Concise Selina Solutions for Class 9 Maths Chapter 23 - Trigonometrical Ratios of Standard Angles

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Exercise 23A Page: 291

1. Find the value of:

(i) sin 30^o cos 30^o

(ii) tan 30^o tan 60^o

(iii) cos² 60^o + sin² 30^o

(iv) cosec² 60^o – tan² 30^o

(v) sin² 30^o + cos² 30^o + cot² 45^o

(vi) cos² 60^o + sec² 30^o + tan² 45^o.

Solution:

(i) Given sin 30^o cos 30^o

By substituting the values, we get

sin 30^o cos 30^o = ½ (√3/2)

= √3/4

(ii) Given tan 30^o tan 60^o

By substituting the values, we get

tan 30^o tan 60^o = 1/√3 (√3)

= 1

(iii) Given cos² 60^o + sin² 30^o

By substituting the values, we get

cos² 60^o + sin² 30^o = (½)² +(½)²

= ¼ + ¼

= ½

(iv) Given cosec² 60^o – tan² 30^o

By substituting the values, we get

cosec² 60^o – tan² 30^o = (2/√3)² – (1/√3)²

= 4/3 – 1/3

= 1

(v) Given sin² 30^o + cos² 30^o + cot² 45^o

By substituting the values, we get

sin² 30^o + cos² 30^o + cot² 45^o = (½)² + (√3/2)² + 1²

= ¼ + ¾ + 1

= 2

(vi) Given cos² 60^o + sec² 30^o + tan² 45^o

By substituting the values, we get

cos² 60^o + sec² 30^o + tan² 45^o = (½)² + (2/√3)² + 1²

= ¼ + 4/3 + 1

= 31/12

2. Find the value of:

(i) tan² 30^o + tan² 45^o + tan² 60^o

(ii) 

(iii) 3 sin² 30^o + 2 tan² 60^o – 5 cos² 45^o.

Solution:

(i) Given tan² 30^o + tan² 45^o + tan² 60^o

By substituting the values, we get

tan² 30^o + tan² 45^o + tan² 60^o = (1//√3)² + 1² + (/√3)²

= 1/3 + 1 + 3

= 13/3

= 4 1/3

(ii) Given

By substituting the values, we get

= ½ + 2/1 – 5/2

= (1 + 4 – 5)/2

= 0

(iii) Given 3 sin² 30^o + 2 tan² 60^o – 5 cos² 45^o.

By substituting the values, we get

3 sin² 30^o + 2 tan² 60^o – 5 cos² 45^o. = 3 (½)² + 2 (√3)² + 5 (1/√3)²

= ¾ + 6 – 5/2

= (3 + 24 – 10)/4

= 4 ¼

3. Prove that:

(i) sin 60^o cos 30^o + cos 60^o. sin 30^o = 1

(ii) cos 30^o. cos 60^o – sin 30^o. sin 60^o = 0

(iii) cosec² 45^o – cot² 45^o = 1

(iv) cos² 30^o – sin² 30^o = cos 60^o.

(v)

(vi) 3 cosec² 60^o – 2 cot² 30^o + sec² 45^o = 0.

Solution:

(i) Given sin 60^o cos 30^o + cos 60^o. sin 30^o

LHS = sin 60^o cos 30^o + cos 60^o. sin 30^o

Now we have to prove that RHS = 1

= (√3/2) (√3/2) + ½ ½

= ¾ + ¼

= 1

= RHS

(ii) Given cos 30^o. cos 60^o – sin 30^o. sin 60^o = 0

LHS = cos 30^o. cos 60^o – sin 30^o. sin 60^o

= (√3/2) ½ – ½ (√3/2)

= (√3/4) – (√3/4)

= 0

= RHS

(iii) Given cosec² 45^o – cot² 45^o = 1

LHS = cosec² 45^o – cot² 45^o = 1

= (√2)² – 1²

= 2 – 1

= 1

= RHS

(iv) Given cos² 30^o – sin² 30^o = cos 60^o.

LHS = cos² 30^o – sin² 30^o = cos 60^o.

= (√3/2)² – (½)²

= ¾ – ¼

= ½

= cos 60^o

= RHS

(vi) Given 3 cosec² 60^o – 2 cot² 30^o + sec² 45^o = 0.

LHS = 3 cosec² 60^o – 2 cot² 30^o + sec² 45^o = 0.

= 3 (2/√3)² – 2 (√3)² + (√2)²

= 4 – 6 + 2

= 0

= RHS

4. Prove that

Solution:

(ii) RHS =

Now substituting the values, we get

= (1 – 1/3)/ (1 + 1/3)

= ½

Consider LHS

cos (2 × 30^o)

= cos 60^o

= ½

Therefore, LHS = RHS

= √3

LHS,

tan (2 × 30^o)

= tan 60^o

= √3

Therefore, LHS = RHS

5. ABC is an isosceles right-angled triangle. Assuming of AB = BC = x, find the value of each of the following trigonometric ratios:

(i) sin 45^o

(ii) cos 45^o

(iii) tan 45^o

Solution:

Given that AB = BC = x

(i) sin 45^o = AB/AC

= x/x√2

= 1/√2

(ii) cos 45^o = BC/AC

= x/x√2

= 1/√2

(iii) tan 45^o = AB/BC

= x/x

= 1

6. Prove that:

(i) sin 60^o = 2 sin 30^o cos 30^o.

(ii) 4 (sin⁴ 30^o + cos⁴ 60^o) – 3 (cos² 45^o – sin² 90^o) = 2

Solution:

(i) LHS = sin 60^o

= √3/2

RHS = 2 sin 30^o cos 30^o

= 2 (√3/2) (½)

= √3/2

Therefore LHS = RHS

(ii) LHS = 4 (sin⁴ 30^o + cos⁴ 60^o) – 3 (cos² 45^o – sin² 90^o)

Now by substituting the values we get

= 4[(½)⁴ + (½)⁴] – 3 [(1/√2)² + 1⁴]

= 4(1/16 + 1/16) – 3 (½ – 1)

= 8/16 + 3/2

= 2

LHS = RHS

7. (i) If sin x = cos x and x is acute, state the value of x.

(ii) If sec A = cosec A and 0^o ≤ A ≤ 90^o, state the value of A.

(iii) If tan θ= cot θ and 0^o ≤ θ ≤ 90^o, state the value of θ.

(iv) If sin x = cos y; write the relation between x and y, if both the angles x and y are acute.

Solution:

(i) The angle, x is acute and hence we have, 0 < x

We know that

Cos²x + sin² x = 1

Since cos x = sin x

Above equation will become

2 sin² x = 1

Sin x = 1/√2

Therefore, x = 45^o

(ii) sec A = cosec A

Cos A = sin A

Cos² A = sin² A

Cos²x + sin² x = 1

Above equation will become

Cos² A = 1 – cos² A

2 cos² A = 1

Cos A = 1/√2

A = 45^o

(iii) tan θ = cot θ

tan θ = 1/tan θ

tan² θ = 1

tan θ = 1

tan θ = tan 45^o

θ = 45^o

(iv) sin x = cos y = sin (90^o – y)

If x and y are acute angles

x = 90^o – y

which implies,

x + y = 90^o

hence x and y are complementary angles.

8. (i) If sin x = cos y, then x + y = 45^o; write true of false.

(ii) sec θ. Cot θ = cosec θ; write true or false.

(iii) For any angle θ, state the value of:

Sin² θ + cos² θ.

Solution:

(i) sin x = cos y = sin (π/2 – y)

If x and y acute angles,

x = (π/2 – y)

x + y = π/2

x + y = 45^o is false

(ii) sec θ. Cot θ = 1/ cos θ. cos θ/ sin θ

= cosec θ

sec θ. Cot θ = cosec θ

is true.

(iii) Sin² θ + cos² θ = Sin² θ + 1 – sin² θ.

= 1

9. State for any acute angle θ whether:

(i) sin θ increases or decreases as θ increases:

(ii) cos θ increases or decreases as θ increases.

(iii) tan θ increases or decreases as θ decreases.

Solution:

(i) For acute angles, remember what sine means: opposite over hypotenuse. If we increase the angle, then the opposite side gets larger. That means “opposite/hypotenuse” gets larger or increases.

(ii)For acute angles, remember what cosine means: base over hypotenuse. If we increase the angle, then the hypotenuse side gets larger. That means “base/hypotenuse” gets smaller or decreases.

(iii)For acute angles, remember what tangent means: opposite over base. If we decrease the angle, then the opposite side gets smaller. That means “opposite /base” gets decreases.

10. If √3 = 1.732, find (correct to two decimal place) the value of each of the following:

(i) sin 60^o 

(ii) 2/ tan 30^o

Solution:

(i) sin 60^o = √3 /2

= 1.732/2

= 0.87

(ii) 2/ tan 30^o = 2/ (1/√3)

= 2√3 

= 2 (1.732)

= 3.46

Exercise 23b Page: 293

1. Given A = 60^o and B = 30^o, prove that:

(i) sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos A cos B – sin A sin B

(iii) cos (A – B) = cos A cos B + sin A sin B

Solution:

(i) Given A = 60^o and B = 30^o

LHS = sin (A + B)

= sin (60^o + 30^o)

= sin 90^o

= 1

RHS = sin A cos B + cos A sin B

= sin 60^o cos 30^o + cos 60^o sin 30^o

= √3/2 (√3/2) + ½ ½

= ¾ + ¼

= 1

LHS = RHS

(ii) LHS = cos (A + B)

= cos (60^o + 30^o)

= cos 90^o

= 0

RHS = cos A cos B – sin A sin B

RHS = cos 60^o cos 30^o – sin 60^o sin 30^o

= ½ (√3/2) – (√3/2) ½

= √¾ – √3/4

= 0

LHS = RHS

(iii) LHS = cos (A – B)

= cos (60^o – 30^o)

= cos 30^o

= √3/2

RHS = cos A cos B + sin A sin B

RHS = cos 60^o cos 30^o + sin 60^o sin 30^o

= ½ (√3/2) + (√3/2) ½

= √¾ + √3/4

= √3/2

LHS = RHS

(iv) LHS = tan (A – B)

= tan (60^o – 30^o)

= tan 30^o

= 1/√3

= (√3 – 1/√3)/ 1 + √3 (1/√3)

= 2/ 2 √3

= 1/√3

Therefore, LHS = RHS

2. If A =30^o, then prove that:

(i) sin 2A = 2sin A cos A = 

(ii) cos 2A = cos²A – sin²A

= 

(iii) 2 cos² A – 1 = 1 – 2 sin²A

(iv) sin 3A = 3 sin A – 4 sin³A.

Solution:

(i) Given A = 30^o

Sin 2A = sin 2(30^o)

= sin 60^o

= √3/2

2 sin A cos A = 2 sin 30^o cos 30^o

= 2 (½) (√3/2)

= √3/2

Now,

(ii) cos 2A = cos 2 (30⁰)

= cos 60^o

= ½

Cos² A – sin² A = cos² 30^o – sin² 30^o

= ¾ – ¼

= ½

= 2/4

= ½

(iii) 2 cos² A – 1 = 2 cos² 30^o – 1

= 2 (¾) – 1

= 3/2 – 1

= ½

1 – 2 sin² A = 1 – 2 sin² 30^o

= 1 – 2 (¼)

= ½

2 cos² A – 1 = 1 – sin² A

(iv) sin 3A = sin 3 (30^o)

= sin 90^o

= 1

3 sin A – 4 sin³ A = 3 sin 30^o – 4 sin³ 30^o

= 3 (½) – 4 (½)³

= 3/2 – ½

= 1

Therefore,

Sin 3A = 3 sin A – 4 sin³ A

3. If A = B = 45^o, show that:

(i) sin (A – B) = sin A cos B – cos A sin B

(ii) cos (A + B) = cos A cos B – sin A sin B

Solution:

Given that A = B = 45^o

(i) LHS = sin (A – B)

= sin (45^o – 45^o)

= sin 0^o

= 0

RHS = sin A cos B – cos A sin B

= sin 45^o cos 45^o – cos 45^o sin 45^o

= 1/√2 (1/√2) – 1/√2 (1/√2)

= 0

Therefore, LHS = RHS

(ii) LHS = cos (A + B)

= cos (45^o + 45^o)

= cos 90^o

= 0

RHS = cos A cos B – sin A sin B

= cos 45^o cos 45^o – sin 45^o sin 45^o

= 1/√2 (1/√2) – 1/√2 (1/√2)

= 0

Therefore, LHS = RHS

4. If A = 30^o; show that:

(i) sin 3 A = 4 sin A sin (60^o – A) sin (60^o + A)

(ii) (sin A – cos A)² = 1 – sin 2A

(iii) cos 2A = cos⁴ A – sin⁴ A

(iv) 

(v) = 2 cos A.

(vi) 4 cos A cos (60^o – A). cos (60^o + A) = cos 3A

(vii) 

Solution:

Given that A = 30^o

(i) According to the question we have

LHS = sin 3A

= sin 3 (30^o)

= sin 90^o

= 1

RHS = 4 sin A sin (60^o – A) sin (60^o + A)

= 4 sin A sin (60^o – 30^o) sin (60^o + 30^o)

= 4 (½) (½) (1)

= 1

LHS = RHS

(ii) According to the question we have

LHS = (sin A – cos A)²

= (sin 30^o – cos 30^o)²

= (½ – √3/2)²

= ¼ + ¾ – √3/2

= 1 – √3/2

= (2 – √3)/2

RHS = 1 – sin 2A

= 1 – sin 2 (30^o)

= 1 – sin 60^o

= 1 – √3/2

= (2 – √3)/2

Therefore, LHS = RHS

(iii) According to the question we have

LHS = cos 2A

= cos 2 (30^o)

= cos 60^o

= ½

RHS = cos⁴ A – sin⁴ A

= cos⁴ 30^o – sin⁴ 30^o

= (√3/2)⁴ – (½)⁴

= 9/16 – 1/16

= ½

LHS = RHS

(iv) According to the question we have

LHS = (1 – cos 2A)/ sin 2A

= 1 – cos 2(30^o)/ sin 2 (30^o)

= 1 – ½/ (√3/2)

= 1/√3

RHS = tan A

= tan 30^o

= 1/√3

LHS = RHS

(v) According to the question we have

= 2√3/2

= √3

RHS = 2 cos A

= 2 cos 30^o

= 2 (√3/2)

= √3

(vi) According to the question we have

LHS = 4 cos A cos (60^o – A) cos (60^o + A)

= 4 cos A cos (60^o – 30^o) cos (60^o + 30^o)

= 4 cos 30^o cos 30^o cos 90^o

= 4 (√3/2) (√3/2) 0

= 0

RHS = cos 3A

= cos 3 (30^o)

= cos 90^o

= 0

LHS = RHS

(vii) According to the question we have

= ¾ + 9/4

= 12/4

= 3

= RHS

Hence the proof

Exercise 23b Page: 297

1. Solve the following equations for A, if:

(i) 2 sin A = 1 

(ii) 2 cos 2 A = 1

(iii) sin 3 A = √3/2 

(iv) sec 2 A = 2

(v) √3 tan A = 1 

(vi) tan 3 A = 1

(vii) 2 sin 3 A = 1 

(viii) √3 cot 2 A = 1

Solution:

(i) Given 2 sin A = 1

Sin A = ½

Sin A = sin 30^o

Therefore, A = 30^o

(ii) Given 2 cos 2A = 1

Cos 2A = ½

Cos 2A = cos 60^o

2A = 60^o

A = 30^o

(iii) Given sin 3A = √3/2

Sin 3A = sin 60^o

3A = 60^o

A = 20^o

(iv) Given sec 2A = 2

Sec 2A = sec 60^o

2A = 60^o

A = 30^o

(v) Given √3 tan A = 1

Tan A = 1/√3

Tan A = tan 30^o

A = 30^o

(vi) Given tan 3A = 1

Tan 3A = tan 45^o

3A = 45^o

A = 15^o

(vii) Given 2 sin 3A = 1

Sin 3A = ½

Sin 3A = sin 30^o

3A = 30^o

A = 10^o

(viii) Given √3 cot 3A = 1

Cot 2A = 1/√3

Cot 2A = cot 60^o

2A = 60^o

A = 30^o

2. Calculate the value of A, if:

(i) (sin A – 1) (2 cos A – 1) = 0

(ii) (tan A – 1) (cosec 3A – 1) = 0

(iii) (sec 2A – 1) (cosec 3A – 1) = 0

(iv) cos 3A. (2 sin 2A – 1) = 0

(v) (cosec 2A – 2) (cot 3A – 1) = 0

Solution:

(i) Given (sin A – 1) (2 cos A – 1) = 0

It can be written as

(sin A – 1) = 0 and (2 cos A – 1) = 0

Sin A = 1 and cos A = ½

Sin A = sin 90^o and cos A = cos 60^o

A = 90^o and A = 60^o

(ii) Given (tan A – 1) (cosec 3A – 1) = 0

It can be written as

(tan A – 1) = 0 and (cosec 3A – 1) = 0

tan A = 1 and cosec 3A = 1

tan A = tan 45^o and cosec 3A = cosec 90^o

A = 45^o and A = 30^o

(iii) Given (sec 2A – 1) (cosec 3A – 1) = 0

It can be written as

(sec 2A – 1) = 0 and (cosec 3A – 1) = 0

sec 2A = 1 and cosec 3A = 1

sec 2A = sec 60^o and cosec 3A = cosec 90^o

A = 0^o and A = 30^o

(iv) Given cos 3A (2 sin 2A – 1) = 0

It can be written as

Cos 3A = 0 and 2 sin 2A – 1 = 0

Cos 3A = cos 90^o and sin 2A = ½

3A = 90^o and sin 2A = sin 30^o

A = 30^o and 2A = 30^o which implies A = 15^o

(v) Given (cosec 2A – 2) (cot 3A – 1) = 0

It can be written as

(cosec 2A – 2) = 0 and (cot 3A – 1) = 0

cosec 2A = 2 and cot 3A = 1

cosec 2A = cosec 30^o and cot 3A = cot 45^o

2A = 30^o and 3A = 45^o

A = 15^o and A = 15^o

3. If 2 sin x^o – 1 = 0 and x^o is an acute angle; find:

(i) sin x^o 

(ii) x^o 

(iii) cos x^o and tan x^o.

Solution:

(i) Given 2 sin x^o – 1 = 0

2 sin x^o = 1

sin x^o = ½

(ii) we have sin x^o = ½

sin x^o = sin 30^o

x^o = 30^o

(iii) we have x^o = 30^o

Cos x^o = cos 30^o = √3/2

Tan x^o = tan 30^o = 1/√3

4. If 4 cos² x^o – 1 = 0 and 0  x^o  90^o, find:

(i) x^o 

(ii) sin² x^o + cos² x^o

(iii) 

Solution:

(i) Given 4 cos² x^o – 1 = 0

4 cos² x^o = 0

cos² x^o = (½)²

cos x^o = ½

cos x^o = cos 60^o

x^o = 60^o

(ii) sin² x^o + cos² x⁰ = sin² 60^o + cos² 60^o

= (√3/2)² + (½)²

= ¾ + ¼

= 1

(iii) Given

= 1/ (½)² – (√3)²

= 4 – 3

= 1

5. If 4 sin² θ – 1= 0 and angle θ is less than 90^o, find the value of θ and hence the value of cos² θ + tan² θ.

Solution:

Given 4 sin² θ – 1= 0

sin² θ = ¼

sin θ = ½

sin θ = sin 30^o

θ = 30^o

cos² θ + tan² θ = cos² 30^o + tan² 30^o

= (√3/2)² + (1/√3)²

= ¾ + 1/3

= (9 + 4)/12

= 13/12

6. If sin 3A = 1 and 0 ≤ A ≤ 90^o, find:

(i) sin A 

(ii) cos 2A

(iii) tan²A – 1/cos² A

Solution:

Given sin 3A = 1

Sin 3A = sin 90^o

3A = 90^o

A = 30^o

(i) according to the question we have,

Sin A = sin 30^o

Sin A = ½

(ii) cos 2A = cos 2(30^o)

= cos 60^o

= ½

(iii) According to the question we have,

= 1/3 – 4/3

= -3/3

= -1

Selina Solutions for Class 9 Maths Chapter 23- Trigonometrical Ratios of Standard Angles

The Chapter 23, Trigonometrical Ratios of Standard Angles is composed of 3 exercises and the solutions given here contain answers to all the questions present in these exercises. Let us have a look at some of the topics that are being discussed in this chapter.

23.1 Trigonometrical Ratios of Angles 30^o and 60^o.

23.2 Trigonometrical Ratios of Angle 45^o

23.3 Solving a trigonometric equation

Selina Solutions for Class 9 Maths Chapter 23- Trigonometrical Ratios of Standard Angles

In Chapter 23 of Class 9, the students are taught about the Trigonometrical Ratios of Standard Angles. The chapter also includes the evaluation of an expression involving trigonometric ratios and problems related to it. Study Chapter 23 of Selina textbook to understand more about Trigonometrical Ratios of Standard Angles. Learn the Selina Solutions for Class 9 effectively to come out with flying colours in the examinations.