# Introduction To Trigonometry Class 10 Notes

## Class 10 Maths Chapter 8 Introduction to Trigonometry Notes

The notes for trigonometry class 10 Maths is provided here. Get the complete concept on trigonometry which is covered in Class 10 Maths. Also, get the various trigonometric ratios for specific angles, the relationship between trigonometric functions, trigonometry tables, various identities given here.

Students can refer to the short notes and MCQ questions along with separate solution pdf of this chapter for quick revision from the links below:

## Trigonometric Ratios

### Opposite & Adjacent Sides in a Right Angled Triangle

In the ΔABC right-angled at B, BC is the side opposite to A, AC is the hypotenuse and AB is the side adjacent to A.

### Trigonometric Ratios

For the right ΔABC, right-angled at B, the trigonometric ratios of the A are as follows:

• sin A=opposite side/hypotenuse=BC/AC
• cosec A=hypotenuse/opposite side=AC/BC

To know more about Trigonometric Ratios, visit here.

### Visualization of Trigonometric Ratios Using a Unit Circle

Draw a circle of the unit radius with the origin as the centre. Consider a line segment OP joining a point P on the circle to the centre which makes an angle θ with the x-axis. Draw a perpendicular from P to the x-axis to cut it at Q.

• sinθ=PQ/OP=PQ/1=PQ
• cosθ=OQ/OP=OQ/1=OQ
• tanθ=PQ/OQ=sinθ/cosθ
• cosecθ=OP/PQ=1/PQ
• secθ=OP/OQ=1/OQ
• cotθ=OQ/PQ=cosθ/sinθ

### 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

### Range of Trigonometric Ratios from 0 to 90 degrees

For 0θ90,

• 0≤sinθ≤1
• 0≤cosθ≤1
• 0≤tanθ<∞
• 1≤secθ<∞
• 0≤cotθ<∞
• 1≤cosecθ<∞

tanθ and secθ are not defined at  90.
cotθ and cosecθ are not defined at 0.

### Variation of trigonometric ratios from 0 to 90 degrees

As θ increases from 0 to 90

• siθ increases from 0 to 1
• coθ decreases from 1 to 0
• taθ increases from 0 to
• coseθ decreases from to 1
• seθ increases from 1 to
• coθ decreases from to 0

### Standard values of Trigonometric ratios

 ∠A 0o 30o 45o 60o 90o sin A 0 1/2 1/√2 √3/2 1 cos A 1 √3/2 1/√2 1/2 0 tan A 0 1/√3 1 √3 not defined cosec A not defined 2 √2 2/√3 1 sec A 1 2/√3 √2 2 not defined cot A not defined √3 1 1/√3 0

To know more about Trigonometric Ratios of Standard Angles, visit here.

## Trigonometric Ratios of Complementary Angles

### Complementary Trigonometric ratios

If θ is an acute angle, its complementary angle is 90θ. The following relations hold true for trigonometric ratios of complementary angles.

• si(90− θcoθ
• co(90− θsiθ
• ta(90− θcoθ
• co(90− θtaθ
• cose(90− θseθ
• se(90− θcoseθ

To know more about Trigonometric Ratios of Complementary Angles, visit here.

## Trigonometric Identities

• sin2θ+cos2θ=1
• 1+cot2θ=coesc2θ
• 1+tan2θ=sec2θ

To know more about Trigonometric Identities, visit here.

### Trigonometry for Class 10 Problems

Example 1:

Find Sin A and Sec A, if 15 cot A = 8.

Solution:

Given that 15 cot A = 8

Therefore, cot A = 8/15.

We know that tan A = 1/ cot A

Hence, tan A = 1/(8/15) = 15/8.

Thus, Side opposite to ∠A/Side Adjacent to ∠A = 15/8

Let BC be the side opposite to ∠A and AB be the side adjacent to ∠A and AC be the hypotenuse of the right triangle ABC respectively.

Hence, BC = 15x and AB = 8x.

Hence, to find the hypotenuse side, we have to use the Pythagoras theorem.

(i.e) AC2 = AB2 + BC2

AC2 = (8x)2+(15x)2

AC2 = 64x2+225x2

AC2 = 289x2

AC = 17x.

Therefore, the hypotenuse AC = 17x.

Finding Sin A:

We know Sin A = Side Opposite to ∠A / Hypotenuse

Sin A = 15x/17x

Sin A = 15/17.

Finding Sec A:

To find Sec A, find cos A first.

Thus, cos A = Side adjacent to ∠A / Hypotenuse

Cos A = 8x/17x

We know that sec A = 1/cos A.

So, Sec A = 1/(8x/17x)

Sec A = 17x/8x

Sec A = 17/8.

Therefore, Sin A = 15/17 and sec A = 17/8.

Example 2:

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°.

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