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Introduction to Trigonometry Chapter Notes - Mathematics (Maths) Class 10

Trigonometry is the branch of mathematics which deals with measurements of angles and the problems related to relationship between sides and angles.

Trigonometric Ratios

  • It is the ratios of two sides of a right angle triangle.
  • Suppose we have a right-angled triangle ABC.
    Introduction to Trigonometry Chapter Notes | Mathematics (Maths) Class 10
  • Thus the trigonometric Ratios are defined as the largest side of a right angle triangle is known as hypotenuse.

Also, by Pythagoras theorem, we know that h2 = b+ p2.
Here, θ is pronounced as theta, ∠A=θ. We write trigonometric ratios according to angle θ.

Introduction to Trigonometry Chapter Notes | Mathematics (Maths) Class 10

Question for Chapter Notes: Introduction to Trigonometry
Try yourself:In a right-angled triangle ABC, the angle ∠A is denoted as θ. Which of the following correctly represents the trigonometric ratio for the angle θ?
View Solution

Relation between Trigonometric Ratios


  • cosec θ =1/sin θ
  • sec θ = 1/cos θ
  • tan θ = sin θ/cos θ
  • cot θ = cos θ/sin θ=1/tan θ

Trigonometric Ratios of Specific Angles

Trigonometric Ratios of 45°

If one of the angles of a right-angled triangle is 45°, then another angle will also be equal to 45°.

Introduction to Trigonometry Chapter Notes | Mathematics (Maths) Class 10

Let us say ABC is a right-angled triangle at B, such that;
∠ A = ∠ C = 45°
Thus, BC = AB = a (say)
Using Pythagoras theorem, we have;
AC2 = AB2 + BC2
= a2 + a2
= 2a2
AC = a√2

Now, from the trigonometric ratios, we have;
sin 45° = (Opp. side to angle 45°)/Hypotenuse = BC/AC = a/a√2 = 1/√2
cos 45° = (Adj. side to angle 45°)/Hypotenuse = AB/AC = a/a√2 = 1/√2
tan 45° = BC/AB = a/a = 1

Similarly,
cosec 45° = 1/sin 45° = √2
sec 45° = 1/cos 45° = √2
cot 45° = 1/tan 45° = 1

Trigonometric Ratios of 30° and 60°

Here, we will consider an equilateral triangle ABC, such that;
AB = BC = AC = 2a
∠A = ∠B = ∠C = 60°
Now, draw a perpendicular AD from vertex A that meets BC at DIntroduction to Trigonometry Chapter Notes | Mathematics (Maths) Class 10

According to the congruency of the triangle, we can say;
Δ ABD ≅ Δ ACD
Hence,
BD = DC
∠ BAD = ∠ CAD (By CPCT)
Now, in triangle ABD, ∠ BAD = 30° and ∠ ABD = 60°
Using Pythagoras theorem
AD2 = AB2 – BD2
= (2a)2 – (a)2
= 3a
AD = a√3

So, the trigonometric ratios for a 30-degree angle will be;
sin 30° = BD/AB = a/2a = 1/2
cos 30° = AD/AB = a√3/2a = √3/2
tan 30° = BD/AD = a/a√3 = 1/√3

Also,
cosec 30° = 1/sin 30 = 2
sec 30° = 1/cos 30 = 2/√3
cot 30° = 1/tan 30 = √3

Similarly, we can derive the values of trigonometric ratios for 60°.
sin 60° = √3/2
cos 60° = 1/2
tan 60° = √3
cosec 60° = 2/√3
sec 60° = 2
cot 60° = 1/√3

Trigonometric Ratios of 0° and 90°

If ABC is a right-angled triangle at B, if ∠A is reduced, then side AC will come near to side AB. So, if ∠ A is nearing 0 degree, then AC becomes almost equal to AB, and BC gets almost equal to 0.

Hence, Sin A = BC /AC = 0
and cos A = AB/AC = 1

tan A = sin A/cos A = 0/1 = 0

Also,
cosec A = 1/sin A = 1/0 = not defined
sec A = 1/cos A = 1/1 = 1
cot A = 1/tan A = 1/0 = not defined

In the same way, we can find the values of trigonometric ratios for a 90-degree angle. Here, angle C is reduced to 0, and the side AB will be nearing side BC such that angle A is almost 90 degrees and AB is almost 0.

Standard Values of Trigonometric Ratios


Introduction to Trigonometry Chapter Notes | Mathematics (Maths) Class 10

Example 1: Evaluate 2 tan2 45° + cos2 30° – sin2 60°.

Solution: Since we know,
tan 45° = 1
cos 30° = √3/2
sin 60° = √3/2
Therefore, putting these values in the given equation:
2(1)2 + (√3/2)2 – (√3/2)2
= 2 + 0
= 2

Question for Chapter Notes: Introduction to Trigonometry
Try yourself:(Sin 30°+cos 60°)-(sin 60° + cos 30°) is equal to:
View Solution

Example 2: If tan (A + B) =√3 and tan (A – B) =1/√3, 0° < A + B ≤ 90°; A > B, find A and B.

Solution: Given,
tan (A + B) = √3
As we know, tan 60° = √3
Thus, we can write;
⇒ tan (A + B) = tan 60°
⇒(A + B) = 60° …… (i)
Now again given;
tan (A – B) = 1/√3
Since, tan 30° = 1/√3
Thus, we can write;
⇒ tan (A – B) = tan 30°
⇒(A – B) = 30° ….. (ii)
Adding the equation (i) and (ii), we get;
A + B + A – B = 60° + 30°
2A = 90°
A= 45°
Now, put the value of A in eq. (i) to find the value of B;
45° + B = 60°
B = 60° – 45°
B = 15°
Therefore A = 45° and B = 15°

Example 3: Show that :
(i) tan 48° tan 23° tan 42° tan 67° = 1

(ii) cos 38° cos 52° – sin 38° sin 52° = 0

Solution:
(i) tan 48° tan 23° tan 42° tan 67°
We can also write the above given tan functions in terms of cot functions, such as;
tan 48° = tan (90° – 42°) = cot 42°
tan 23° = tan (90° – 67°) = cot 67°
Hence, substituting these values, we get
= cot 42° cot 67° tan 42° tan 67°
= (cot 42° tan 42°) (cot 67° tan 67°)
= 1 × 1 [since cot A.tan A = 1]
= 1
(ii) cos 38° cos 52° – sin 38° sin 52°
We can also write the given cos functions in terms of sin functions.
cos 38° = cos (90° – 52°) = sin 52°
cos 52°= cos (90° – 38°) = sin 38°
Hence, putting these values in the given equation, we get;
sin 52° sin 38° – sin 38° sin 52° = 0

Example 4: If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

Solution: Given,
tan 2A = cot (A – 18°)
As we know by trigonometric identities,
tan 2A = cot (90° – 2A)
Substituting the above equation in the given equation, we get;
⇒ cot (90° – 2A) = cot (A – 18°)
Therefore,
⇒ 90° – 2A = A – 18°
⇒ 108° = 3A
A = 108° / 3
Hence, the value of A = 36°

Example 5:  If A, B and C are interior angles of a triangle ABC, then show that sin [(B + C)/2] = cos A/2.

Solution:
As we know, for any given triangle, the sum of all its interior angles is equals to 180°.
Thus,
A + B + C = 180° ….(1)
Now we can write the above equation as;
⇒ B + C = 180° – A
Dividing by 2 on both the sides;
⇒ (B + C)/2 = (180° – A)/2
⇒ (B + C)/2 = 90° – A/2
Now, put sin function on both sides.
⇒ sin (B + C)/2 = sin (90° – A/2)
Since,
sin (90° – A/2) = cos A/2
Therefore,
sin (B + C)/2 = cos A/2

Trigonometric Identities


Identities are equations which are valid for any conditions. Trigonometric identities are equations involving trigonometric ratios.

Some Trigonometric Identities are:

(i) sin2θ + cos2⁡θ = 1

(ii) sec2θ - tan2θ =1

(iii) cosec2θ - cot2⁡θ = 1

Question for Chapter Notes: Introduction to Trigonometry
Try yourself:If cos X = ⅔ then tan X is equal to:
View Solution

Example 6: Prove the identities:
(i) √[1 + sinA/1 – sinA] = sec A + tan A

(ii) (1 + tan2A/1 + cot2A) = (1 – tan A/1 – cot A)2 = tan2A

Solution:
(i) Given:√[1 + sinA/1 – sinA] = sec A + tan A

Introduction to Trigonometry Chapter Notes | Mathematics (Maths) Class 10
(ii) Given: (1 + tan2A/1 + cot2A) = (1 – tan A/1 – cot A)2 = tan2A
LHS:
= (1+tan²A) / (1+cot²A)
Using the trigonometric identities we know that 1+tan²A = sec²A and 1+cot²A= cosec²A
= sec²A/ cosec²A
On taking the reciprocals we get

= sin²A/cos²A

= tan²A

RHS:

=(1-tanA)²/(1-cotA)²

Substituting the reciprocal value of tan A and cot A we get,

= (1-sinA/cosA)²/(1-cosA/sinA)²

= [(cosA-sinA)/cosA]²/ [(sinA-cos)/sinA)²] = [(cosA-sinA)²×sin²A] /[cos²A. /(sinA-cosA)²] =  sin²A/cos²A

= tan2A

The values of LHS and RHS are the same.

Hence proved.

Example 7: If sin θ + cos θ = √3, then prove that tan θ + cot θ = 1.

Solution:

Given,

sin θ + cos θ = √3

Squaring on both sides,

(sin θ + cos θ)= (√3)2

sin2θ + cos2θ + 2 sin θ cos θ = 3

Using the identity sin2A + cos2A = 1,

1 + 2 sin θ cos θ = 3

2 sin θ cos θ = 3 – 1

2 sin θ cos θ = 2

sin θ cos θ = 1

sin θ cos θ = sin2θ + cos2θ

⇒ (sin2θ + cos2θ)/(sin θ cos θ) = 1

⇒ [sin2θ/(sin θ cos θ)] + [cos2θ/(sin θ cos θ)] = 1

⇒ (sin θ/cos θ) + (cos θ/sin θ) = 1

⇒ tan θ + cot θ = 1

Hence proved.

Example 8: Express cot 85° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Solution:

cot 85° + cos 75° 

= cot (90° – 5°) + cos (90° – 15°) 

We know that cos(90° – A) = sin A and cot(90° – A) = tan A

= tan 5° + sin 15°

Trigonometric Ratios for Complementary Angles (Deleted from NCERT)

(i) sin (90 – θ) = cosθ

(ii) cos (90 – θ) = sinθ

(iii) tan (90 – θ) = cotθ

(iv) cot (90 – θ) = tanθ

(v) sec (90 – θ) = cosecθ

(vi) cosec (90 – θ) = secθ

The document Introduction to Trigonometry Chapter Notes | Mathematics (Maths) Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10.
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