Introduction: Trigonometric Ratios | Mathematics (Maths) Class 10 PDF Download
Introduction to Trigonometric Ratios of a Triangle
Trigonometry concerns the relationships between the angles and sides of a right-angled triangle. In a right-angled triangle there are three sides with respect to a given acute angle:
- Hypotenuse - the side opposite the right angle; it is the longest side.
- Opposite side - the side opposite the angle under consideration.
- Adjacent side - the side next to the angle under consideration (other than the hypotenuse).
The ratio of the lengths of these sides, taken with respect to an acute angle, give the basic trigonometric ratios.
Trigonometric ratios
There are six primary trigonometric ratios, each defined for an acute angle θ of a right-angled triangle. These ratios are:
- Sine
- Cosine
- Tangent
- Cosecant
- Secant
- Cotangent
Sine (sin)
The sine of an angle θ is the ratio of the length of the side opposite θ to the hypotenuse.
sin θ = Opposite / Hypotenuse
Cosine (cos)
The cosine of an angle θ is the ratio of the length of the side adjacent to θ to the hypotenuse.
cos θ = Adjacent / Hypotenuse
Tangent (tan)
The tangent of an angle θ is the ratio of the length of the side opposite θ to the side adjacent to θ.
tan θ = Opposite / Adjacent
Cosecant (csc)
The cosecant is the reciprocal of sine. It is the ratio of the hypotenuse to the side opposite θ.
csc θ = Hypotenuse / Opposite = 1 / sin θ
Secant (sec)
The secant is the reciprocal of cosine. It is the ratio of the hypotenuse to the side adjacent to θ.
sec θ = Hypotenuse / Adjacent = 1 / cos θ
Cotangent (cot)
The cotangent is the reciprocal of tangent. It is the ratio of the side adjacent to θ to the side opposite θ.
cot θ = Adjacent / Opposite = 1 / tan θ
| Trigonometric Ratio | Abbreviation | Formula |
|---|---|---|
| sine | sin | Opposite / Hypotenuse |
| cosine | cos | Adjacent / Hypotenuse |
| tangent | tan | Opposite / Adjacent |
| cosecant | csc | Hypotenuse / Opposite |
| secant | sec | Hypotenuse / Adjacent |
| cotangent | cot | Adjacent / Opposite |
Using trigonometry to find a side of a right triangle
Trigonometry is commonly used to determine the length of a side of a right-angled triangle when one side and one acute angle are known, or when two sides are known and an angle is required.
When one side and one angle are known
Steps
- Identify the known side and classify it as hypotenuse, opposite or adjacent with respect to the given angle.
- Choose the trigonometric ratio that relates the known side and the unknown side.
- Set up the equation using the appropriate ratio and solve algebraically for the unknown side.
Example: In a right angled ΔABC ∠B = 30° length of side AB is 4 find length of BC. given tan30 = 1/√3
Sol.
AB is given as 4 and ∠B = 30°.
Identify sides with respect to the angle: for ∠B, AB is the side adjacent or opposite depending on labelling; in the figure AB is the side opposite or adjacent as shown in the images.
Use the trigonometric ratio tan 30° = Opposite / Adjacent = 1/√3.
Form the equation relating given side (4) and unknown BC as per the figure and solve for BC.
When two sides are known
Steps
- Label the known sides as adjacent, opposite or hypotenuse with respect to one of the acute angles.
- Choose the trigonometric ratio that uses the two known sides or that relates a known side to the unknown angle.
- Find the required angle (if needed) using inverse trigonometric functions, then use the ratio to find the unknown side.
- Otherwise, use algebra to find the unknown side directly from the chosen ratio.
Trigonometric ratios of some specific angles
We commonly require exact trigonometric values for angles 0°, 30°, 45°, 60° and 90°. These can be obtained using simple geometric constructions.
Trigonometric ratios of angle C (generic)
With reference to the triangle in the figure, using angle C:
- sin C = BA / AC
- cos C = BC / AC
- tan C = BA / BC
- csc C = AC / BA
- sec C = AC / BC
- cot C = BC / BA
Values at 0° and 90°
When an acute angle A = 0° the side opposite A has length 0 and the adjacent side equals the hypotenuse. When A = 90° the hypotenuse equals the side opposite the other acute angle. Some ratios become undefined where division by zero occurs.
Note: tan 90° and sec 90° are undefined. csc 0° and cot 0° are undefined.
Values at 30° and 60°
Consider an equilateral triangle of side length 2a. Splitting it by a perpendicular from one vertex gives two congruent right triangles with angles 30° and 60° and side lengths a (shorter leg), a√3 (longer leg), and 2a (hypotenuse of the original triangle becomes 2a across the equilateral triangle - for right triangle the hypotenuse is 2a in that construction). From that construction we obtain the standard values.
Values at 45°
In an isosceles right-angled triangle both acute angles are 45°. If each of the equal legs has length a, the hypotenuse has length a√2. Thus:
Result - Standard exact values
The standard exact trigonometric values (useful to memorise) are:
- sin 0° = 0, cos 0° = 1, tan 0° = 0
- sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
- sin 45° = √2/2, cos 45° = √2/2, tan 45° = 1
- sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3
- sin 90° = 1, cos 90° = 0, tan 90° is undefined
Trigonometric identities
An identity is an equation involving trigonometric functions that is true for all permissible values of the variable. The most important fundamental identities follow from definitions on a right triangle and the Pythagorean theorem.
Prove the following identities for any acute angle θ:
- (i) tan θ = sin θ / cos θ
- (ii) cot θ = cos θ / sin θ
- (iii) tan θ · cot θ = 1
- (iv) sin² θ + cos² θ = 1
- (v) 1 + tan² θ = sec² θ
- (vi) 1 + cot² θ = cosec² θ
Proof.
Consider a right-angled triangle ABC with ∠B = 90°. Let AB = x, BC = y and AC = r (hypotenuse).
tan θ = y / x = (y / r) / (x / r)
Therefore tan θ = sin θ / cos θ.
cot θ = x / y = (x / r) / (y / r)
Therefore cot θ = cos θ / sin θ.
tan θ · cot θ = (sin θ / cos θ) · (cos θ / sin θ)
Therefore tan θ · cot θ = 1.
By Pythagoras' theorem, x² + y² = r².
sin² θ + cos² θ = (y / r)² + (x / r)²
= (y² / r²) + (x² / r²)
= (x² + y²) / r²
= r² / r² = 1.
Therefore sin² θ + cos² θ = 1.
1 + tan² θ = 1 + (y / x)²
= (x² + y²) / x²
= r² / x²
= (r / x)² = sec² θ.
Therefore 1 + tan² θ = sec² θ.
1 + cot² θ = 1 + (x / y)²
= (x² + y²) / y²
= r² / y²
= (r / y)² = cosec² θ.
Therefore 1 + cot² θ = cosec² θ.
Applications of trigonometric identities
Application 1: Prove that (1 - sin² θ) sec² θ = 1
Proof:
LHS = (1 - sin² θ) sec² θ
= cos² θ · sec² θ
= cos² θ · (1 / cos² θ)
= 1 = RHS.
Therefore LHS = RHS.
Application 2: Prove that (1 + tan² θ) cos² θ = 1
Proof:
LHS = (1 + tan² θ) cos² θ
= sec² θ · cos² θ
= (1 / cos² θ) · cos² θ
= 1 = RHS.
Therefore LHS = RHS.
Application 3: Prove that (cosec² θ - 1) tan² θ = 1
Proof:
LHS = (cosec² θ - 1) tan² θ
= (1 + cot² θ - 1) tan² θ
= cot² θ · tan² θ
= (1 / tan² θ) · tan² θ
= 1 = RHS.
Therefore LHS = RHS.
Application 4: Prove that sec⁴ θ - sec² θ = tan² θ + tan⁴ θ
Proof:
LHS = sec⁴ θ - sec² θ
= sec² θ (sec² θ - 1)
= sec² θ · tan² θ
= (1 + tan² θ) · tan² θ
= tan² θ + tan⁴ θ = RHS.
Therefore LHS = RHS.
Application 5: Prove that √(sec² θ + cosec² θ) = tan θ + cot θ
Proof:
LHS = √(sec² θ + cosec² θ)
= √((1 + tan² θ) + (1 + cot² θ))
= √(tan² θ + cot² θ + 2)
= √(tan² θ + cot² θ + 2 tan θ · cot θ)
= √(tan θ + cot θ)²
= tan θ + cot θ = RHS.
Therefore LHS = RHS.
Summary
Trigonometric ratios connect the sides of a right-angled triangle with its angles. The six primary ratios (sin, cos, tan, csc, sec, cot) and the fundamental identities derived from them are essential tools for solving problems involving right triangles and for simplifying trigonometric expressions. Exact values at 0°, 30°, 45°, 60°, and 90° are frequently used in problems and should be memorised for quick reference.
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FAQs on Introduction: Trigonometric Ratios - Mathematics (Maths) Class 10
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