CBSE Class 10 Maths Revision Notes Chapter 8

Class 10 Mathematics revision notes for Introduction to Trigonometry of Chapter 8 

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Once going through the Chapter 8 Mathematics Class 10 Notes and revising the important mathematical formulas students should also look into CBSE previous year question papers and CBSE extra questions. This would be a good way to revise all the class 10 Mathematics Notes Chapter 8. Students can access the detailed revision notes for Introduction to Trigonometry and practice. They can regularly revise the chapter and master the concepts. 

Class 10 Mathematics revision notes for Introduction to Trigonometry of Chapter 8 

Trigonometry Notes Class 10 – Brief Overview.

The Greek terms “tri” (meaning three), “gon” (meaning sides), and “metron” are the roots of the word “trigonometry” (meaning measure). The branch of mathematics known as trigonometry deals with finding relations between a triangle’s length of sides and its angles. It principally has numerous uses in the field of oceanography, sound waves, and music. The earliest known work on trigonometry was recorded in Egypt and Babylon when astronomers were trying to find the distances of the stars and planets from the earth. The use of trigonometry has found its place even today in some of the most advanced mechanical and technological backgrounds. Students must therefore have a good grip on the topic of trigonometry which would be useful in their future education and career ahead. Extramarks presents a detailed yet simple learning method for the students to have a glimpse of some of the most important Class 10 Mathematics Chapter 8 Notes to prepare ahead for the CBSE exams. Trigonometric ratios of the angle, or ratios of a right triangle’s sides to its sharp angles, will be studied in this chapter. We will just address acute angles in this session. 

Some revisions before diving into deeper trigonometric concepts:

• Trigonometry is based on the study of right-angled triangles.
• The three sides of a triangle represent its height, base, and hypotenuse.
• The Pythagoras theorem states that the square of the hypotenuse of a right triangle equals the sum of the squares of the other two sides i.e hypotenuse squared equals height squared + base squared.

Note: Learning through diagrammatic representation is an effective method to solidify these concepts. It escalates revision and practice speed. The use of such diagrams also helps fetch extra marks in the examination. 

Trigonometry Class 10 Notes – Revision Notes.

Extramarks has provided detailed revision notes to help students prepare adequately before exams. These notes are curated after taking into account the learning potential of different students and hence are accessible to all of them. Revising through these notes gives a better picture of the class 10 trigonometry chapter.

The trigonometric ratios:

There are six trigonometric ratios defined for a right angles triangle. They are listed below:

• Sine
• Cosine
• Tangent
• Cotangent
• Secant
• Cosecant 

In a given right-angled triangle ABC, the ratios are defined as follows:

• Sine of ∠ A = side opposite to angle A/hypotenuse = BC / AC.
• Cosine of ∠ A = side adjacent to angle A / hypotenuse = AB / AC.

Remarks: Since the hypotenuse is the longest side in a right triangle, the value of sin A or cos A is always less than 1 (or, in particular, equal to 1)

• Tangent of ∠ A = side opposite to angle A/side adjacent to angle A = BC / AB.
• Cosecant of ∠ A = 1 / Sine of angle A = hypotenuse / side opposite to angle A = AC / BC.
• secant of ∠ A = 1 / Cosine of angle A = hypotenuse / side adjacent to angle A = AC / AB.
• cotangent of ∠ A = 1 / Tangent of angle A = side adjacent to angle A / side opposite to angle A = AB / BC.

Relationship 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

Solved practice question on the relation property of trigonometric ratios.

Suppose a right-angled triangle ABC, right-angled at B such that hypotenuse AC = 5cm, base BC = 3cm and perpendicular AB = 4cm. Also, ∠ACB = θ. Find the trigonometric ratios tan θ, sin θ and cos θ.

Solution: Given, in ∆ABC,

Hypotenuse, AC = 5cm

Base, BC = 3cm

Perpendicular, AB = 4cm

Then, by the trigonometric ratios, we have:

tan θ = Perpendicular/Base = 4 / 3

Sin θ = Perpendicular/Hypotenuse = AB/AC = 4 / 5

Cos θ = Base/Hypotenuse = BC/AC = 3 / 5

Note

1. The values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle if the angle remains the same. 
2. If we know any one of the ratios, can we obtain the other ratios as well by using the above formulas along with the Pythagoras theorem?

Important trigonometric ratios of some specific angles:

Range of trigonometric ratios for 0 deg ≤ θ ≤ 90 deg.

• 0 ≤ sin θ ≤ 1
• 0 ≤ cos θ ≤ 1
• 0 ≤ tan θ < ∞
• 1 ≤ sec θ < ∞
• 0 ≤ cot θ < ∞
• 1 ≤ cosec θ < ∞

:tan θ and sec θ are not defined at 90∘.

:cot θ and cosec θ are not defined at 0∘.

Oscillating trigonometric values.

As θ increases from 0 deg to 90 deg.

• sin θ increases from 0 to 1
• cos θ decreases from 1 to 0
• tan θ increases from 0 to ∞
• cosec θ decreases from ∞ to 1
• sec θ increases from 1 to ∞
• cot θ decreases from ∞ to 0

Solved practice questions on important trigonometric angles.

If tan (A+ B) =√3, tan (A-B) = 1/√3, then find A and B. [Given that  0° <A+B ≤ 90°; A>B ]

Solution:

Given that 

Tan (A+B) = √3.

We know that tan 60 = √3.

Thus, tan (A+B) = tan 60° = √3.

Hence A+B= 60° …(1)

Similarly, given that,

Tan (A-B) = 1/√3.

We know that tan 30° = 1/√3.

Thus, tan (A-B) = tan 30° = 1/√3.

Hence, A-B = 30° …(2)

Now, adding the equations (1) and (2), we get

A+B+A-B = 60° + 30°

2A = 90°

A = 45°.

Now, substitute A = 45° in equation (1), we get

45° +B = 60°

B = 60°- 45°

B = 15°

Hence, A = 45 and B = 15°.

Trigonometric ratios of complementary angles:

Under this section, students must remember the following relationships. Derivations are however not important from an exam point of view.

• sin (90° – A) = cos A
• cos (90° – A) = sin A
• tan (90° – A) = cot A
• cot (90° – A) = tan A
• sec (90° – A) = cosec A
• cosec (90° – A) = sec A

Solved practice questions on complementary angle property:

1.Evaluate tan 65° cot 25°.

Solution: We know that: cot A = tan (90° – A).

So, cot 25° = tan (90° – 25°) = tan 65°

i.e.,  tan 65° /  cot 25° = tan 65° / tan 65 = 1

2.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°).

 = tan 5° + sin 15°.

The trigonometric identities:

Trigonometric identities are basically the equations involving trigonometric ratios of an angle. The three primary trigonometric identities are:

• Sin2 A + cos2 A = 1
• 1 + tan2 A = sec2 A
• 1 + cot2 A = cosec2 A

Solved practice questions on trigonometric identities:

1.Prove that sec A (1 – sin A)(sec A + tan A) = 1.

Solution : 

LHS = sec A (1 – sin A)(sec A + tan A) = ( 1 / Cos A ) ( 1 – sin A ) ( 1 / Cos A + Sin A / Cos A )

= [( 1 – Sin A) ( 1 + Sin A )] / Cos2 A.

= ( 1 – Sin2 A ) / Cos2 A.

=  Cos2 A /  Cos2 A = 1

= RHS. ( Hence proved )

2.Example 12: Express the ratios cos A, tan A, and sec A in terms of sin A. 

Solution: Since Cos2 A + Sin2 A = 1, therefore,

Cos2 A = 1 – Sin2 A , i.e., cos A = ± √1 – Sin2 A.

This gives Cos A = √1 – Sin2 A.

Hence, Tan A = Sin A / Cos A

= Sin A / √1 – Sin2 A

 and Sec A = 1 / Cos A.

= 1 / √1 – Sin2 A.