ∆ ABC

sin θ = perpendicular/hypotenuse

So we get

sin θ = AB/AC = 5x/13x = 5/13

In right angled ∆ ABC

cos θ = base/hypotenuse

So we get

cos θ = BC/AC

Substituting the values

cos θ = 12x/13x = 12/13

11. If sin θ = 6/10, find the value of cos θ + tan θ.

Solution

Consider ∆ ABC be right angled at B and ∠ACB = θ

It is given that

sin θ = AB/AC

⇒ sin θ = 6/10

Take AB = 6x then AC = 10x

In right angled ∆ ABC

Using Pythagoras theorem

AC² = AB² + BC²

Substituting the values

(10x)² = (6x)² + BC²

By further calculation

BC² = 100x² – 36x² = 64x²

So we get

BC² = (8x)²

⇒ BC = 8x

In right angled ∆ ABC

cos θ = base/hypotenuse

⇒ cos θ = BC/AC

Substituting the values

cos θ = 8x/10x = 4/5

In right angled ∆ ABC

tan θ = perpendicular/base

⇒ tan θ = AB/BC

Substituting the values

tan θ = 6x/8x = ¾

Here

cos θ + tan θ = 4/5 + ¾

Taking LCM

= (16 + 15)/ 20

= 31/20

= 1 11/20

12. If tan = 4/3, find the value of sin θ + cos θ (both sin θ and cos θ are positive).

Solution

Let ∆ABC be a right angled

∠ACB = θ

Given that, tan θ = 4/3

(AB/BC = 4/3)

Give that, tan θ = 4/3

(AB/BC = 4/3)

Let AB = 4x,

then BC = 3x

In right angled ∆ABC

By Pythagoras theorem, we get

AC² = AB² + BC²

⇒ AC² = AB² + BC²

⇒ AC² = AB² + BC²

⇒ AC² = AB² + BC²

⇒ (AC² = (4x)² + (3x)²

⇒ AC² = 16x² + 9x²

⇒ AC² = 25x²

⇒ AC² = (5x)²

⇒ AC = 5x

Sin θ = perpendicular/Hypotenuse

= AB/AC

= 4x/5x

= 4/5

cos θ = Base/Hypotenuse

= BC/AC

= 3x/5x

= 3/5

sin θ + cos θ

= 4/5 + 3/5

= (4 + 3)/5

= 7/5

Hence, Sin θ + cos θ = 7/5 = 1 2/5

13. 1f cosec = √5 and θ is less than 90°, find the value of cot θ – cos θ.

Solution

Given cosec θ = √5/1 = OP/PM

OP = √5 and PM = 1

Now OP² = OM² + PM² using Pythagoras theorem

(√5)² = OM² + 1²

⇒ 5 = OM² + 1

⇒ OM² = 5 – 1

⇒ OM² = 4

⇒ OM = 2

Now cot θ = OM/PM

= 2/1

= 2

cos θ = OM/OP

= 2/√5

Now,

cot θ – cos θ = 2 – (2/√5)

= 2 (√5 – 1)/ √5

14. Given sin θ = p/q, find cos θ + sin θ in terms of p and q.

Solution

Given that sin θ = p/q

Which implies,

AB/AC = p/q

Let AB = px

And then AC = qx

In right angled triangle ABC

By Pythagoras theorem,

We get

AC² = AB² + BC²

⇒ BC² = AC² – AB²

⇒ BC² = q²x² – p²x²

⇒ BC² = (q² – p²)x²

⇒ BC = √( q² – p²)x

In right angled triangle ABC,

cos θ = base/hypotenuse

= BC/AC

= √( q² – p²)x/qx

= √( q² – p²)/q

Now,

Sin θ + cos θ = p/q + √(q²–p²)/q

= [p + √(q²–p²)]/q

15. If θ is an acute angle and tan = 8/15, find the value of sec θ + cosec θ.

Solution

Given tan θ = 8/15

θ is an acute angle

in the figure triangle OMP is a right angled triangle,

∠M = 90^o and ∠Q = θ

tanθ = PM/OL = 8/15

Therefore, PM = 8, OM = 15

But OP² = OM² + PM² using Pythagoras theorem,

= 15² + 8²

= 225 + 64

= 289

= 17²

Therefore, OP = 17

sec θ = OP/OM = 17/15

cosec θ = OP/PM = 17/8

Now,

sec θ + cosec θ = (17/15) + (17/8)

= (136 + 255)/120

= 391/120

= 3 31/120

16. Given A is an acute angle and 13 sin A = 5, Evaluate:

(5 sin A – 2 cos A)/ tan A.

Solution

Let triangle ABC be a right angled triangle at B and A is an acute angle

Given that 13 sin A = 5

Sin A = 5/13

AB/AC = 5/13

Let AB = 5x

AC = 13 x

In right angled triangle ABC,

Using Pythagoras theorem,

We get

AC² = AB² + BC²

⇒ BC² = AC² – BC²

⇒ BC² = (13x)² – (5x)²

⇒ BC² = 169x² – 25x²

⇒ BC² = 144x²

⇒ BC = 12x

⇒ sin A = 5/13

⇒ cos A = base/ hypotenuse

= BC/AC

= 12x/ 13x

= 12/13

Tan A = perpendicular/ base

= AB/BC

= 5x/ 12x

= 5/ 12

Now,

(5 sin A – 2 cos A)/tan A = [(5) (5/13) – (2) (12/13)]/(5/12)

= (1/13)/(5/12)

= 12/65

Hence (5 sin A – 2 cos A)/tan A = 12/65

17. Given A is an acute angle and cosec A = √2, find the value of

(2 sin² A + 3 cot² A)/ (tan² A – cos² A).

Solution

Let triangle ABC be a right angled at B and A is a acute angle.

Given that cosec A = √2

Which implies,

AC/BC = √2/1

Let AC = √2x

Then BC = x

In right angled triangle ABC

By using Pythagoras theorem,

We get

AC² = AB² + BC²

⇒ (√2x)² = AB² + x²

⇒ AB² = 2x² – x²

⇒ AB = x

sin A = perpendicular/ hypotenuse

= BC/AC

= 1/ √2

cot A = base/ perpendicular

= x/x

= 1

Tan A = perpendicular/ base

= BC/AB

= x/x

= 1

cos A = base/ hypotenuse

= AB/AC

= x/ √2x

= 1/√2

Substituting these values we get

2 sin²A + 3 cot²A/(tan²A – cos²A) = 8

18. The diagonals AC and BD of a rhombus ABCD meet at O. If AC = 8 cm and BD = 6 cm, find sin ∠OCD.

Solution

It is given that

Diagonals AC and BD of rhombus ABCD meet at O

AC = 8 cm and BD = 6 cm

O is the mid point of AC

We know that

AO = OC = AC/2 = 8/2 = 4 cm

O is the mid point of BD

BO = OD = BD/2 = 6/2 = 3 cm

In right angled ∆COD

CD² = OC² + OD²

Substituting the values

CD² = 4² + 3²

So we get

CD² = 16 + 9 = 25

⇒ CD² = 5²

⇒ CD = 5 cm

In right angled ∆COD

sin ∠OCD = perpendicular/ hypotenuse

So we get

sin ∠OCD = OD/CD = 3/5

19. If tan θ = 5/12, find the value of (cos θ + sin θ)/(cos θ – sin θ).

Solution

Consider ∆ABC be right angled at B and ∠ACB = θ

It is given that

tan θ = AB/BC = 5/12

Take AB = 5x then BC = 12x

In right angled ∆ABC,

Using Pythagoras theorem

AC² = AB² + BC²

Substituting the values

AC² = (5x)² + (12x)²

By further calculation

AC² = 25x² + 144x² = 169x²

So we get

AC² = (13x)²

⇒ AC = 13x

In right angled ∆ABC

cos θ = base/hypotenuse

cos θ = BC/AC

Substituting the values

cos θ = 12x/13x = 12/13

In right angled ∆ABC

sin θ = perpendicular/hypotenuse

⇒ sin θ = AB/AC

Substituting the values

sin θ = 5x/13x = 5/13

Here

(cos θ + sin θ)/(cos θ – sin θ) = [12/13 + 5/13]/ [12/13 – 5/13]

Taking LCM

= [(12 + 5)/13]/[(12 – 5)/13]

So we get

= (17/13)/(7/13)

= 17/13 × 13/7

= 17/7

Therefore, (cos θ + sin θ)/(cos θ – sin θ) = 17/7

20. Given 5 cos A – 12 sin A = 0, find the value of (sin A + cos A)/ (2 cos A – sin A).

Solution

It is given that

5 cos A – 12 sin A = 0

We can write it as

5 cos A = 12 sin A

So we get

sin A/cos A = 5/12

We know that sin A/ cos A = tan A

tan A = 5/12

Consider ∆ABC right angled at B and ∠A is acute angle

Here,

tan A = BC/AB = 5/12

Take BC = 5x then AB = 12x

In right angled ∆ABC

Using Pythagoras theorem

AC² = BC² + AB²

Substituting the values

AC² = (5x)² + (12x)²

⇒ AC² = 25x² + 144x² = 169x²

So we get

AC² = (13x)²

⇒ AC = 13x

In right angled ∆ABC

sin A = perpendicular/hypotenuse

So we get

sin A = BC/AC = 5x/13x = 5/13

In right angled ∆ABC

cos A = base/hypotenuse

So we get

⇒ cos A = AB/AC = 12x/13x = 12/13

Here

(sin A + cos A)/(2 cos A – sin A) = [(5/13) + (12/13)]/[(2× 12/13) – 5/13]

By further calculation

= [(5+12)/13]/[24/13 – 5/13]

So we get

= [(5+12)/13]/[(24 – 5)/13]

= (17/13)/(19/13)

= 17/13 × 13/19

= 17/19

Therefore, (sin A + cos A)/ (2 cos A – sin A) = 17/19

21. If tan θ = p/q, find the value of (p sin θ – q cos θ)/ (p sin θ + q cos θ).

Solution

It is given that

tan θ = p/q

Consider ∆ABC be right angled at B and ∠BCA = θ

tan θ = BC/AB = p/q

BC = px then AB = qx

In right angled ∆ABC

Using Pythagoras theorem

AC² = BC² + AB²

Substituting the values

AC² = (px)² + (qx)²

⇒ AC² = p²x² + q²x²

⇒ AC² = x² (p² + q²)

So we get

AC = √x² (p² + q²)

⇒ AC = x(√p² + q²)

In right angled ∆ABC

sin θ = perpendicular/hypotenuse

⇒ sin θ = BC/AC

Substituting the values

sin θ = px/x(√p² + q²)

So we get

sin θ = p/(√p² + q²)

In right angled ∆ABC

cos θ = base/hypotenuse

⇒ cos θ = AB/AC

Substituting the values

cos θ = qx/x(√p² + q²)

So we get

cos θ = q/(√p² + q²)

Here,

22. If 3 cot θ = 4, find the value of (5 sinθ – 3 cosθ)/(5 sinθ + 3 cosθ).

Solution

It is given that

3 cot θ = 4

⇒ cot θ = 4/3

Consider ∆ABC be right angled at B and ∠ACB = θ

cot θ = BC/AB = 4/3

Take BC = 4x then AB = 3x

In right angled ∆ABC

Using Pythagoras theorem

AC² = AB² + BC²

Substituting the values

AC² = (3x)² + (4x)²

⇒ AC² = 9x² + 16x² = 25x²

So we get

AC² = (5x)²

⇒ AC = 5x

In right angled ∆ABC

sin θ = perpendicular/hypotenuse

⇒ sin θ = AB/AC

Substituting the values

sin θ = 3x/5x = 3/5

In right angled ∆ABC

cos θ = base/hypotenuse

⇒ cos θ = BC/AC

Substituting the values

cos θ = 4x/5x = 4/5

23. (i) If 5 cosθ – 12 sinθ = 0, find the value of (sin θ + cos θ)/(2 cosθ – sinθ).

(ii) If cosecθ = 13/12, find the value of (2 sinθ – 3 cosθ)/(4 sinθ – 9 cosθ).

Solution

(i) It is given that

5 cosθ – 12 sinθ = 0

We can write it as

5 cosθ = 12 sinθ

⇒ sin θ/cos θ = 5/12

⇒ tan θ = 5/12

Dividing both numerator and denominator by cos θ

(ii) It is given that

cosec θ = 13/12

We know that cosec θ = 1/sin θ

1/sin θ = 13/12

⇒ sin θ = 12/13

Here cos² θ = 1 – sin² θ

Substituting the values

= 1 – (12/13)²

By further calculation

= 1 – 144/169

Taking LCM

= (169 – 144)/ 169

= 25/169

So we get

= (5/13)²

cos θ = 5/13

Here

24. If 5 sin θ = 3, find the value of (secθ – tanθ)/(secθ + tanθ).

Solution

Consider ∆ABC be right angled at B and ∠ACB = θ

It is given that

5 sin θ = 3

sin θ = AB/AC = 3/5

Take AB = 3x then AC = 5x

In right angled ∆ABC

Using Pythagoras theorem

AC² = AB² + BC²

⇒ BC² = AC² – AB²

Substituting the values

BC² = (5x)² – (3x)²

So we get

BC² = 25x² – 9x² = 16x²

⇒ BC² = (4x)²

⇒ BC = 4x

In right angled ∆ABC

sec θ = hypotenuse/base

⇒ sec θ = AC/BC = 5x/4x = 5/4

In right angled ∆ABC

tan θ = perpendicular/base

⇒ tan θ = AB/BC = 3x/4x = ¾

25. If θ is an acute angle and sin θ = cos θ, find the value of 2 tan² θ + sin² θ – 1.

Solution

Consider ∆ABC be right angled at B and ∠ACB = θ

It is given that

sin θ = cos θ

⇒ sin θ/cos θ = 1

tan θ = AB/BC = 1

Take AB = x then BC = x

In right angled ∆ABC,

Using Pythagoras theorem

AC² = AB² + BC²

⇒ AC² = x² + x² = 2x²

So we get

AC = √2x²

⇒ AC = (√2)x

In right angled ∆ABC

sin θ = perpendicular/hypotenuse

sin θ = AB/AC = x/√2x = 1/√2

Here

2 tan² θ + sin² θ – 1 = 2×(1)² + (1/√2)² – 1

By further calculation

= (2×1) + ½ – 1

= 2 + ½ – 1

= 1+ ½

Taking LCM

= (2 + 1)/2

= 3/2

Therefore, 2 tan² θ + sin² θ – 1 = 3/2.

26. Prove the following:

(i) cos θ tan θ = sin θ

(ii) sin θ cot θ = cos θ

(iii) sin² θ/ cos θ + cos θ = 1/ cos θ.

Solution

(i) cos θ tan θ = sin θ

LHS = cos θ tan θ

We know that tan θ = sin θ/cos θ

= cos θ (sin θ/cos θ)

So we get

= 1× sin θ/1

= sin θ

= RHS

Therefore, LHS = RHS.

(ii) sin θ cot θ = cos θ

LHS = sin θ cot θ

We know that cot θ = cos θ/sin θ

= sin θ (cos θ/sin θ)

= 1× cos θ/1

= cos θ

= RHS

Therefore, LHS = RHS.

(iii) sin²θ/cosθ + cosθ = 1/cosθ

LHS = sin²θ/cosθ + cosθ/1

Taking LCM

= (sin²θ + cos²θ)/cosθ

We know that,

sin²θ + cos²θ = 1

= 1/cos θ

= RHS

Therefore, LHS = RHS.

27. If in ∆ABC, ∠C = 90° and tan A = ¾, prove that sin A cos B + cos A sin B = 1.

Solution

It is given that

tan A = BC/AC = ¾

Using Pythagoras theorem

AB² = AC² + BC²

Substituting the values

= 4² + 3²

= 16 + 9

= 25

= 5²

So we get AB = 5

Here

sin A = BC/AC = 3/5

cos A = AC/AB = 4/5

cos B = BC/AB = 3/5

sin B = AC/AB = 4/5

LHS = sin A cos B + cos A sin B

Substituting the values

= (3/5 × 3/5) + (4/5 × 4/5)

By further calculation

= 9/25 + 16/25

= (9 + 16)/ 25

= 25/25

= 1

= RHS

Therefore, LHS = RHS.

28. (a) In figure (1) given below, ∆ABC is right-angled at B and ∆BRS is right-angled at R. If AB = 18 cm, BC = 7.5 cm, RS = 5 cm, ∠BSR = x° and ∠SAB = y°, then find:

(i) tan x°

(ii) sin y°.

(b) In the figure (2) given below, ∆ABC is right angled at B and BD is perpendicular to AC. Find

(i) cos ∠CBD

(ii) cot ∠ABD.

Solution

(a) ∆ABC is right-angled at B, ∆BSC is right-angled at S and ∆BRS is right-angled at R

It is given that

AB = 18 cm, BC = 7.5 cm, RS = 5 cm, ∠BSR = x° and ∠SAB = y°

By Geometry ∆ARS and ∆ABC are similar

AR/AB = RS/BC

Substituting the values

AR/18 = 5/7.5

By further calculation

AR = (5×18)/7.5 = (1×18)/1.5

Multiply both numerator and denominator by 10

AR = (18×10)/15

⇒ AR = (10×6)/5

⇒ AR = (2×6)/1 = 12

So we get

RB = AB – AR

⇒ RB = 18 – 12 = 6

In right angled ∆ABC

Using Pythagoras theorem

AC² = AB² + BC²

Substituting the values

AC² = 18² + 7.5²

By further calculation

AC² = 324 + 56.25 = 380.25

⇒ AC = √380.25 = 19.5 cm

(i) In right angled ∆BSR

tan x° = perpendicular/base

tan x° = RB/RS = 6/5

(ii) In right angled ∆ASR

sin y° = perpendicular/hypotenuse

Using Pythagoras theorem

AS² = 12² + 5²

By further calculation

AS² = 144 + 25 = 169

⇒ AS = √169 = 13 cm

So we get

sin y° = RS/AS = 5/13

(b) We know that

∆ABC is right angled at B and BD is perpendicular to AC

In right angled ∆ABC

Using Pythagoras theorem

AC² = AB² + BC²

Substituting the values

AC² = 12² + 5²

By further calculation

AC² = 144 + 25 = 169

So we get

AC² = (13)²

⇒ AC = 13

By Geometry ∠CBD = ∠A and ∠ABD = ∠C

(i) cos ∠CBD = cos ∠A = base/hypotenuse

In right angled ∆ABC

cos ∠CBD = cos ∠A = AB/AC = 12/13

(ii) cos ∠ABD = cos ∠C = base/perpendicular

In right angled ∆ABC

cos ∠ABD = cos ∠C = BC/AB = 5/12

29. In the adjoining figure, ABCD is a rectangle. Its diagonal AC = 15 cm and ∠ACD = α. If cot α = 3/2, find the perimeter and the area of the rectangle.

Solution

In right ∆ADC

cot α = CD/AD = 3/2

Take CD = 3x then AD = 2x

Using Pythagoras theorem

AC² = CD² + AD²

Substituting the values

(15)² = (3x)² + (2x)²

By further calculation

13x² = 225

⇒ x² = 225/13

So we get

x = √225/13 = 15/√13

Length of rectangle (l) = 3x = (3 × 15)/ √13 = 45/√13 cm

Breadth of rectangle (b) = 2x = (2×15)/√13 = 30/√13 cm

(i) Perimeter of rectangle = 2 (l + b)

Substituting the values of l and b

= 2 (45/√13 + 30/√13)

So we get

= 2 × 75/√13

= 150/√13 cm

(ii) Area of rectangle = l × b

Substituting the values of l and b

= 45/√13 × 30/√13

So we get

= 1350/13

= 103 (11/13) cm²

30. Using the measurements given in the figure alongside,

(a) Find the values of:

(i) sin ϕ

(ii) tan θ.

(b) Write an expression for AD in terms of θ.

Solution

From the figure

BC = 12, BD = 13

In right angled ∆BCD

Using Pythagoras theorem

BD² = BC² + CD²

It can be written as

CD² = BD² – BC²

Substituting the values

CD² = (13)² – (12)²

⇒ CD² = 169 – 144 = 25

So we get

CD = √25 = 5

Construct BE perpendicular to AB

CD = BE = 5 and EA = AE = 14 – 5 = 9

(a) (i) sin ϕ = perpendicular/hypotenuse

In right angled ∆BCD

sin ϕ = CD/BD = 5/13

(ii) tan θ = perpendicular/hypotenuse

In right angled ∆AED

tan θ = ED/AE = BC/AE = 12/9 = 4/3 (Since ED = BC)

(b) In right angled ∆AED

sin θ = perpendicular/hypotenuse

cos θ = base/perpendicular

We can write it as

sin θ = ED/AD or cos θ = AE/AD

AD = ED/sin θ or AD = AE/cos θ

Substituting the values

AD = 12/sin θ or AD = 9/cos θ

Therefore, AD = 12/ sin θ or AD = 9/cos θ.

31. Prove the following:

(i) (sin A + cos A)² + (sin A – cos A)² = 2

(ii) cot² A – 1/sin² A + 1 = 0

(iii) 1/(1 + tan² A) + 1/(1 + cot² A) = 1

Solution

(i) (sin A + cos A)² + (sin A – cos A)² = 2

LHS = (sin A + cos A)² + (sin A – cos A)²

Using the formula

(a + b)² = a² + b² + 2ab and (a – b)² = a² + b² – 2ab

= [(sin A)² + (cos A)² + 2 sin A cos A] + [(sin A)² + (cos A)² – 2 sin A cos A]

By further calculation

= sin² A + cos² A + 2 sin A cos A + sin² A + cos² A – 2 sin A cos A

= sin² A + cos² A + sin² A + cos² A

= 2 sin² A + 2 cos² A

We know that sin² A + cos² A = 1

= 2 (sin² A + cos² A)

= 2 (1)

= 2

= RHS

Therefore, LHS = RHS.

(ii) cot² A – 1/sin² A + 1 = 0

LHS = cot² A – 1/sin² A + 1

We know that

1/sin A = cosec A

= cot² A – cosec² A + 1

= (1 + cot² A) – cosec² A

We know that 1 + cot² A = cosec² A

= cosec² A – cosec² A

= 0

= RHS

Therefore, LHS = RHS.

(iii) 1/(1 + tan² A) + 1/(1 + cot² A) = 1

LHS = 1/(1 + tan² A) + 1/(1 + cot² A)

We know that

sec² A – tan² A = 1

⇒ sec² A = 1 + tan² A

⇒ cosec² A – cot² A = 1

⇒ cosec² A = 1 + cot² A

So we get

= 1/sec² A + 1/cosec² A

Here,

1/sec A = cos A and 1/cosec A = sin A

= cos² A + sin² A

= 1

= RHS

Therefore, LHS = RHS.

32. Simplify

Solution

We know that 1 = sin² θ + cos² θ

= √cos² θ/sin² θ

= cos θ/ sin θ

Here cos θ/sin θ = cot θ

= cot θ

Therefore,

 = cot θ.

33. If sin θ + cosec θ = 2, find the value of sin² θ + cosec² θ.

Solution

It is given that

sin θ + cosec θ = 2

⇒ sin θ + 1/sin θ = 2

By further calculation

sin² θ + 1 = 2 sin θ

⇒ sin² θ – 2 sin θ + 1 = 0

So we get,

(sin θ – 1)² = 0

⇒ sin θ – 1 = 0

⇒ sin θ = 1

Here,

sin² θ + cosec² θ = sin² θ + 1/sin² θ

Substituting the values

= 1² + 1/1²

= 1 + 1/1

= 1 + 1

= 2

34. If x = a cos θ + b sin θ and y = a sin θ – b cos θ, prove that x² + y² = a² + b².

Solution

It is given that

x = a cos θ + b sin θ …(1)

y = a sin θ – b cos θ …(2)

By squaring and adding both the equations

x² + y² = (a cos θ + b sin θ)² + (a sin θ – b cos θ)²

Using the formula

(a + b)² = a² + b² + 2ab and (a – b)² = a² + b² – 2ab

= [(a cosθ)² + (b sinθ)² + 2 (a cosθ) (b sinθ)] + [(a sinθ)² + (b cosθ)² – 2 (a sinθ) (b cosθ)]

By further calculation

= a² cos²θ + b² sin²θ + 2 ab sinθ cosθ + a² sin²θ + b² cos²θ – 2 ab sinθ cosθ

= a² cos²θ + b² sin²θ + a² sin²θ + b² cos²θ

So we get,

= a² (cos²θ + sin²θ) + b² (sin²θ + cos²θ)

Here sin²θ + cos²θ = 1

= a² (1) + b² (1)

= a² + b²

Therefore, x² + y² = a² + b².

Chapter test

1. (a)From the figure (i) given below, calculate all the six t-ratios for both acute……

(b)From the figure (ii) given below, find the values of x and y in terms of t-ratios

Solution

(a) From right angled triangle ABC,

By Pythagoras theorem, we get

AC² = AB² + BC²

⇒ AB² = AC² – BC²

⇒ AB² = (3)² – (2)²

⇒ AB² = 9 – 4

⇒ AB² = 5

⇒ AB = √5

(i) sin A = perpendicular/ hypotenuse

= BC/AC

= 2/3

(ii) cos A = base/ hypotenuse

= AB/AC

= √5/3

(iii) tan A = perpendicular/ base

= BC/AB

= 2/ √5

(iv) cot A = base/perpendicular

= AB/ BC

= √5/2

(v) sec A = hypotenuse/ base

= AC/AB

= 3/ √5

(vi) cosec A = hypotenuse/perpendicular

= AC/BC

= 3/2

(b) From right angled triangle ABC,

∠BAC = θ

Then we know that,

cotθ = base/ perpendicular

= AB/ BC

= x/ 10

x = 10 cotθ

Also,

cosecθ = hypotenuse/ perpendicular

= AC/ BC

= y/ 10

y = 10 cosec θ

Therefore, x = 10 cot θ and y = 10 cosec θ.

2. (a) From the figure (1) given below, find the values of:

(i) sin ∠ABC

(ii) tan x – cos x + 3 sin x.

(b) From the figure (2) given below, find the values of:

(i) 5 sin x

(ii) 7 tan x

(iii) 5 cos x – 17 sin y – tan x.

Solution

(a) From the figure

BC = 12, CD = 9 and BC = 20

In right angled ∆ABC,

Using Pythagoras theorem

AB² = AC² + BC²

It can be written as

AC² = AB² – BC²

Substituting the values

AC² = (20)² – (12)²

By further calculation

AC² = 400 – 144 = 256

So we get

AC² = (16)²

⇒ AC = 16

In right angled ∆BCD

Using Pythagoras theorem

BD² = BC² + CD²

Substituting the values

BD² = 12² + 9²

By further calculation

BD² = 144 + 81 = 225

So we get

BD² = (15)²

⇒ BD = 15

(i) In right angled ∆BCD

sin ∠ABC = perpendicular/hypotenuse

So we get

sin ∠ABC = AC/AB = 16/20 = 4/5

(ii) In right angled ∆BCD

tan x = perpendicular/base

So we get

tan x = BC/CD = 12/9 = 4/3

In right angled ∆BCD

cos x = base/hypotenuse

So we get

cos x = CD/BD = 9/15 = 3/5

In right angled ∆BCD

sin x = perpendicular/hypotenuse

So we get

sin x = BC/BD = 12/15 = 4/5

⇒ tan x – cos x + 3 sin x = 4/3 – 3/5 + (3× 4/5)

By further calculation

= 4/3 – 3/5 + 12/5

Taking LCM

= [(4×5) – (3×3) + (12×3)]/15

So we get

= (20 – 9 + 36)/15

= (56 – 9)/15

= 27/15

= 3 (2/15)

Therefore, tan x – cos x + 3 sin x = 3 2/15.

(b) In the figure

AC = 17, AB = 25, AD = 15

In right angled ∆ACD

Using Pythagoras theorem

AC² = AD² + CD²

Substituting the values

(17)² = (15)² + (CD)²

By further calculation

CD² = (17)² – (15)²

⇒ CD² = 289 – 225 = 64

So we get

CD² = 8²

⇒ CD = 8

In right angled ∆ABD

Using Pythagoras theorem

AB² = AD² + BD²

Substituting the values

(25)² = (15)² + BD²

By further calculation

BD² = (25)² – (15)²

⇒ BD² = 625 – 225 = 400

So we get

BD² = (20)²

⇒ BD = 20

(i) In right angled ∆ABD

5 sin x = 5 (perpendicular/hypotenuse)

So we get

= 5 (AD/AB)

= 5 × 15/25

= 15/5

= 3

(ii) In right angled ∆ABD

7 tan x = 7 (perpendicular/base)

So we get

= 7 (AD/AB)

= 7× 15/20

= 7× ¾

= 21/4

= 5 ¼

(iii) In right angled ∆ABD

cos x = base/hypotenuse

So we get

cos x = BD/AB = 20/25 = 4/5

In right angled ∆ACD

sin y = perpendicular/hypotenuse

So we get

sin y = CD/AC = 8/17

In right angled ∆ABD

tan x = perpendicular/base

So we get

tan x = AD/BD = 15/20 = ¾

5 cosx – 17 siny – tanx = (5× 4/5) – (17× 8/17) – ¾

It can be written as

= 4/1 – 8/1 – ¾

Taking LCM

= (16 – 32 – 3)/4

= (16 – 35)/4

So we get

= –19/4

= –4 ¾

Therefore, 5 cos x – 17 sin y – tan x = – 4 ¾.

3. If q cosθ = p, find tanθ – cotθ in terms of p and q.

Solution

Consider ABC as a triangle right angled at B and ∠ACB = θ

It is given that

q cos θ = p

cos θ = BC/AC = p/q

Take BC = px then AC = qx

In right angled ∆ABC

Using Pythagoras theorem

AC² = AB² + BC²

It can be written as

AB² = AC² – BC²

Substituting the values

AB² = (qx)² – (px)²

⇒ AB² = q²x² – p²x²

Taking out the common terms

AB² = (q² – p²)x²

So we get

AB = √(q² – p²) x²

⇒ AB = (√q² – p²)x

In right angled ∆ABC

tan θ = perpendicular/base

So we get

tan θ = AB/BC = [(√q² – p²)x]/px

⇒ tan θ = (√q² – p²)/p

In right angled ∆ABC

cot θ = base/perpendicular

So we get

cot θ = BC/AB = px/[(√q² – p²)x]

⇒ cot [(√q² – p²)x] = p/(√q² – p²)

4. Given 4 sin θ = 3 cos θ, find the values of:

(i) sin θ

(ii) cos θ

(iii) cot² θ – cosec² θ.

Solution

It is given that

4 sin θ = 3 cos θ

⇒ sin θ/cos θ = ¾

⇒ tan θ = ¾

Consider ∆ABC right angled at B and ∠ACB = θ

tan θ = perpendicular/base

Substituting the values

¾ = AB/BC

⇒ AB/BC = ¾

Take AB = 3x then BC = 4x

In right angled ∆ABC

Using Pythagoras theorem

AC² = AB² + BC²

Substituting the values

AC² = (3x)² + (4x)²

By further calculation

AC² = 9x² + 16x² = 25x²

So we get

AC² = (5x)²

⇒ AC = 5x

(i) In right angled ∆ABC

sin θ = perpendicular/hypotenuse

So we get

sin θ = AB/AC = 3x/5x = 3/5

(ii) In right angled ∆ABC

cos θ = base/hypotenuse

So we get

cos θ = BC/AC = 4x/5x = 4/5

(iii) In right angled ∆ABC

cot θ = base/perpendicular

So we get

cot θ = BC/AB = 4x/3x = 4/3

In right angled ∆ABC

cosec θ = hypotenuse/perpendicular

So we get

cosec θ = AC/AB = 5x/3x = 5/3

Here

cot² θ – cosec² θ = (4/3)² – (5/3)²

By further calculation

= 16/9 – 25/9

= (16 – 25)/9

= -9/9

= -1

Therefore, cot² θ – cosec² θ = -1.

5. If 2 cos θ = √3, prove that 3 sin θ – 4 sin³ θ = 1.

Solution

It is given that

2 cos θ = √3

⇒ cos θ = √3/2

We know that

sin² θ = 1 – cos² θ

Substituting the values

= 1 – (√3/2)²

= 1 – ¾

= ¼

sin θ = √ ¼ = ½

Consider

LHS = 3 sin θ – 4 sin³ θ

It can be written as

= sinθ (3 – 4sin² θ)

Substituting the values

= ½ [3 – (4 × ¼)]

= ½ (3 – 1)

= ½ × 2

= 1

= RHS

Therefore, proved.

6. If (sec θ – tan θ)/ (sec θ + tan θ) = ¼, find sin θ.

Solution:

We know that

By cross multiplication

4 – 4 sin θ = 1 + sin θ

We get

4 – 1 = sin θ + 4 sin θ

⇒ 3 = 5 sin θ

⇒ sin θ = 3/5

7. If sin θ + cosec θ = 3 1/3, find the value of sin² θ + cosec² θ.

Solution

It is given that

sin θ + cosec θ = 3 1/3 = 10/3

By squaring on both sides

(sin θ + cosec θ)² = (10/3)²

Expanding using formula (a + b)² = a² + b² + 2ab

sin² θ + cosec² θ + 2 sin θ cosec θ = 100/9

We know that sin θ = 1/cosec θ

sin² θ + cosec² θ + (2 sinθ× 1/sin θ) = 100/9

By further calculation

sin² θ + cosec² θ + 2 = 100/9

⇒ sin² θ + cosec² θ = 100/9 – 2

Taking LCM

sin² θ + cosec² θ = (100 – 18)/9 = 82/9

So we get

sin² θ + cosec² θ = 9 1/9

8. In the adjoining figure, AB = 4 m and ED = 3 m.

If sin α = 3/5 and cos β = 12/13, find the length of BD.

Solution

It is given that

sin α = AB/AC = 3/5

AB = 3 and AC = 5

Using Pythagoras theorem

AC² = AB² + BC²

Substituting the values

5² = 3² + BC²

By further calculation

25 = 9 + BC²

⇒ BC² = 25 – 9 = 16

So we get

BC² = 4²

⇒ BC = 4

We know that

tan α = AB/BC = 4/5

cos β = CD/CE = 12/13

CD = 12 and CE = 13

Using Pythagoras theorem

CE² = CD² + ED²

Substituting the values

13² = 12² + ED²

By further calculation

ED² = 13² – 12²

⇒ ED² = 169 – 144 = 25

So we get

ED² = (5)²

⇒ ED = 5

⇒ tan β = ED/CD = 5/12

From the figure

tan α = AB/BC = 4/BC

So we get

¾ = 4/BC

⇒ BC = (4×4)/3 = 16/3 m

tan β = ED/CD = 3/CD

⇒ 5/12 = 3/CD

So we get

CD = (12 × 3)/5 = 36/5 m

Here,

BD = BC + CD

Substituting the values

= 16/3 + 36/5

Taking LCM

= (80 + 108)/15

= 188/15 m

= 12 8/15 m