CBSE Class 10 Mathematics Introduction to Trigonometry Notes Set 21

Trigonometric Ratios

In a right triangle with an acute angle θ, with sides named opposite (to θ), adjacent (to θ) and hypotenuse:
sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent,
cosec θ = hypotenuse/opposite, sec θ = hypotenuse/adjacent, cot θ = adjacent/opposite.
Reciprocals: cosec θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ. Also tan θ = sin θ/cos θ and cot θ = cos θ/sin θ.
The value of a ratio depends only on the angle, not on the size of the triangle. Since the hypotenuse is the longest side, sin θ and cos θ are never greater than 1 for an acute angle, while sec θ and cosec θ are never less than 1.

Ratios of Standard Angles

Memory aid: sin of 0°, 30°, 45°, 60°, 90° is √0/2, √1/2, √2/2, √3/2, √4/2; cos is the same list in reverse order.

Complementary Angles

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.
Consequences: tan 1° · tan 2° · ... · tan 89° = 1, and sin² 1° + sin² 2° + ... + sin² 89° = 44.5.

Trigonometric Identities

1. sin² θ + cos² θ = 1 (so sin² θ = 1 − cos² θ and cos² θ = 1 − sin² θ).
2. 1 + tan² θ = sec² θ (so sec² θ − tan² θ = 1 and (sec θ − tan θ)(sec θ + tan θ) = 1).
3. 1 + cot² θ = cosec² θ (so cosec² θ − cot² θ = 1).
Strategy for proving identities: start with the more complicated side, convert everything to sin and cos, and simplify; use the identities above to replace squares.

Worked Examples

Example 1. If tan A = 3/4, find sin A and sec A.
Solution: Take opposite = 3 and adjacent = 4, so the hypotenuse is 5. sin A = 3/5 and sec A = 5/4.

Example 2. Evaluate sin 60° cos 30° + sin 30° cos 60°.
Solution: (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1.

Example 3. Evaluate tan 35° tan 40° tan 45° tan 50° tan 55°.
Solution: tan 35° tan 55° = 1 and tan 40° tan 50° = 1 because the angles are complementary, and tan 45° = 1. The value is 1.

Example 4. Prove that (1 + tan² A)/(1 + cot² A) = tan² A.
Solution: The left side is sec² A/cosec² A = (1/cos² A)(sin² A) = tan² A.

Example 5. If sin θ + cos θ = √2 cos θ, show that cos θ − sin θ = √2 sin θ.
Solution: sin θ = (√2 − 1) cos θ, so cos θ = (√2 + 1) sin θ and cos θ − sin θ = √2 sin θ.