Introduction
The word trigonometry comes from the Greek words trigon (triangle) and metron (measure). As the name suggests, it is the study of relationships between the angles and sides of a triangle. In Class 10, we deal only with right-angled triangles and acute angles (angles between 0° and 90°). The key idea is that for a given acute angle, the ratios of the sides of a right-angled triangle containing that angle are always the same, no matter how big or small the triangle is. Those ratios are called trigonometric ratios.
Trigonometric Ratios
Consider a right-angled triangle ABC, right-angled at B. For the acute angle at A, we name the sides like this:
- Hypotenuse, the side opposite the right angle (AC). It is always the longest side.
- Opposite side (or perpendicular), the side opposite to angle A (BC).
- Adjacent side (or base), the side that forms angle A together with the hypotenuse (AB).
If we look at the other acute angle C instead, then AB becomes the opposite side and BC becomes the adjacent side. The opposite and adjacent are always with respect to the angle you are talking about.
For the acute angle A, six ratios are defined:
sin A = opposite / hypotenuse = BC / AC
cos A = adjacent / hypotenuse = AB / AC
tan A = opposite / adjacent = BC / AB
cosec A = hypotenuse / opposite = AC / BC = 1 / sin A
sec A = hypotenuse / adjacent = AC / AB = 1 / cos A
cot A = adjacent / opposite = AB / BC = 1 / tan A
For 0° < A < 90°, all six values are positive. The ratios depend only on the angle A, not on the size of the triangle.
Three facts are used very often, so it helps to remember them directly:
tan A = sin A / cos A and cot A = cos A / sin A (cos A ≠ 0, sin A ≠ 0)
Also, sin A and cos A are always between 0 and 1 for an acute angle A (they can be 0 or 1 only at 0° or 90°, as we will see below), because in a right-angled triangle the hypotenuse is the longest side, so opposite/hypotenuse and adjacent/hypotenuse are always less than 1. But sec A and cosec A are always ≥ 1, and tan A and cot A can be any positive number.
In ΔABC, right-angled at B, AB = 3 cm and BC = 4 cm. Find sin A, cos A and tan A.
First find the hypotenuse AC by Pythagoras: AC2 = AB2 + BC2 = 32 + 42 = 9 + 16 = 25, so AC = 5 cm.
For angle A, opposite = BC = 4, adjacent = AB = 3, hypotenuse = 5.
sin A = 4/5, cos A = 3/5, tan A = 4/3
You can check that tan A = sin A / cos A = (4/5) ÷ (3/5) = 4/3.
If sin A = 1/2, find cos A and tan A, where A is acute.
Draw a right-angled triangle with opposite = 1 and hypotenuse = 2. Let adjacent = x. Then by Pythagoras, x2 + 12 = 22, so x2 = 3 and x = √3.
cos A = √3 / 2, tan A = 1 / √3
Trigonometric Ratios of Some Specific Angles
For most angles we need a calculator, but for 0°, 30°, 45°, 60° and 90° the ratios can be found exactly with simple geometry. NCERT expects you to know this table by heart.
45° comes from an isosceles right-angled triangle (both legs = 1, hypotenuse = √2).
30° and 60° come from an equilateral triangle of side 2, cut in half (sides 1, √3, 2).
0° and 90° are the limiting cases as the angle shrinks to 0° or grows to 90° in a right-angled triangle.
| ∠ A | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin A | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos A | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan A | 0 | 1/√3 | 1 | √3 | not defined (∞) |
| cosec A | not defined (∞) | 2 | √2 | 2/√3 | 1 |
| sec A | 1 | 2/√3 | √2 | 2 | not defined (∞) |
| cot A | not defined (∞) | √3 | 1 | 1/√3 | 0 |
Notice that sin A increases from 0 to 1 as A goes from 0° to 90°, while cos A decreases from 1 to 0. Tan A increases from 0 and becomes not defined at 90° because cos 90° = 0 and tan A = sin A / cos A would mean division by zero.
Evaluate: 2 sin2 60° + 3 cos2 45° − sec2 30°.
Substitute from the table: sin 60° = √3/2, cos 45° = 1/√2, sec 30° = 2/√3.
2(√3/2)2 + 3(1/√2)2 − (2/√3)2 = 2·(3/4) + 3·(1/2) − (4/3) = 3/2 + 3/2 − 4/3 = 3 − 4/3 = 5/3
Trigonometric Ratios of Complementary Angles
Two angles are complementary when they add to 90°. In a right-angled triangle, the two acute angles are always complementary. This gives a direct link between the ratios of an angle and its complement.
For any acute angle A (0° < A < 90°):
sin(90° − A) = cos A , cos(90° − A) = sin A
tan(90° − A) = cot A , cot(90° − A) = tan A
sec(90° − A) = cosec A , cosec(90° − A) = sec A
In words: the sine of an angle equals the cosine of its complement, tangent and cotangent swap, and secant and cosecant swap.
Evaluate sin 48° − cos 42° without a calculator.
Since 42° = 90° − 48°, we have cos 42° = cos(90° − 48°) = sin 48°. So sin 48° − cos 42° = sin 48° − sin 48° = 0.
Evaluate tan 26° / cot 64°.
cot 64° = cot(90° − 26°) = tan 26°. So tan 26° / cot 64° = tan 26° / tan 26° = 1.
Trigonometric Identities
An identity is an equation that is true for all values of the angle for which both sides are defined. There are three fundamental identities that NCERT expects you to know and to be able to prove. All other 90°-type problems in the exercises reduce to these three with basic algebra.
For an acute angle A (0° ≤ A ≤ 90°) where the expressions are defined:
(1) sin2 A + cos2 A = 1
(2) 1 + tan2 A = sec2 A (for 0° ≤ A < 90°)
(3) 1 + cot2 A = cosec2 A (for 0° < A ≤ 90°)
Equivalently, (2) can be written as sec2 A − tan2 A = 1, and (3) as cosec2 A − cot2 A = 1.
Why (1) is true: In right ΔABC with ∠B = 90° and ∠A = A, let AB = x, BC = y, AC = r. Then sin A = y/r, cos A = x/r, and by Pythagoras x2 + y2 = r2. So sin2 A + cos2 A = y2/r2 + x2/r2 = (x2 + y2)/r2 = 1. Identities (2) and (3) follow by dividing (1) by cos2 A and sin2 A respectively.
A very useful rearrangement of (1) is:
sin2 A = 1 − cos2 A , cos2 A = 1 − sin2 A
These let you find one of sin A or cos A if you know the other.
If sin A = 3/5, find cos A and verify sin2 A + cos2 A = 1.
Using the identity: cos2 A = 1 − sin2 A = 1 − (3/5)2 = 1 − 9/25 = 16/25. Since A is acute, cos A is positive, so cos A = 4/5. Check: (3/5)2 + (4/5)2 = 9/25 + 16/25 = 25/25 = 1.
Prove that (1 + tan2 A) / (1 + cot2 A) = tan2 A.
Left-hand side = sec2 A / cosec2 A [using 1 + tan2 = sec2 and 1 + cot2 = cosec2]
= (1/cos2 A) / (1/sin2 A) = sin2 A / cos2 A = tan2 A = RHS
Tip for board exams: start from one side (usually the more complicated one) and use the identities to reach the other side. Avoid cross-multiplying unless you are sure neither side is zero.
Some identities are easier if you write everything in terms of sin A and cos A first. For example, to simplify (sec A + tan A)(1 − sin A), write sec A = 1/cos A and tan A = sin A / cos A, then combine.
Summary
- Right-angled triangle: for acute angle A, sin A = opposite/hypotenuse, cos A = adjacent/hypotenuse, tan A = opposite/adjacent; cosec A = 1/sin A, sec A = 1/cos A, cot A = 1/tan A.
- Relations: tan A = sin A / cos A, cot A = cos A / sin A; ratios depend only on the angle, not the size of the triangle.
- Specific values for 0°, 30°, 45°, 60°, 90° as in the table above, to be memorised. In particular: sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2; cos 30° = √3/2, cos 45° = 1/√2, cos 60° = 1/2; tan 30° = 1/√3, tan 45° = 1, tan 60° = √3.
- Complementary angles: sin(90° − A) = cos A, cos(90° − A) = sin A, tan(90° − A) = cot A, cot(90° − A) = tan A, sec(90° − A) = cosec A, cosec(90° − A) = sec A.
- Identities: sin2 A + cos2 A = 1; 1 + tan2 A = sec2 A; 1 + cot2 A = cosec2 A.
- For acute A, 0 ≤ sin A ≤ 1 and 0 ≤ cos A ≤ 1, while sec A, cosec A ≥ 1.
Practice Questions
- In ΔPQR, right-angled at Q, PQ = 5 cm and PR = 13 cm. Find sin P, cos P and tan P. Also find sin R and cos R and verify that sin P = cos R.
- If cos A = 12/13, find sin A and tan A, where A is an acute angle.
- Evaluate without a calculator: 2 tan2 45° + cos2 30° − sin2 60°.
- Evaluate: sin 25° cos 65° + cos 25° sin 65°.
- Prove the identity: (cosec A − sin A)(sec A − cos A) = 1 / (tan A + cot A).
- Prove that (sin4 A − cos4 A) / (sin2 A − cos2 A) = 1.