TRIGONOMETRY

# Phase I: Ratios & Identities# Phase II: Trigonometric Equation# Phase III: Properties of Triangle

Ratio & IdentitiesProblem: Show that:

1.tan A+sec A−1tan A−sec A+1

= 1+sin A

cos A2. √sec A2+csc A2 = ±(cot A+ tan A )

Extended definition of Trigonometric Function [Circular Function]

In cycle of unity raids sine all

sin θ = x (-, +) P(x, y) (+, +)

cosθ = y r=1

tanθ =yx

(-, -) (+, -)

tangent cosine

Convention:

i) If θ is measured in anticlockwise then it is assume positive.ii) If θ is measured in clockwise then it is assume negative.

Angle Measurements System

(i) British System (Degree)Right angle = 90°1° = 60 min1min = 60sece.g.: 43°23’37”

(ii) French System (grades)Right angle = 100g

100g = 100 min1min = 100sece.g.: 83g63’97”(iii) Circular System

Right angle = π2

radian

e.g.: π2

c

NOTE: - If nothing is mentioned, consider it to be in radians.

T-Ratios of Allied Angles: (90°±θ, 180°±θ, 270°±θ etc.)sin(90 °−θ) = cosθ p” x P(x,y)

Proof: 90°-θ cosθ = x y yIn ΔOPP”; θ

sin(90 °−θ) = PP} over {OP ¿ x P’ ⇒sin(90 °−θ) = x ⇒sin(90 °−θ) = cosθ

Working Rule:Step1: First of all decide sing of T-Ratio from which quadrant (n×90° ±θ ) is sutatute.

e.g.: 180°+θ is in third quadrant where θ∈RStep2: #If n is even ⇒ NO CHANGE#If n is odd then interchanges as followssin ⟷ costan ⟷ cotsec ⟷ cosecProblem: Prove that:

i) sec(270 °−A) . sec(90 °−A)−tan(270°−A ). tan(90 °−A) = -1

ii) sin3π5

+sin4 π5

+sin6 π5

+sin7 π5

= 0

iii) cotπ12

cot3π12

cot5 π12

= 0

iv)cos(360°−A )sin(270°+A)

+ cot❑❑