Introduction to Trigonometry

CBSE Class 10 · Mathematics · Notes, formulas and practice questions

Learn the six trigonometric ratios of an acute angle in a right triangle, their exact values at 0°, 30°, 45°, 60° and 90°, and the identities used to prove trigonometric relations.

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What this chapter covers

This chapter is about connecting an acute angle of a right triangle to the ratios of its sides. If you fix one acute angle A, the three sides are named with respect to that angle: hypotenuse, opposite side and adjacent side. Taking two sides at a time gives six trigonometric ratios — sine, cosine, tangent, cosecant, secant and cotangent. These ratios depend only on the size of the angle, not on the size of the triangle, because similar right triangles have proportional sides.

The chapter then fixes five particular angles: 0°, 30°, 45°, 60° and 90°. The values at 30° and 60° are obtained by splitting an equilateral triangle into two right triangles; the value at 45° comes from an isosceles right triangle; 0° and 90° are limiting cases. You are expected to know the exact values, for example sin 30° = 1/2, sin 45° = √2/2 and sin 60° = √3/2. Rote memorisation helps, but so does remembering which triangle each value came from.

The third idea is trigonometric identities. From Pythagoras's theorem for a right triangle, sin²A + cos²A = 1. Dividing this by cos²A gives 1 + tan²A = sec²A, and dividing by sin²A gives 1 + cot²A = cosec²A. These three identities hold for all values of A for which the terms are defined. They are the main tool for proving that a complicated-looking expression is equal to a simpler one, a standard examination skill in this chapter.

Most mistakes in this chapter come from three places: using the wrong pair of sides for an acute angle, confusing √3/2 with 1/√3, and losing terms while proving identities. Always name the reference angle first, then mark the opposite and adjacent sides relative to it. After substituting standard values, check your result numerically. Once the ratios and identities feel automatic, this is one of the highest-scoring portions of the Class 10 syllabus.

Key terms

Trigonometric ratio: For a fixed acute angle in a right triangle, any of the six ratios obtained by dividing two sides of the triangle. These ratios depend only on the angle and remain the same for every similar right triangle drawn with that angle.
Hypotenuse: The side opposite the right angle and the longest side in a right triangle. It is the denominator when writing sin A, cos A, cosec A and sec A.
Opposite side: For an acute angle A in a right triangle, the side that does not touch A and is not the hypotenuse. In triangle ABC with the right angle at B, the opposite side to angle A is BC.
Adjacent side: For an acute angle A, the side that joins A to the right angle; it touches angle A but is not the hypotenuse. In triangle ABC right-angled at B, the adjacent side to angle A is AB.
Sine and cosine: For an acute angle A, sin A = opposite/hypotenuse and cos A = adjacent/hypotenuse. Their values are between 0 and 1 for acute angles.
Tangent: For an acute angle A, tan A = opposite/adjacent, so it equals sin A/cos A. tan 45° = 1, tan 30° = 1/√3 and tan 60° = √3.
Cosecant, secant and cotangent: Reciprocals of the three main ratios: cosec A = 1/sin A, sec A = 1/cos A and cot A = 1/tan A. Their names appear mostly with identities and while simplifying expressions.
Standard angles: The angles 0°, 30°, 45°, 60° and 90° whose trigonometric ratios are exact numbers, such as cos 60° = 1/2. These values come from the 30°-60°-90° and 45°-45°-90° right triangles and from the limiting cases at 0° and 90°.
Trigonometric identity: An equation such as sin²A + cos²A = 1 that holds for every value of A for which both sides are defined. To prove one, keep changing one side using known identities until it matches the other side.

Formula sheet

What	Formula	Notes
Sine ratio	`sin A = opposite/hypotenuse`	A is an acute angle of a right triangle and opposite is the side facing A.
Cosine ratio	`cos A = adjacent/hypotenuse`	Adjacent is the side joining A to the right angle; it is not the hypotenuse.
Tangent ratio	`tan A = opposite/adjacent`	Tan A is also equal to sin A/cos A, and cos A is not zero.
Cosecant ratio	`cosec A = hypotenuse/opposite = 1/sin A`	Defined when sin A is not zero; for the boundary value A = 0° it is not defined.
Secant ratio	`sec A = hypotenuse/adjacent = 1/cos A`	Defined when cos A is not zero; for the boundary value A = 90° it is not defined.
Cotangent ratio	`cot A = adjacent/opposite = 1/tan A`	Defined when tan A is not zero; for A = 0°, cot A is not defined.
Quotient relation	`tan A = sin A/cos A, cot A = cos A/sin A`	These relations connect the main ratios and hold for angles for which the denominators are non-zero.
Fundamental Pythagorean identity	`sin²A + cos²A = 1`	A is an angle for which sine and cosine are defined; it follows directly from Pythagoras's theorem.
First derived identity	`1 + tan²A = sec²A`	Obtained by dividing sin²A + cos²A = 1 by cos²A, so it requires cos A ≠ 0.
Second derived identity	`1 + cot²A = cosec²A`	Obtained by dividing sin²A + cos²A = 1 by sin²A, so it requires sin A ≠ 0.
Exact values for standard angles	`sin 0° = 0, sin 30° = 1/2, sin 45° = √2/2, sin 60° = √3/2, sin 90° = 1; cos 0° = 1, cos 30° = √3/2, cos 45° = √2/2, cos 60° = 1/2, cos 90° = 0; tan 0° = 0, tan 30° = 1/√3, tan 45° = 1, tan 60° = √3, tan 90° is not defined`	All angles are in degrees. These exact values should be remembered; the identity sin²A + cos²A = 1 can be used to check each row.

Practice questions with answers

1. Define cosec A and cot A in terms of the sides of a right triangle.

For an acute angle A in a right triangle, cosec A = hypotenuse/opposite side and cot A = adjacent side/opposite side. Equivalently, cosec A = 1/sin A and cot A = 1/tan A.

2. In a right triangle ABC, right-angled at B, AB = 24 cm and BC = 7 cm. Find sin A and cos A.

Hypotenuse AC = √(24² + 7²) = √625 = 25 cm. With respect to angle A, the opposite side is BC = 7 cm and the adjacent side is AB = 24 cm, so sin A = 7/25 and cos A = 24/25.

3. Find the value of 2 tan² 45° + cos² 30° − sin² 60°.

tan 45° = 1, cos 30° = √3/2 and sin 60° = √3/2. Substituting gives 2(1)² + (√3/2)² − (√3/2)² = 2 + 3/4 − 3/4 = 2.

4. Given that θ is acute and sin θ = 3/5, find cos θ and tan θ.

Using sin²θ + cos²θ = 1, we get cos²θ = 1 − 9/25 = 16/25. Since θ is acute, cos θ = 4/5, and tan θ = sin θ/cos θ = (3/5)/(4/5) = 3/4.

5. If tan A = 3/4, find the value of sin A cos A.

Take opposite side = 3k and adjacent side = 4k; then the hypotenuse is 5k. So sin A = 3/5 and cos A = 4/5, giving sin A cos A = 12/25.

6. Prove that (1 + tan²A) cos²A = 1.

Since 1 + tan²A = sec²A and sec A = 1/cos A, the left side becomes sec²A cos²A = (1/cos²A) cos²A = 1. Hence the identity is proved.

7. Prove that tan²A − sin²A = tan²A sin²A.

Write tan²A as sin²A/cos²A. Then the left side is sin²A/cos²A − sin²A = sin²A(1 − cos²A)/cos²A = sin²A(sin²A)/cos²A = tan²A sin²A, which equals the right side.

8. Evaluate: sin 30° cos 60° + cos 30° sin 60°.

sin 30° = 1/2, cos 60° = 1/2, cos 30° = √3/2 and sin 60° = √3/2. Therefore the value is (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1.

9. State the values of tan 0°, tan 45° and tan 60°.

tan 0° = 0, tan 45° = 1 and tan 60° = √3. These follow from sin/cos values at the same angles.

10. If θ is acute and tan θ = cot θ, find θ.

cot θ = 1/tan θ, so the equation gives tan θ = 1/tan θ, or tan²θ = 1. Since θ is acute, tan θ is positive, so tan θ = 1 and θ = 45°.