IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~7 min read HL only

Reciprocal Trigonometric Functions

Three new functions: sec, cosec, and cot — the reciprocals of cos, sin, and tan. They come with two new Pythagorean identities (1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ) that turn equations involving sec², cosec², or cot² into clean quadratics.

📘 What you need to know

sec θ = 1/cos θ cosec θ = 1/sin θ cot θ = 1/tan θ = cos θ/sin θ.
Memory trick: the third letter of each reciprocal points to its partner. sec → cos, cosec → sin, cot → tan.
Two new Pythagorean identities: 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ.
Both identities and sec/cosec are in the formula booklet.
Don’t confuse with inverse trig: sin⁻¹x ≠ 1/sin x = cosec x.
Graphs: y = sec x and y = cosec x have asymptotes (where cos = 0 or sin = 0) and range |y| ≥ 1; y = cot x looks like the tan graph but flipped.
For equations with sec² and tan (or cosec² and cot), use the new Pythagorean identity to reduce to one trig function.

The three reciprocal functions

Definitions sec θ = 1cos θ cosec θ = 1sin θ cot θ = 1tan θ = cos θsin θ

Look at the third letter of each reciprocal name: sec → c → cos; cosec → s → sin; cot → t → tan. Quick way to remember which goes with which.

The graphs at a glance

Function	Period	Asymptotes at	Range
y = sec x	360° (2π)	cos x = 0 → odd multiples of 90° (π/2)	y ≤ −1 or y ≥ 1
y = cosec x	360° (2π)	sin x = 0 → multiples of 180° (π)	y ≤ −1 or y ≥ 1
y = cot x	180° (π)	tan x = 0 → multiples of 180° (π)	y ∈ ℝ (any real value)

To sketch any of them by hand: draw the parent graph first (cos for sec, sin for cosec, tan for cot), mark its zeros — those become the asymptotes — and flip near the peaks/troughs.

Two new Pythagorean identities

Pythagorean identities 1 + tan²θ = sec²θ 1 + cot²θ = cosec²θ

Both come from the original sin²θ + cos²θ = 1. Divide every term by cos²θ to get the first; divide every term by sin²θ to get the second.

When to use them: any equation with sec² and tan (or cosec² and cot) with at least one squared term. Reduce to one trig function and factor as a quadratic.

🧭 Recipe — solve a reciprocal trig equation

Spot the pair: sec² with tan? cosec² with cot? Use the matching Pythagorean identity.
Substitute sec²θ = 1 + tan²θ (or cosec²θ = 1 + cot²θ) and rearrange to a quadratic in tan θ (or cot θ).
Factor or use the quadratic formula to find values of tan θ or cot θ.
Convert cot back: cot θ = k → tan θ = 1/k.
Solve each tan equation in the given interval (principal + 180° periodicity).

Worked examples

WE 1

Find exact values of sec and cot

Without using a calculator, find the exact values of (a) sec(π/3) and (b) cot 120°.

(a) sec → reciprocal of cos sec(π/3) = 1/cos(π/3) = 1/(1/2) = 2 (a) sec(π/3) = 2 (b) cot → reciprocal of tan tan 120° = tan(180° − 60°) = −tan 60° = −√3 cot 120° = 1/tan 120° = −1/√3 = −√3/3 (b) cot 120° = −√33

WE 2

Solve a basic reciprocal trig equation

Solve the equation cosec x = 2 for 0° ≤ x ≤ 360°.

Step 1: Take the reciprocal of both sides 1/sin x = 2 → sin x = 1/2 Step 2: Solve sin x = 1/2 in [0°, 360°] x = 30° or x = 180° − 30° = 150° x = 30°, 150° cosec is just 1/sin — flip and solve

WE 3

Solve sec²x − tan x − 3 = 0

Solve the equation sec²x − tan x − 3 = 0 for 0° ≤ x ≤ 360°. Give answers to 3 s.f. where necessary.

Mixed sec² and tan → use 1 + tan²x = sec²x Step 1: Substitute (1 + tan²x) − tan x − 3 = 0 tan²x − tan x − 2 = 0 Step 2: Factor (tan x − 2)(tan x + 1) = 0 tan x = 2 or tan x = −1 Step 3: Solve each — tan repeats every 180° tan x = 2 → x = tan⁻¹(2) ≈ 63.43°, 243.43° tan x = −1 → x = 135°, 315° x ≈ 63.4°, 135°, 243°, 315° (3 s.f.)

WE 4

Solve 2 cosec²x − 5 cot x = 5

Solve the equation 2 cosec²x − 5 cot x = 5 for 0° ≤ x ≤ 360°. Give answers to 3 s.f.

Mixed cosec² and cot → use 1 + cot²x = cosec²x Step 1: Substitute and simplify 2(1 + cot²x) − 5 cot x = 5 2 cot²x − 5 cot x − 3 = 0 Step 2: Factor (2 cot x + 1)(cot x − 3) = 0 cot x = −1/2 or cot x = 3 Step 3: Convert to tan x cot x = −1/2 → tan x = −2 cot x = 3 → tan x = 1/3 Step 4: Solve each in [0°, 360°] tan x = −2 → x ≈ 116.57°, 296.57° tan x = 1/3 → x ≈ 18.43°, 198.43° x ≈ 18.4°, 117°, 198°, 297° (3 s.f.)

WE 5

Solve sin x · cot x = 1/2

Solve the equation sin x · cot x = 1/2 for 0° ≤ x ≤ 360°.

Step 1: Replace cot x with cos x / sin x sin x · (cos x / sin x) = 1/2 cos x = 1/2 Step 2: Solve cos x = 1/2 in [0°, 360°] x = 60° or x = 360° − 60° = 300° x = 60°, 300° when reciprocals are mixed, rewriting in sin and cos often collapses the equation

WE 6

Prove (sec θ − 1)(sec θ + 1) = tan²θ

Prove that (sec θ − 1)(sec θ + 1) = tan²θ for all valid θ.

Step 1: Expand the LHS — difference of two squares (sec θ − 1)(sec θ + 1) = sec²θ − 1 Step 2: Apply 1 + tan²θ = sec²θ sec²θ − 1 = tan²θ Step 3: LHS = RHS proved “valid θ” means cos θ ≠ 0, so sec θ is defined

💡 Top tips

Always try the new Pythagorean identities first when you see sec² with tan, or cosec² with cot. It’s faster than rewriting in sin and cos.
Use the third-letter rule: sec ↔ cos; cosec ↔ sin; cot ↔ tan. Saves confusion.
Sketch the parent graph first — cos before sec, sin before cosec, tan before cot.
For mixed reciprocals, rewriting in terms of sin and cos often reveals a clean cancellation.
To find exact values, just compute the parent (cos, sin, or tan) at the angle and take the reciprocal.

⚠ Common mistakes

Confusing reciprocal with inverse. sin⁻¹x means arcsin x, NOT 1/sin x. Reciprocal is cosec x.
Wrong identity sign. It’s 1 + tan²θ = sec²θ — not 1 − tan²θ. Same for cosec.
Forgetting cot x = 1/tan x ≠ tan x. cot 60° = 1/√3, not √3.
Wrong period for cot. Cot has period 180° (like tan), not 360°.
Solving for tan from cot incorrectly. cot x = k means tan x = 1/k — flip, don’t change sign.

Next note: Inverse Trigonometric Functions. arcsin, arccos, arctan — what they really are, the restricted domains they need to be functions, what their graphs look like, and how to use them carefully when solving equations.

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