# 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

From right angled ∆ABD,

By Pythagoras angled ∆ABD

By Pythagoras theorem we get

⇒ AB= 400

⇒ AB= (20)2

⇒ AB = 20

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

= 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)

= 16/12

= 4/3

cosec2 y – coty = (5/3)2 – (4/3)2

= (25/9) – (16/9)

= (25-16)/9

= 9/9

= 1

Hence, cosec2 y – coty = 1

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

= 12/20

= 3/5

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

= 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 = 90o

= 144

= (12)2

= (12)+ (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

⇒ AB= (4)2 + (3)2

⇒ AB= 16 + 9

⇒ AB2 = 25

⇒ AB2 = (5)2

⇒ AB = 5

In right angled triangle ACD

By Pythagoras theorem,

⇒ CD2 = AC2 – AD2

⇒ CD2 = (12)– (4)2

⇒ CD= 144 – 16

⇒ CD2 = 128

⇒ CD = √128

⇒ CD = √64 × 2 CD

= 8√2

(i) sin x = perpendicular/Hypotenuse

= 4/5

(ii) cot x = Base/Perpendicular

= ¾

(iii) cot x = Base/Perpendicular

= 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)

= 12/(8 √2)

= 3/(2 √2)

cot y = Base/Hypotenuse

= 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/cosy

= (1/2√2)2 – 1/(2√2/3)2

= ¼ × – ¼ × 2

= (1/8) – (9/8)

= (1-9)/8

= -8/8

= -1

tany – 1/cosy = –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 = 90o

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)+ (12x)2

⇒ AB= 25x2 + 144x2

⇒ AB= 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

⇒ AD2 = (15)2 – (9)2

⇒ AD2 = 225 – 81

(i) cos ∠ABC = Base/Hypotenuse

(In right angled ∆ABD, ∠ABC = ∠ABD)

= BD / AB

= 9/15

= 3/5

(ii) sin ∠ACB = sin ∠ACD

= perpendicular/ Hypotenuse

= 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

D is the mid point of BC

So BD = CD

Here

BD = CD = 6/2 = 3 cm

In right angled ∆ABD

Using Pythagoras theorem

We can write it as

Substituting the values

By further calculation

AD2 = 25 – 9 = 16

So we get

(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

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,

We can write it as

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

Using Pythagoras theorem

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

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)+ BC2

⇒ BC2 = (5x)– (3x)2

⇒ BC= (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 = AB+ BC2

⇒ AC= AB+ BC2

⇒ AC2 = AB2 + BC2

⇒ AC= AB+ BC2

⇒ (AC= (4x)2 + (3x)2

⇒ AC= 16x+ 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 = 90o 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 sinA + 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 = AB+ 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

cot α = CD/AD = 3/2

Take CD = 3x then AD = 2x

Using Pythagoras theorem

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

Substituting the values

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,

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

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

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 × 15/25

= 15/5

= 3

(ii) In right angled ∆ABD

7 tan x = 7 (perpendicular/base)

So we get

= 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