Chapter 11.5 — Exercise 11.4 — Trigonometric Identities
Trigonometric identities and their applications. This is Lesson 5 of 5 in Chapter 11: Trigonometry.
One Theorem, Three Identities
Every identity in this exercise, however different they look on the page, traces back to the exact same single fact: Pythagoras' theorem applied to a right triangle's three sides, then divided through in three different ways. Every problem that follows is really just those same three identities, applied, combined, and disguised inside increasingly elaborate algebra.
Deriving the Three Pythagorean Identities
In right triangle ABC (right angle at B), Pythagoras' theorem gives BC²+AB²=AC² directly. Dividing every single term through by AC²:
(BC/AC)² + (AB/AC)² = 1 ⟹ sin²A + cos²A = 1Dividing that same identity through by cos²A once, then again separately by sin²A, produces the remaining two identities directly:
tan²A + 1 = sec²A 1 + cot²A = cosec²AAll three of these identities hold for every angle a right triangle can possibly contain — not approximately, not for specific convenient values, but exactly, for every single angle at once. That's what makes them fundamentally different from the specific-angle values in Exercise 11.2: those were true only at five particular angles, while these three are true everywhere. It's also worth noticing that only one of the three identities needed a fresh derivation from Pythagoras — the other two came for free, just by dividing the first one through by cos²A or sin²A respectively, rather than starting over from the triangle each time.
Two Evaluate Problems That Are Products, Not Fractions
Two separate problems in this exercise each present a pair of bracketed expressions side by side. It's tempting to read them as one fraction stacked over another — but the algebra only works out if they're read as a straightforward product instead.
(1+tanθ+secθ)(1+cotθ−cosecθ) = 2
(sec²θ−1)(cosec²θ−1) = 1The first expands by converting every term to sinθ and cosθ, combining each bracket into a single fraction, then applying the difference-of-squares pattern (a+b)(a−b)=a²−b² to the product of the two combined fractions — the sin²θ+cos²θ=1 identity clears out everything except a clean 2sinθcosθ over sinθcosθ, leaving exactly 2. The second is quicker: sec²θ−1 is just tan²θ and cosec²θ−1 is just cot²θ by the identities derived above, so the product collapses to (tanθ·cotθ)²=1² =1 immediately.
Reading the source material for these two problems takes real care: the way the original printed expression is laid out on the page, with one bracket stacked above the other, looks visually identical to how a fraction is normally written. Only working through the actual algebra reveals that a fraction reading gives an answer that changes depending on θ (never a valid answer to an "evaluate" question), while the product reading gives a genuine constant — 2 and 1 respectively — confirmed independently by substituting a specific test angle into both possible readings and checking which one stays fixed.
A Genuine Sum of Squares
(sinθ+cosθ)² + (sinθ−cosθ)² = 2Expanding both squares gives (sin²θ+cos²θ+2sinθcosθ) + (sin²θ+cos²θ−2sinθcosθ). The two 2sinθcosθ cross-terms are opposite in sign and cancel completely, leaving 1+1=2 from the two Pythagorean-identity substitutions alone. Unlike the two problems just above, this one genuinely is a sum — no product-versus-fraction ambiguity here, since the "+" between the two squared brackets is explicit and unambiguous.
Five Proofs, Compressed
| Show that | Key step |
|---|---|
| (cosecθ−cotθ)² = (1−cosθ)/(1+cosθ) | LHS = [(1−cosθ)/sinθ]², then sin²θ=(1−cosθ)(1+cosθ) cancels one factor |
| (1−tan²A)/(cot²A−1) = tan²A | cot²A−1 rewrites as (1−tan²A)/tan²A, so the whole fraction inverts and cancels |
| 1/cosθ − cosθ = tanθ·sinθ | Combine into (1−cos²θ)/cosθ = sin²θ/cosθ = sinθ·tanθ |
| secA(1−sinA)(secA+tanA) = 1 | secA−sinA·secA = secA−tanA, giving (secA−tanA)(secA+tanA) = sec²A−tan²A = 1 |
| (sinA+cosecA)²+(cosA+secA)² = 7+tan²A+cot²A | Expand both squares; sin²A+cos²A=1, plus cosec²A=1+cot²A and sec²A=1+tan²A, plus four cross-terms each equal to 1 |
Every one of these five proofs follows an identical overall strategy, even though the individual algebra differs: convert everything to sinθ and cosθ (or spot a Pythagorean identity hiding in disguise), combine over a common denominator if needed, then simplify using sin²+cos²=1 or one of its two derived forms. There's genuinely no other technique required anywhere in this exercise beyond that one repeated combination.
The fourth row is worth a second glance, since it's the only one of the five that runs the substitution backward — instead of converting toward sinθ and cosθ, it converts cot²A−1 into a fraction of tan²A terms and lets the whole expression invert and cancel. Recognising when working "backward" toward tan and cot, rather than the usual default of sin and cos, actually shortens the proof is a judgment call worth practising deliberately rather than always defaulting to the same conversion.
A Root Over the Whole Expression
Show that √[(1+sinA)/(1−sinA)] = secA+tanA. Multiplying both the numerator and the denominator inside the root by (1+sinA):
√[(1+sinA)²/((1−sinA)(1+sinA))] = √[(1+sinA)²/cos²A] = (1+sinA)/cosA = 1/cosA + sinA/cosA = secA+tanAThe square root has to sit over the entire fraction, not just one piece of it — squaring both sides first (getting (1+sinA)/(1−sinA) alone) would prove a genuinely different, stronger statement than the one actually asked for here. Multiplying by (1+sinA)/(1+sinA) inside the root is the same rationalising move used throughout this whole chapter whenever a fraction's denominator needs clearing away.
A Third Corrected Product
Simplify (1−cosθ)(1+cosθ)(1+cot²θ)
= (1−cos²θ)(1+cot²θ) [difference of squares]
= sin²θ · cosec²θ [since 1−cos²θ=sin²θ and 1+cot²θ=cosec²θ]
= 1This is the third and final "expression that looks like it could be a fraction but is really a product" in this exercise — the same pattern already seen twice above. Once spotted, all three resolve the identical way: expand or simplify each bracket first using a Pythagorean identity, then let two matching reciprocal-style terms cancel to 1. All three also share one more feature worth noticing: every single one evaluates to a genuine constant (2, 1, and 1) rather than an expression still containing θ — a strong, quick clue in itself that the intended reading really was a product all along, since a true ratio built from these same brackets would never stay constant as θ varies.
Two Elegant Reciprocal Results
- If secθ+tanθ=p, find secθ−tanθ: since sec²θ−tan²θ=1 always, (secθ+tanθ)(secθ−tanθ)=1. Substituting p for the first factor gives p·(secθ−tanθ)=1, so secθ−tanθ = 1/p.
- If cosecθ+cotθ=k, prove cosθ=(k²−1)/(k²+1): the identical trick gives cosecθ−cotθ=1/k. Adding the two equations: 2cosecθ=k+1/k=(k²+1)/k, so sinθ=2k/(k²+1). Then cosθ=√(1−sin²θ) simplifies, via (k²+1)²−4k²=(k²−1)², to exactly (k²−1)/(k²+1).
Both results share the identical opening move — multiplying a sum and a difference of the same two ratios together to trigger sec²−tan²=1 or cosec²−cot²=1 — proving once again that this whole exercise really only has one underlying trick, reused again and again in every problem under a slightly different disguise.
The Chapter, Complete
Across four exercises, Trigonometry built up from essentially nothing: the introduction defined all six ratios and proved they're constant for a given angle, Exercise 11.1 practised computing them from raw side lengths, Exercise 11.2 pinned down exact values at five standard angles, Exercise 11.3 connected any angle's ratios to its complement's, and this exercise closed with algebraic identities holding for every angle at once, without a single specific numeric angle value needed anywhere in any of the ten problems worked through above. The next chapter, Applications of Trigonometry, finally points every single one of these ratios at real, physical heights and distances — the practical payoff this entire chapter has been quietly building toward from its very first page.