Trigonometry is the branch of mathematics that deals with the sides and angles of a right-angled triangle. It is derived from the Greek words ‘tri’ which means three, ‘gon’ which means sides, ‘metron’ which means measure. It was used by early astronomers and in Egypt and Babylon. It is used today in various subjects like Architecture, Engineering, Physical Science. It is in class 10 that students have an **introduction to Trigonometry.** In this blog, we will study class 10 trigonometry and trigonometric ratios for different angles of measure ( 0 to 90 degrees) in detail.

**Must Read: Probability Class 10 Maths Study Notes**

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## Trigonometric Ratios

The first part in class 10 trigonometry is Trigonometric Ratios. Trigonometric ratios are the **ratio of sides** of a right-angle triangle and the acute angles.

There are 6 Trigonometric ratios. **Sine, Cosine, Tangent, Cotangent, Secant, Cosecant.**

In the right angle triangle given above **∠****B is 90 degrees. **

We will define the Trigonometric Ratios of ∠A.

**Sine of****∠****A = SinA = Side opposite to Angle A / Hypotenuse = BC / AC****Cosine of****∠****A = CosA = Side adjacent to Angle A / Hypotenuse = AB / AC****Tangent of****∠****A = TanA = Side opposite to Angle A / Side adjacent to Angle A = BC / AB****Cotangent of****∠****A = CotA= 1/TanA = Side Adjacent to Angle A / Side opposite to Angle A = AB/BC****Cosecant of****∠****A = CosecA = 1 / SinA = Hypotenuse / Side opposite to Angle A = AC/BC****Secant of****∠****A = SecA = 1/CosA = Hypotenuse/Side adjacent to Angle A = AC/AB**

Observe that TanA is SinA / CosA and CosecA, SecA, CotA are **reciprocals **of SinA, CosA, TanA respectively.

It is important to understand class 10 trigonometry through examples.

**Example 1:** In triangle ABC, ∠B = 90 degrees, tan A = 6/5, Find other trigonometric ratios of Angle A

**Solution**:

Given: ∠B = 90. So AC is hypotenuse

TanA = BC/AB =6/5 = 6k/5k

for k having positive value.

So BC= 6k and AB = 5k

Using Pythagoras Theorem

AB^{2} + BC^{2 }= AC^{2 }

(5k)^{2} + (6k)^{2} = 25k^{2} + 36k^{2}

= 61k^{2 } = AC^{2}

Square rooting both sides**√**61k = AC

Now we have a length of all sides.

- SinA = BC/AC = 6k/
**√**61k = 6/**√**61 - CosA = AB/AC = 5k/
**√**61k = 5/**√**61 - TanA = 6/5 (GIVEN)
- CotA = 1/TanA = 5/6
- CosecA = 1/SinA =
**√**61/6 - SecA = 1/COsA =
**√**61/5

**Example 2**:

In triangle ABC, right-angled at C, AB= 5 units and BC= 4 units, and ∠B = θ. Find the value of

- Cos
^{2}θ + Sin^{2}θ - Cos
^{2}θ – Sin^{2.}θ

**Solutions**:

Using pythagoras theorem, AC^{2} = AB^{2} – BC^{2}

AC^{2 .}= 5^{2 }– 4^{2} = ( 5-4) ( 5 + 4 )

AC^{2 }= (1) (9) = 9

AC = 3 units

Sinθ = AC/AB = 3/5 , Cosθ = BC/AB = 4/5

Now

- Cos
^{2}θ + Sin^{2}θ = (4/5)^{2 }+ (3/5)^{2 }= 16+9/25 = 25/25 = 1 - Cos
^{2}θ – Sin^{2}θ = (4/5)^{2 }– (3/5)^{2 }= 16 – 9/25 = 7/25

**Also Read: Statistics Class 10 Study Notes**

## Trigonometric ratios of Some Specific Angles

θ | 0 | 30 | 45 | 60 | 90 |

Sinθ | 0 | 1/2 | 1/√2 | √3/2 | 1 |

Cosθ | 1 | √3/2 | 1/√2 | 1/2 | 0 |

Tanθ | 0 | 1/√3 | 1 | √3 | Not Defined |

Cosecθ | Not Defined | 2 | √2 | 2/√3 | 1 |

Secθ | 1 | 2/√3 | √2 | 2 | Not Defined |

Cotθ | Not Defined | √3 | 1 | 1/√3 | 0 |

Using the above information let’s solve the following questions as it is easier to understand class 10 trigonometry:

**Q1**. In an triangle ABC, right-angled at B, AB = 7 and ∠ACB = 30 degrees. Find the lengths of sides BC and AC.

**Solution**:

Given: ∠ACB = 30 degrees, AB=7 cm,

Let BC = x cm and AC = y cm

TanC = AB/BC = 7/x

TanC = Tan30 = 1/√3

7/x = 1/√3

x = 7√3

BC = x = 7√3

To find AC

Sin30 = AB/AC

1/2 = 7/y

AC = y = 14

Alternatively, we could have used the Pythagoras theorem to find AC

I.e. AC^{2 }= AB^{2 }+ BC^{2 }= 7^{2 }+ (7√3)^{2}

= 49 + 147

= 196

AC= 14 cm

**Q2**. If tan(A+B)= √3 and tan(A–B)= 1/√3 ;0°<A+B≤90° ; A>B, find A and B.

**Solution**.

tan(A+B) = √3

Since tan60° = √3

tan(A+B) = tan60°

Therefore A+B = 60 –(i)

tan(A-B) = 1/√3

Since tan30° = 1/√3

tan(A-B) = tan30°

Therefore A-B = 30 –(ii)

Now adding (i) and (ii).

A + B + A – B = 60 + 30

2A = 90

A = 45

Substitute this value of A in (i)

A + B = 60

45 + B = 60

B = 15

Therefore A = 45°, B = 15°

**Explore: Pair of Linear Equations in Two Variables Class 10**

## Trigonometric Ratios of Complementary Angles

We move on next to trigonometric ratios of complementary angles in class 10 trigonometry.

If θ is an acute angle, its **complementary angle is 90-θ. **The trigonometric ratios of complementary angles.

**Sin( 90-****θ****) = Cos****θ****Cos (90-θ) = Sinθ****Tan(90-θ) = Cotθ****Cot(90-θ) = Tanθ****Cosec(90-θ) = Secθ****Sec(90-θ) = Cosecθ**

Let us understand this concept of class 10 trigonometry through these practice questions:

**Q.1** Evaluate Cos55 / Sin35

**Solution**: Sin(35) = Cos (90 – 35) = Cos(55)

Cos(55)/Cos(55) = 1

**Q.2** If Sin 3A = Cos (A-30), where 3A is an acute angle, Find the value of A.

**Solution**: Given Sin3A = Cos(A-30)

Since Sin3A = Cos(90-3A)

So Cos(90-3A) = Cos(A-30)

90-3A = A – 30

120 = 4A

A = 30

**Q3.** Express Tan75 + Cos65 in terms of trigonometric ratios of angles between 0 and 45

**Solution**: Tan75 + Cos65 = Tan(90-15) + Cos(90-25)

= Cot15 + Sin25

## Trigonometric Identities

We further move on to trigonometric identities in class 10 trigonometry. An equation involving trigonometric ratios of an angle is called a** trigonometric identity. **

- Cos
^{2}A + Sin^{2}A = 1 - 1 + Tan
^{2}A = Sec^{2}A - Cot
^{2}A + 1 = Cosec^{2}A

We can understand this section of class 10 trigonometry better with these practice questions:

**Q.1** Prove that CosecA (1-CosA) (CosecA + CotA) = 1

**Solution**:

CosecA (1-CosA) (CosecA + CotA) = (1/SinA) (1-CosA) ( 1/SinA + CosA/SinA)

(1-CosA)(1+CosA) / Sin^{2}A = 1-Cos^{2}A / Sin^{2}A

Sin^{2}A / Sin^{2}A = 1

**Q.2** Prove that Cosθ − Sinθ + 1 / Cosθ + Sinθ -1 = 1 / Cosecθ – Cotθ,

using the identity Sec^{2}θ = 1 + Tan^{2}θ

**Solution**: LHS = Cosθ – Sinθ + 1 / Cosθ + Sinθ – 1

Divide both LHS and RHS by Sinθ

= Cotθ – 1 + Cosecθ / Cotθ + 1 – Cosecθ

= (Cotθ + Cosecθ) – 1 / (Cotθ – Cosecθ) + 1

Multiplying both numerator and denominator by Cotθ – Cosecθ

= {(Cotθ + Cosecθ) – 1 } (Cotθ – Cosecθ) / {(Cotθ – Cosecθ) + 1} (Cotθ – Cosecθ)

= ( Cot^{2}θ – Cosec^{2}θ) – (Cotθ – Cosecθ) / (Cotθ – Cosecθ) + 1) (Cotθ – Cosecθ)

Now, Cot^{2}θ + 1 = Cosec^{2}θ

Cot^{2}θ – Cosec^{2}θ = -1

= -1 -Cotθ + Cosecθ / (Cotθ – Cosecθ) + 1) (Cotθ – Cosecθ)

= -1 (Cotθ – Cosecθ + 1 ) / (Cotθ – Cosecθ + 1) (Cotθ – Cosecθ)

= -1 / Cotθ – Cosecθ

= 1 / Cosecθ – Cotθ

Which is the RHS of the identity.

So Hence Proved.

**Explore: Surface Area And Volume Class 10 Maths**

Thus, we come to the end of our blog. We hope this blog helped you understand and solve trigonometric questions of class 10 trigonometry. For more admission, related queries contact our experts at **Leverage Edu.** Sign up for a free session today!