Trigonometry

high importance~3 Q in Tier 119 formulas⚡ 11 shortcuts5 subtopics

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Subtopic 3 of 5·← Fundamental identities·Value-putting and given-ratio questions →

Complementary angles

For angles that add to $90^\circ$ : $\sin(90^\circ-\theta)=\cos\theta,\quad \tan(90^\circ-\theta)=\cot\theta,\quad \sec(90^\circ-\theta)=\text{cosec}\theta$ and similarly in reverse. This converts "odd" angles ( $67\tfrac12^\circ$ , $72^\circ$ , ...) into their partners and makes paired products equal 1: $\tan\theta\cdot\tan(90^\circ-\theta)=1$ .

An equation like $\sin(A+\theta)=\cos(B+\theta)$ forces $(A+\theta)+(B+\theta)=90^\circ$ .

📖 Notes 🧩 4 question types ⚡ Formulas 🎯 14 practice Q

Detailed notes

The complementary pair rules

For any angle $\theta$ : $\sin(90^\circ - \theta) = \cos\theta$ , $\tan(90^\circ - \theta) = \cot\theta$ , $\sec(90^\circ - \theta) = \cosec\theta$ — and each reversed. The switch only happens between the three PAIRS (sin↔cos, tan↔cot, sec↔cosec); sin to sin at $90^\circ - \theta$ never changes function. $\sin\theta = \cos(90^\circ - \theta),\quad \tan\theta = \cot(90^\circ - \theta),\quad \sec\theta = \cosec(90^\circ - \theta)$

The working algorithm

Scan the expression for angles that sum to $90^\circ$ in pairs ( $20^\circ$ & $70^\circ$ , $35^\circ$ & $55^\circ$ , $5^\circ$ & $85^\circ$ ...).
Convert ONE angle of each pair so both terms share an angle: $\sec 55^\circ = \cosec 35^\circ = \frac{1}{\sin 35^\circ}$ .
Fractions collapse ( $x \cdot \frac1x$ ), products of the form $\tan\theta\cot\theta$ become 1, and the expression evaluates to a small constant.

Product chains

Products like $\tan 5^\circ \tan 25^\circ \tan 45^\circ \tan 65^\circ \tan 85^\circ$ pair into $\tan\theta \cdot \tan(90^\circ - \theta) = \tan\theta\cot\theta = 1$ with $\tan 45^\circ = 1$ left over. The full run $\tan 1^\circ \tan 2^\circ \cdots \tan 89^\circ$ is 1 for the same reason — 44 cancelling pairs around the middle term.

Finding the angle itself

Equations pairing different functions at related angles resolve by forcing the SAME function: $\sin 3A = \cos(A - 10^\circ)$ becomes $\sin 3A = \sin(90^\circ - (A - 10^\circ))$ , so the angles match: $3A = 100^\circ - A$ . Equivalently, complementary-function arguments must sum to $90^\circ$ : $\sec 2\theta = \cosec(\theta + 30^\circ)$ gives $2\theta + \theta + 30^\circ = 90^\circ$ .

The tan 45° anchor and half-chain variants

$\tan 45^\circ = 1$ is the middle term of every full chain and the most common "unpaired" value in short products. Half-chains behave the same: $\tan 10^\circ \tan 20^\circ \tan 70^\circ \tan 80^\circ = (10,80)(20,70) = 1$ , and cot chains like $\cot 15^\circ \cot 35^\circ \cot 55^\circ \cot 75^\circ$ collapse identically because complementary cots are still reciprocals. One caution: the collapse needs PAIRS summing to $90^\circ$ — if a product holds $\tan 20^\circ \tan 50^\circ \tan 40^\circ$ , the $20/70$ pairing is absent, so convert through cot and simplify before trusting a pattern.

Quick revision

Only three switches exist: sin↔cos, tan↔cot, sec↔cosec, argument → $90^\circ - x$ .
Hunt pairs summing to $90^\circ$ ; convert, then cancel.
$\tan\theta \cdot \tan(90^\circ - \theta) = 1$ ; full chains $\tan 1^\circ \cdots \tan 89^\circ = 1$ .
Angle equations: set complementary arguments to sum to $90^\circ$ , solve linearly.

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Pair conversion to collapse a sumvery common3 practice Q

How to spot it:

A short sum/product whose angles pair to 90 degrees (35 & 55, 20 & 70...).

$\sin\theta = \cos(90^\circ-\theta)$ etc.; convert one of each pair and cancel.

List the angles; find the 90-degree pairs.
Convert the odd-pair function: $\sec 55^\circ = \frac{1}{\cos 55^\circ} = \frac{1}{\sin 35^\circ}$ .
Each term becomes $x \cdot \frac1x = 1$ ; read the sum.

Why: the exam numbers never share a function until you flip one — that flip is the entire question.

Example: The value of $\sin 35^\circ \sec 55^\circ + \cos 35^\circ \cosec 55^\circ$ is:

$\sec 55^\circ = \frac{1}{\sin 35^\circ}$ and $\cosec 55^\circ = \frac{1}{\cos 35^\circ}$ : sum $= 1 + 1 = 2$ .

Type 2: Product chains pairing to 1common4 practice Q

How to spot it:

Long products of tangents/cots at many small angles, with 45 in the middle or symmetric ends.

$\tan\theta \cdot \tan(90^\circ - \theta) = 1$; pair from the ends inward.

Pair $\theta$ with $90^\circ - \theta$ from both ends.
Each pair contributes 1; any middle $\tan 45^\circ$ also contributes 1.
The product is 1 unless an unpaired angle remains — rare, and then it is stated.

Why: complementary tangents are reciprocals, so the chain self-destructs into a product of ones.

Example: The value of $\tan 5^\circ \tan 25^\circ \tan 45^\circ \tan 65^\circ \tan 85^\circ$ is:

Pairs: $\tan 5 \tan 85 = 1$ , $\tan 25 \tan 65 = 1$ , $\tan 45 = 1$ . Product $= 1$ .

Type 3: Finding the angle from complementary equalityvery common3 practice Q

How to spot it:

'If sin 3A = cos(A − 10), find A' — different functions at related angles, one unknown angle.

Force the same function: $\cos x = \sin(90^\circ - x)$; then equate arguments. Shortcut: complementary arguments sum to 90.

Rewrite so both sides carry the same function.
Equate the arguments (or set them to sum to $90^\circ$ for complementary pairs).
Solve the resulting linear equation; verify the angle keeps every argument in range.

Why: equality of different functions is meaningless until one side is converted — then it is class-8 algebra.

Example: If $\sin 3A = \cos(A - 10^\circ)$ , the value of $A$ is:

$\cos(A - 10^\circ) = \sin(90^\circ - A + 10^\circ)$ , so $3A = 100^\circ - A$ → $A = 25^\circ$ .

Type 4: Difference of complementary twins = 0common2 practice Q

How to spot it:

'Find cos 50° − sin 40°' style — two terms that LOOK unrelated.

$\sin(90^\circ - \theta) = \cos\theta$: the two terms are identical; the difference is 0.

Check the arguments: $50 + 40 = 90$ → twins.
$\sin 40^\circ = \cos 50^\circ$ ; the expression is zero.
Same trick for $\cosec 68^\circ - \sec 22^\circ$ , $\tan 81^\circ - \cot 9^\circ$ .

Why: the question only wins if you don't check for the 90-degree sum — checking it ends the question instantly.

Example: The value of $\cosec 68^\circ - \sec 22^\circ$ is:

$\sec 22^\circ = \cosec(90^\circ - 22^\circ) = \cosec 68^\circ$ ; difference $= 0$ .

Formulas

Complementary pairs

\sin(90-\theta)=\cos\theta,\ \tan(90-\theta)=\cot\theta,\ \sec(90-\theta)=\text{cosec}\theta

Paired products

\tan\theta\,\tan(90^\circ-\theta)=1,\quad \sin\theta\,\text{cosec}\theta=1

Angle-equation rule

\sin(A+\theta)=\cos(B+\theta)\Rightarrow A+B+2\theta=90^\circ

Shortcut tricks

⚡ Pair the angles that add to 90°

In a long product, look for $\theta$ and $90-\theta$ pairs first — each pair collapses to 1. The classic $\tan1^\circ\tan2^\circ\cdots\tan89^\circ$ collapses completely.

Example: Find the value of $\tan15^\circ\tan25^\circ\tan45^\circ\tan65^\circ\tan75^\circ$ .

$(\tan15\tan75)(\tan25\tan65)(\tan45)=1\times1\times1=1$ .

⚡ Convert everything to one angle

Rewrite each term at its complementary partner until only one angle remains; a $\sin^2+\cos^2$ pair may then collapse.

Example: Find the value of $\sin^265^\circ+\sin^225^\circ$ .

$\sin65^\circ=\cos25^\circ$ , so the sum $=\cos^225^\circ+\sin^225^\circ=1$ .

⚡ Equation → sum to 90°

When $\sin(\ldots)=\cos(\ldots)$ , set the two bracketed angles to sum to $90^\circ$ and solve.

Example: If $\sin(40^\circ+\theta)=\cos(30^\circ+\theta)$ and both angles are acute, find $\theta$ .

$40+\theta+30+\theta=90\Rightarrow2\theta=20\Rightarrow\theta=10^\circ$ .

Where students lose marks

Complementary conversion applied to the wrong function ( $\sec(90-\theta)=\sec\theta$ is wrong).
Setting $\sin A=\cos B\Rightarrow A+B=180^\circ$ — it is $90^\circ$ .
Products paired as $\tan\theta\cdot\cot\theta=1$ but evaluated before converting to the same angle.

Practice sets — 14 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 10 questions

Suggested time 6 min · wrong answers go to your mistake notebook automatically.