**Trigonometric
Ratios**

**Trigonometry** deals with the measurement of triangles,
that is, it deals with the relationship between the sides and angles of
triangles.

Let us consider a right-angled triangle ABC in which ∠ABC
= 90°. This triangle consists of a right angle, two acute angles and three
sides. These are called **elements** of a right-angled triangle.

One of the acute angle is considered as the **angle of reference**.
This angle of reference is used for naming the sides of a right-angled
triangle.

The side opposite to the angle of reference is called **perpendicular**
and it is denoted by the letter **p**. The side opposite to the right angle is
called **hypotenuse** and it is denoted by the letter **h**. The remaining side is **base**
and it is denoted by **b**. Clearly, the base is the side between right angle and the
angle of reference.

If ∠ACB = θ is taken as the angle of reference, then AB, AC and BC are perpendicular, hypotenuse and base of the right angled triangle ABC.

If ∠BAC = φ is the angle of reference, then BC, AC and AB respectively are perpendicular, hypotenuse and base of the same right angled triangle ABC.

So, the name of the
sides of a right angled triangle depends on the choice of the angle of reference.

With the help of three sides of a right angled
triangle, six ratios can be derived taking any one acute angle as the angle of reference.

**Six ratios are:**

(i)
AB/BC

(ii)
AB/CA

(iii)
BC/AB

(iv)
BC/CA

(v)
CA/AB

(vi)
CA/BC

These ratios are commonly known as **trigonometric
ratios**.

If ∠ACB = θ is taken as angle of reference, then

AB = perpendicular (p)

CA = hypotenuse (h)

BC = base (b)

Now, the six trigonometric ratios of DABC
right angled at B with angle of reference θ are as follows:

(i)
AB/CA = p/h = Sine of angle θ = Sinθ

(ii)
CA/AB = h/p = Cosecant of angle θ = Cosecθ

(iii)
BC/CA = b/h = Cosine of angle θ = Cosθ

(iv)
CA/BC = h/b = Secant of angle θ = Secθ

(v)
AB/BC = p/b = Tangent of angle θ = Tanθ

(vi)
BC/AB = b/p = Cotangent of angle θ = Cotθ

Traditionally, trigonometric ratios are defined on the
angles of a triangle as above. But, now-a-days these trigonometric ratios are
defined on the angles of any magnitude as below.

Let us consider an angle θ placed in the standard
position. Draw circle with centre at O and radius r. Let the circle intersect
the terminal line at some point P(x, y) as shown in the figure. Draw
perpendicular PM from P to x-axis. Then, OM = x, PM = y and OP = r. For any
angle θ, the six trigonometric ratios can be defined in terms of x-coordinate
‘x’, y-coordinate ‘y’ and the radius ‘r’ by the formula.

Sinθ = y/r Cosθ = x/r Tanθ = y/x

Cosecθ = r/y Secθ
= r/x Cotθ
= x/y

**Fundamental
Relations of Trigonometric Ratios**

There are generally 3 fundamental relations of
trigonometric ratios.

**(a) Reciprocal Relations**

**(b) Quotient Relations**

**(c) Pythagorean
Relations**

Let us consider a circle centered at the origin O and
having radius r. Let P(x, y) be any point on the circle. Join OP and let ∠XOP
= θ. Draw perpendicular PM from P to x-axis. Then OM = x, PM = y and OP = r.

Now, Sinθ = y/r Cosθ
= x/r Tanθ
= y/x

Cosecθ =
r/y Secθ = r/x Cotθ = x/y

**(a) ****Reciprocal Relations**: Out
of six ratios, there are 3 pairs which form reciprocal relations. These 3 pairs
are: sinθ and cosecθ, cosθ and secθ and tanθ and cotθ.

(1) We know, y/r × r/y = 1

or, sinθ × cosecθ = 1 ……….. (i)

or, sinθ = 1/cosecθ …………. (ii)

or, cosecθ = 1/sinθ …………. (iii)

(2) x/r × r/x = 1

or, cosθ × secθ = 1 ………….. (iv)

or, cosθ = 1/secθ …………….. (v)

or, secθ = 1/cosθ …………….. (vi)

(3) y/x × x/y = 1

or, tanθ × cotθ = 1 …………… (vii)

or, tanθ = 1/cotθ …………..… (viii)

or, cotθ = 1/tanθ …………….. (ix)

Hence, the reciprocal
relations are:

sinθ
. cosecθ = 1 |
sinθ
= 1/cosecθ |
cosecθ
= 1/sinθ |

cosθ
. secθ = 1 |
cosθ
= 1/secθ |
secθ
= 1/cosθ |

tanθ
. cotθ = 1 |
tanθ
= 1/cotθ |
cotθ
= 1/tanθ |

**(b) ****Quotient Relations**

(1)We have,

tanθ
= y/x = (y/r)/(x/r) = sinθ/cosθ = secθ/cosecθ

(2)Again,

cotθ
= x/y = (x/r)/(y/r) = cosθ/sinθ = cosecθ/secθ

Hence, the quotient
relations are:

tanθ
= sinθ/cosθ |
cotθ
= cosθ/sinθ |

tanθ
= secθ/cosecθ |
cotθ
= cosecθ/secθ |

**(c) ****Pythagorean
Relations**

(1) By Pythagoras theorem,

x^{2}
+ y^{2} = r^{2}

Dividing
both sides by r^{2}, we get

(x/r)2
+ (y/r)^{2} = 1

or, (cosθ)^{2} + (sinθ)^{2}
= 1

or, sin^{2}θ + cos^{2}θ = 1
………………. (i)

or, sin^{2}θ = 1 – cos^{2}θ
………………. (ii)

or, cos^{2}θ = 1 – sin^{2}θ
………………. (iii)

(2) r^{2} – y^{2}
= x^{2}

Dividing
both sides by x^{2}, we get

(r/x)^{2}
– (y/x)^{2} = 1

or, (secθ)^{2} – (tanθ)^{2}
= 1

or, sec^{2}θ – tan^{2}θ
= 1 ………………. (iv)

or, sec^{2}θ = 1 + tan^{2}θ
………………. (v)

or, tan^{2}θ = sec^{2}θ – 1
…………….... (vi)

(3) r^{2} – x^{2}
= y^{2}

Dividing
both sides by y^{2}, we get

(r/y)^{2}
– (x/y)^{2} = 1

or, (cosecθ)^{2} – (cotθ)^{2}
= 1

or, cosec^{2}θ – cot^{2}θ =
1 ……………. (vii)

or, cosec^{2}θ = 1 + cot^{2}θ
…………… (viii)

or, cot^{2}θ = cosec^{2}θ –
1 ……….….. (ix)

Hence, the Pythagorean
relations are:

sin |
sec |
cosec |

sin |
sec |
cosec |

cos |
tan |
cot |

**Derived
Relations:**

From the above relations, it is clear that the
following relations are also true.

(1) sinθ . cosecθ = 1. So, sin^{2}θ
. cosec^{2}θ = 1, sin^{3}θ . cosec^{3}θ = 1 and so on.

(2) sinθ = 1/cosecθ. So, sin^{2}θ
= 1/cosec^{2}θ, sin^{3}θ = 1/cosec^{3}θ and so on.

(3) cosecθ = 1/sinθ. So, cosec^{2}θ
= 1/sin^{2}θ, cosec^{3}θ = 1/sin^{3}θ and so on.

(4) cosθ . secθ = 1. So, cos^{2}θ
. sec^{2}θ = 1, cos^{3}θ . sec^{3}θ = 1 and so on.

(5) cosθ = 1/secθ. So, cos^{2}θ
= 1/sec^{2}θ, cos^{3}θ = 1/sec^{3}θ and so on.

(6) secθ = 1/cosθ. So, sec^{2}θ
= 1/cos^{2}θ, sec^{3}θ = 1/cos^{3}θ and so on.

(7) tanθ . cotθ = 1. So, tan^{2}θ
. cot^{2}θ = 1, tan^{3}θ . cot^{3}θ = 1 and so on.

(8) tanθ = 1/cotθ. So, tan^{2}θ
= 1/cot^{2}θ, tan^{3}θ = 1/cot^{3}θ and so on.

(9) cotθ = 1/tanθ. So, cot^{2}θ
= 1/tan^{2}θ, cot^{3}θ = 1/tan^{3}θ and so on.

(10) tanθ = sinθ/cosθ. So, tan^{2}θ =
sin^{2}θ/cos^{2}θ, tan^{3}θ = sin^{3}θ/cos^{3}θ
and so on.

(11) tanθ = secθ/cosecθ. So, tan^{2}θ =
sec^{2}θ/cosec^{2}θ, tan^{3}θ = sec^{3}θ/cosec^{3}θ
and so on.

(12) cotθ = cosθ/sinθ. So, cot^{2}θ =
cos^{2}θ/sin^{2}θ, cot^{3}θ = cos^{3}θ/sin^{3}θ
and so on.

(13) cotθ = cosecθ/secθ. So, cot^{2}θ =
cosec^{2}θ/sec^{2}θ, cot^{3}θ = cosec^{3}θ/sec^{3}θ
and so on.

(14) sin^{2}θ = 1 – cos^{2}θ.
So, sinθ = √(1 – cos^{2}θ)

(15) cos^{2}θ = 1 – sin^{2}θ.
So, cosθ = √(1 – sin^{2}θ)

(16) sec^{2}θ = 1 + tan^{2}θ.
So, secθ = √(1 + tan^{2}θ)

(17) tan^{2}θ = sec^{2}θ – 1.
So, tanθ = √( sec^{2}θ – 1)

(18) cosec^{2}θ = 1 + cot^{2}θ.
So, cosecθ = √(1 + cot^{2}θ)

(19) cot^{2}θ = cosec^{2}θ – 1.
So, cotθ = √( cosec^{2}θ – 1)

Follow: Trigonometric Identities for worked out examples.

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