Class 10 — Mathematics (NCERT)

Introduction to Trigonometry

Class 10

✓By the end of this lesson students will be able to define and calculate trigonometric ratios for acute angles in a right-angled triangle.
✓By the end of this lesson students will be able to recall and apply the values of trigonometric ratios for specific angles (0°, 30°, 45°, 60°, 90°).
✓By the end of this lesson students will be able to state and use the fundamental trigonometric identities to simplify expressions and prove other identities.
✓By the end of this lesson students will be able to solve problems involving trigonometric ratios and identities.

Key concepts

Trigonometric Ratios

In a right-angled triangle, the trigonometric ratios of an acute angle are defined as the ratio of the lengths of its sides. Consider a right-angled triangle ABC, right-angled at B. For the acute angle A:\n\n1. Sine of angle A (sin A) = Side opposite to angle A / Hypotenuse = BC / AC\n2. Cosine of angle A (cos A) = Side adjacent to angle A / Hypotenuse = AB / AC\n3. Tangent of angle A (tan A) = Side opposite to angle A / Side adjacent to angle A = BC / AB\n\nThe reciprocals of these ratios are:\n\n4. Cosecant of angle A (cosec A) = 1 / sin A = Hypotenuse / Side opposite to angle A = AC / BC\n5. Secant of angle A (sec A) = 1 / cos A = Hypotenuse / Side adjacent to angle A = AC / AB\n6. Cotangent of angle A (cot A) = 1 / tan A = Side adjacent to angle A / Side opposite to angle A = AB / BC\n\nAlso, tan A = sin A / cos A and cot A = cos A / sin A.

Trigonometric Ratios of Specific Angles

The values of trigonometric ratios for certain specific angles (0°, 30°, 45°, 60°, 90°) are important and should be memorised. These values are derived from special right-angled triangles (e.g., isosceles right triangle for 45°, equilateral triangle for 30° and 60°).

Angle (A) | sin A | cos A | tan A | cosec A | sec A | cot A\n---|---|---|---|---|---|---\n0° | 0 | 1 | 0 | Not defined | 1 | Not defined\n30° | 1/2 | √3/2 | 1/√3 | 2 | 2/√3 | √3\n45° | 1/√2 | 1/√2 | 1 | √2 | √2 | 1\n60° | √3/2 | 1/2 | √3 | 2/√3 | 2 | 1/√3\n90° | 1 | 0 | Not defined | 1 | Not defined | 0

Trigonometric Identities

An equation involving trigonometric ratios of an angle is called a trigonometric identity if it is true for all values of the angle for which the trigonometric ratios are defined. The three fundamental trigonometric identities are:\n\n1. sin²A + cos²A = 1\n2. 1 + tan²A = sec²A\n3. 1 + cot²A = cosec²A\n\nThese identities can be derived from the Pythagorean theorem and are extremely useful for simplifying trigonometric expressions and proving other identities.

sin²A + cos²A = 1\n1 + tan²A = sec²A\n1 + cot²A = cosec²A

Key facts to remember

1Trigonometric ratios relate the sides of a right-angled triangle to its acute angles.
2sin A = Opposite/Hypotenuse, cos A = Adjacent/Hypotenuse, tan A = Opposite/Adjacent.
3cosec A = 1/sin A, sec A = 1/cos A, cot A = 1/tan A.
4tan A = sin A / cos A and cot A = cos A / sin A.
5The values of trigonometric ratios for 0°, 30°, 45°, 60°, 90° must be memorised.
6The three fundamental trigonometric identities are: sin²A + cos²A = 1, 1 + tan²A = sec²A, 1 + cot²A = cosec²A.

Worked examples

Example 1

If sin A = 3/5, find the values of cos A and tan A.

IGiven sin A = 3/5.

IIWe know that sin²A + cos²A = 1 (Trigonometric Identity).

IIISubstitute the value of sin A: (3/5)² + cos²A = 1.

IV9/25 + cos²A = 1.

Vcos²A = 1 - 9/25.

VIcos²A = (25 - 9) / 25.

VIIcos²A = 16/25.

VIIIcos A = √(16/25) = 4/5 (Since A is an acute angle, cos A is positive).

9Now, to find tan A, we use the relation tan A = sin A / cos A.

10tan A = (3/5) / (4/5).

11tan A = 3/4.

Answer

cos A = 4/5, tan A = 3/4

Alternatively, one could draw a right-angled triangle, label the opposite side as 3k and hypotenuse as 5k, then use Pythagoras theorem to find the adjacent side as 4k, and then write the ratios.

Example 2

Evaluate: (sin 30° + tan 45° - cosec 60°) / (sec 30° + cos 60° + cot 45°)

IRecall the values of the trigonometric ratios for specific angles:

IIsin 30° = 1/2

IIItan 45° = 1

IVcosec 60° = 2/√3

Vsec 30° = 2/√3

VIcos 60° = 1/2

VIIcot 45° = 1

VIIISubstitute these values into the expression:

9Numerator = (1/2 + 1 - 2/√3)

10Numerator = (3/2 - 2/√3) = (3√3 - 4) / (2√3)

11Denominator = (2/√3 + 1/2 + 1)

12Denominator = (2/√3 + 3/2) = (4 + 3√3) / (2√3)

13Now, divide the Numerator by the Denominator:

14Expression = [(3√3 - 4) / (2√3)] / [(4 + 3√3) / (2√3)]

15Expression = (3√3 - 4) / (4 + 3√3)

16To rationalise the denominator, multiply the numerator and denominator by the conjugate of the denominator (4 - 3√3):

17Expression = [(3√3 - 4) * (4 - 3√3)] / [(4 + 3√3) * (4 - 3√3)]

18Numerator = 12√3 - (3√3)² - 16 + 12√3 = 12√3 - 27 - 16 + 12√3 = 24√3 - 43

19Denominator = 4² - (3√3)² = 16 - 27 = -11

20Expression = (24√3 - 43) / (-11)

21Expression = (43 - 24√3) / 11

Answer

(43 - 24√3) / 11

Careful calculation and rationalisation of the denominator are crucial steps.

Example 3

Prove the identity: (1 + tan²A) / (1 + cot²A) = tan²A

IStart with the L.H.S.: (1 + tan²A) / (1 + cot²A)

IIWe know the trigonometric identities: 1 + tan²A = sec²A and 1 + cot²A = cosec²A.

IIISubstitute these identities into the L.H.S.: L.H.S. = sec²A / cosec²A

IVRecall that sec A = 1/cos A and cosec A = 1/sin A.

VSo, sec²A = 1/cos²A and cosec²A = 1/sin²A.

VISubstitute these into the expression: L.H.S. = (1/cos²A) / (1/sin²A)

VIIL.H.S. = (1/cos²A) * (sin²A/1)

VIIIL.H.S. = sin²A / cos²A

9We know that tan A = sin A / cos A, so tan²A = sin²A / cos²A.

10Therefore, L.H.S. = tan²A.

11This is equal to the R.H.S.

12Hence proved.

Answer

L.H.S. = R.H.S. = tan²A

Always start with one side (usually the more complex one) and transform it into the other side using known identities and algebraic manipulation.

Common mistakes

✗Confusing the ratios, e.g., writing cos A as Opposite/Hypotenuse instead of Adjacent/Hypotenuse.
✗Incorrectly recalling or applying the values of trigonometric ratios for specific angles.
✗Making algebraic errors when simplifying expressions or solving equations.
✗Not squaring the entire ratio when using identities like sin²A (e.g., writing sin A² instead of (sin A)²).
✗Trying to prove identities by manipulating both sides simultaneously, which is not a valid proof method.

Exam tips

★Memorise the trigonometric ratios and their reciprocal relationships thoroughly. A mnemonic like 'Pandit Badri Prasad Har Har Bole, Sona Chandi Tole' can be helpful.
★Practise drawing the table for specific angle values quickly and accurately to avoid errors during exams.
★When proving identities, always start with one side (usually the more complicated one) and transform it step-by-step into the other side. Show all intermediate steps clearly.
★Be careful with algebraic manipulations, especially when dealing with fractions and square roots.

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