Trigonometry is a discipline of mathematics that studies the relationships between the lengths of the sides and angles of a right-angled triangle. Trigonometric functions, also known as goniometric functions, angle functions, or circular functions, are functions that establish the relationship between an angle to the ratio of two of the sides of a right-angled triangle. The six main trigonometric functions are sine, cosine, tangent, cotangent, secant, or cosecant.
Angles defined by the ratios of trigonometric functions are known as trigonometry angles. Trigonometric angles represent trigonometric functions. The value of the angle can be anywhere between 0-360°.
Right angled triangle
As given in the above figure in a right-angled triangle:
- Hypotenuse: The side opposite to the right angle is the hypotenuse, It is the longest side in a right-angled triangle and opposite to the 90° angle.
- Base: The side on which angle C lies is known as the base.
- Perpendicular: It is the side opposite to angle C in consideration.
Trigonometric Functions
Trigonometry has 6 basic trigonometric functions, they are sine, cosine, tangent, cosecant, secant, and cotangent. Now let’s look into the trigonometric functions. The six trigonometric functions are as follows,
- sine: It is defined as the ratio of perpendicular and hypotenuse and It is represented as sin θ
- cosine: It is defined as the ratio of base and hypotenuse and it is represented as cos θ
- tangent: It is defined as the ratio of sine and cosine of an angle. Thus the definition of tangent comes out to be the ratio of perpendicular and base and is represented as tan θ
- cosecant: It is the reciprocal of sin θ and is represented as cosec θ.
- secant: It is the reciprocal of cos θ and is represented as sec θ.
- cotangent: It is the reciprocal of tan θ and is represented as cot θ.
According to the above image, Trigonometric Ratios are
Sin θ = Perpendicular / Hypotenuse = AB/AC
Cosine θ = Base / Hypotenuse = BC / AC
Tangent θ = Perpendicular / Base = AB / BC
Cosecant θ = Hypotenuse / Perpendicular = AC/AB
Secant θ = Hypotenuse / Base = AC/BC
Cotangent θ = Base / Perpendicular = BC/AB
Reciprocal Identities
Sin θ = 1/ Cosec θ OR Cosec θ = 1/ Sin θ
Cos θ = 1/ Sec θ OR Sec θ = 1 / Cos θ
Cot θ = 1 / Tan θ OR Tan θ = 1 / Cot θ
Cot θ = Cos θ / Sin θ OR Tan θ = Sin θ / Cos θ
Tan θ.Cot θ = 1
Some other identities are
sin2x + cos2x = 1
1 + tan2x = sec2x
1 + cot2x = cosec2x
Trigonometric Identities of Complementary and Supplementary Angles
- Complementary Angles: Pair of angles whose sum is equal to 90°
- Supplementary Angles: Pair of angles whose sum is equal to 180°
Identities of Complementary angles are
sin (90° – θ) = cos θ
cos (90° – θ) = sin θ
tan (90° – θ) = cot θ
cot (90° – θ) = tan θ
sec (90° – θ) = cosec θ
cosec (90° – θ) = sec θ
Identities of supplementary angles
sin (180° – θ) = sin θ
cos (180° – θ) = – cos θ
tan (180° – θ) = – tan θ
cot (180° – θ) = – cot θ
sec (180° – θ) = – sec θ
cosec (180° – θ) = – cosec θ
Values of Trigonometric Ratios
| 0° | 30° | 45° | 60° | 90° | |
|---|---|---|---|---|---|
| sin θ | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos θ | 1 | √3/2 | 1/√2 | 1 | 0 |
| tan θ | 0 | 1/√3 | 1 | √3 | Not Defined |
| sec θ | Not Defined | 2 | √2 | 2/√3 | 1 |
| cosec θ | 1 | 2/√3 | √2 | 2 | Not Defined |
| cot θ | Not Defined | √3 | 1 | 1/√3 | 0 |
Quadrants of trigonometry
Prove that 1/sec A – tan A – 1/cosA = 1/cos A – 1/sec A + tan A
Solution:
We have 1/sec A – tan A – 1/cosA = 1/ cos A -1/sec A + tan A
First take LHS
1/ (sec A – tan A) – 1/cosA
multiply and divide 1/ (sec A – tan A) by (sec A + tan A)
so now we can write as
= {1/ (sec A – tan A) × (sec A + tan A)/(sec A + tan A)} – 1/cosA
= {(sec A + tan A)/ (sec2 A – tan2 A)} – 1/cosA
= {(sec A + tan A)/ 1} – sec A {1 + tan2x = sec2x or sec2x – tan2x = 1}
= sec A + tan A – sec A {Sec θ = 1 / Cos θ }
= tan A
Now RHS
1/ cos A -1/sec A + tan A
multiply and divide {1/ (sec A + tan A)} by (sec A – tan A)
Now we can write as
= 1/ cos A – [{1/ (sec A + tan A)} × (sec A – tan A )/(sec A – tan A)]
= 1/ cos A – [(sec A – tan A) / (sec2 A – tan2 A)] {1 + tan2x = sec2x or sec2x – tan2x = 1}
= sec A – [(sec A – tan A) / 1] {Sec θ = 1 / Cos θ}
= sec A – sec A + tan A
= tan A
Therefore LHS = RHS
So, 1/sec A – tan A – 1/cos A = 1/ cos A – 1/sec A + tan A
Hence proved
Similar Questions
Question 1: Prove cot2θ – 1/sin2θ = -1
Solution:
We have cot2θ – 1/sin2θ
= cot2θ – cosec2θ { Cosec2θ = 1/ Sin2 θ }
= – (cosec2θ – cot2θ)
= -1
= RHS
Hence Proved
Question 2: Prove (1 + cot2θ) (1 – cos θ)(1 + cos θ) = 1
Solution:
We have LHS
= (1 + cot2θ) (1 – cos θ)(1 + cos θ)
= (1 + cot2θ) (1 – cos2 θ)
= cosec2θ sin2θ {1 + cot2θ = cosec2θ and 1 – cos2 θ = sin2θ}
= (1/sin2θ) × sin2θ {Cosec2θ = 1/ Sin2θ}
= 1
= RHS
Hence Proved