Notes on Trigonometry | Grade 9 > Optional Mathematics > Trigonometry | KULLABS.COM

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The area of mathematics relating to the study of trigonometric function in relation to measurement in triangles is known as trigonometry. Trigonometry is the branch of mathematics which literally consists of three words, 'Tri' referring to 'three', 'gona' meaning 'angle' and 'metry' which means 'measurement'. Hence, trigonometry refers to the measurement of triangles. We shall only deal with right angled triangles.

#### Reference Angles

An angle that is taken into consideration before finding out perpendicular and base in a right angled triangle are known as the reference angle. The reference angles are usually denoted by Greek Alphabets. The side in front of the reference angle is called perpendicular and the remaining side is called base. Perpendicular and base are denoted by p and b respectively. So, in short p, b and h are called the elements of a right angled triangle.

In figure, let ∠C = θ be the reference angle. Then,

AB = Perpendicular (p)

BC = Base (b)

AC = Hypotenuse (h)

Similarly, if $$\angle$$A becomes the reference angle then,

AB = Base (b)

BC = Perpendicular (p)

AC = Hypotenuse (h)

#### Trigonometric Identities

Trigonometric identities are equalities that involve trigonometric functions and are true for every single value of the occurring variables where both sides of the equality are defined. Geometrically, these are identities involving certain functions of one or more angles.Geometrically, these are identities involving certain functions of one or more angles. For examples,

tan x sin x + cos x = sec x

Solution: We will only use the fact that sin2 x + cos2 x = 1 for all values of x.

L.H.S = tan x sin x + cos x

= $$\frac{sin x}{cos x}$$ × sin x + cos x = $$\frac{sin^2 x}{cos x}$$ + cos x

= $$\frac{sin^2 x}{cos x}$$ + cos x

= $$\frac{sin^2 x}{cos x}$$ + $$\frac{cos^2 x}{cos x}$$

= $$\frac{sin^2 x + cos^2 x}{cos x}$$

= $$\frac{1}{cos x}$$

= R.H.S

∴ L.H.S = R.H.S proved.

Proving Trigonometric Identities

Showing the both sides of an identity equal to each other by using various techniques is known as proving trigonometric identity. For example,

sinθ × cotθ = cosθ

Here,

L.H.S = sinθ × cotθ

= sinθ × $$\frac{cosθ}{sinθ}$$

= cosθ

= R.H.S

∴ L.H.S = R.H.S proved.

Techniques for proving Trigonometric Identities

1. Start with left-hand side (L.H.S) and reduce it to the right-hand side (R.H.S) if L.H.S is complicated.
2. Start with right-hand side (R.H.S) and reduce it to the left-hand side (L.H.S) if R.H.S is complicated.
3. If left-hand side and right-hand side are equivalent, reduce both of them to the lowest term.
4. If complicated, transpose or apply the method of cross multiplication to change the form of identity. Then prove the new LHS = new RHS, whatever more priority is given to the proving of the original question.

• The side in front of the reference angle is called perpendicular and the remaining side is called base.
• Perpendicular and base are denoted by p and b respectively.
• Perpendicular, Base and Hypotenuse are called the elements of a right angled triangle.
.

### Very Short Questions

Solution

(sinθ + cosθ) (sinθ + cosθ)

= sin$$\theta$$(sinθ + cosθ) (sinθ + cosθ) + cos$$\theta$$(sinθ + cosθ) (sinθ + cosθ)

= sin2$$\theta$$ + sin$$\theta$$ × cos$$\theta$$ + cos$$\theta$$ × sin$$\theta$$ + cos2$$\theta$$

= sin2$$\theta$$ + sin$$\theta$$ × cos$$\theta$$ + cos$$\theta$$ × sin$$\theta$$ + cos2$$\theta$$

= sin2$$\theta$$ + 2 sin$$\theta$$ × cos$$\theta$$ + cos2$$\theta$$

Solution

(1- tan$$\theta$$) (1+tan$$\theta$$) (1+tan2$$\theta$$) (1+tan4$$\theta$$)

= (1 - tan2$$\theta$$)(1 + tan2$$\theta$$)(1 + tan4$$\theta$$)

= (1 - tan4$$\theta$$)(1 + tan4$$\theta$$)

1 - tan8$$\theta$$

Solution

$$\frac{1}{1 + cosθ}$$ + $$\frac{1}{1-cosθ}$$

=  $$\frac{1}{1 + cosθ}$$ + $$\frac{1}{1-cosθ}$$

= $$\frac{1-cosθ +1+cosθ}{(1-cosθ)(1-cosθ)}$$

= $$\frac{2}{1-cosθ}$$ = $$\frac{2}{sin^2θ}$$ =  2 cosec2θ

Solution

= $$\frac{1}{secA - tanA}$$ - $$\frac{1}{secA + tanA}$$

= $$\frac{secA + tanA - (secA - tanA)}{sec^2A - tan^2A}$$

= $$\frac{secA + tanA - secA + tanA}{1}$$

= 2 tanA

L.H.S = $$\frac{1 - sin\alpha}{cos\alpha}$$

= $$\frac{1 - sin\alpha}{cos\alpha}$$ × $$\frac{1 + sin\alpha}{1 + sin\alpha}$$

= $$\frac{1 - sin\alpha}{cos\alpha(1 + sin\alpha)}$$

= $$\frac{cos\alpha}{(1 + sin\alpha)cos\alpha}$$

= $$\frac{cos\alpha}{1 +sin\alpha}$$ = R.H.S

$$\therefore$$ LHS = RHS proved.

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• ### The area of mathematics relating to the study of trigonometric function in relation to measurement in triangles is known as _________.

trigonometry
measurement of triangles
co-ordinate geometry
reference nagle
• ### An angle that is taken into consideration before finding out perpendicular and base in a right angled triangle are known as ______.

reference angle
co-ordinate geometry
trigonometry
measurement of triangls
• ### The reference angles are usually denoted by ______.

Greek Alphabets
Latin Alphabets
Roman Alphabets
Italic Alphabets
• ### The side in front of the reference angle is called ______.

parallel
perpendicular
base
hypotenuse
• ### ______ are equalities that involve trigonometric functions and are true for every single value of the occurring variables where both sides of the equality are defined.

trigonometric order
trigonometric angles
Trigonometric identities
trigonometric ratio
• ### Trigonometry refers to the ______.

measurement of angles
measurement of square
measurement of triangles
measurement of cube

three
thirteen
teen
thirty

cube
square
angle
triangle

geometry
measurement
angle
triangles

## ASK ANY QUESTION ON Trigonometry

Forum Time Replies Report

##### Rojan Rasaily

CosA-sinA 1/cosA SinA-1=cotA cosecA

##### Buddhiram kewat

Cosecπ 1/cosecπ-1=tan^2π/sec^2-1