Trigonometry

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

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Subtopic 2 of 5·← Ratios and standard values·Complementary angles →

Fundamental identities

Three identities drive almost every simplification: $\sin^2\theta+\cos^2\theta=1,\qquad 1+\tan^2\theta=\sec^2\theta,\qquad 1+\cot^2\theta=\text{cosec}^2\theta$ Each splits into handy parts: $\sin^2\theta=1-\cos^2\theta$ , $\sec^2\theta-\tan^2\theta=1$ , $\text{cosec}^2\theta-\cot^2\theta=1$ . Expressions with $\sin^6+\cos^6$ or $\sin^4+\cos^4$ reduce through these — treat $x=\sin^2\theta$ so $x+(1-x)=1$ .

📖 Notes 🧩 4 question types ⚡ Formulas 🎯 17 practice Q

Detailed notes

The three identity engines

$\sin^2\theta + \cos^2\theta = 1,\qquad 1 + \tan^2\theta = \sec^2\theta,\qquad 1 + \cot^2\theta = \cosec^2\theta$ Every identity question runs one of three engines:

Divide by $\cos^2$ (or $\sin^2$ ): $\frac{\sin^2}{\cos^2} + 1 = \frac{1}{\cos^2}$ is the second identity — so any $\sec^2 \pm \tan^2$ combination converts to a single ratio.
Pair products: $(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1$ , $(\cosec\theta + \cot\theta)(\cosec\theta - \cot\theta) = 1$ — conjugate pairs whose product is 1.
Squares of sums: $(\sin\theta \pm \cos\theta)^2 = 1 \pm 2\sin\theta\cos\theta$ — links "sin+cos = m" questions to "sin·cos" questions.

The sin+cos / sin−cos web

Given $\sin\theta + \cos\theta = m$ : squaring gives $\sin\theta\cos\theta = \frac{m^2 - 1}{2}$ , and $(\sin\theta - \cos\theta)^2 = 2 - m^2$ . Given $\sec\theta + \tan\theta = m$ , the conjugate gives $\sec\theta - \tan\theta = \frac1m$ — then adding/subtracting the pair returns $\sec\theta$ and $\tan\theta$ individually (this 2-step unlocks $\sin\theta$ via a triangle).

Convert-to-sin-cos for anything composite

For fractions mixing tan/sin/cos with an unknown coefficient ("if $\tan\theta = \frac34$ , find $\frac{4\sin\theta - \cos\theta}{4\cos\theta + \sin\theta}$ "): divide numerator and denominator by $\cos\theta$ — everything becomes $t = \tan\theta$ , one substitution, one fraction. Squared combos like $\tan^2\theta + \cot^2\theta$ come from squaring $\tan\theta + \cot\theta$ and subtracting 2.

The fourth- and sixth-power chains

Two reduced forms answer a whole question family: $\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta, \qquad \sin^6\theta + \cos^6\theta = 1 - 3\sin^2\theta\cos^2\theta$ Both come from squaring/cubing $\sin^2 + \cos^2 = 1$ . The classic combination $2(\sin^6 + \cos^6) - 3(\sin^4 + \cos^4) + 1$ then collapses: $2 - 6s^2c^2 - 3 + 6s^2c^2 + 1 = 0$ for every $\theta$ — the cross terms cancel by design. When a question mixes powers, reduce each bracket to these two forms first and only then simplify; almost always the $s^2c^2$ terms cancel and a constant survives.

Quick revision

$\sin^2 + \cos^2 = 1$ ; divide it by $\cos^2$ or $\sin^2$ for the other two.
$(\sec + \tan)(\sec - \tan) = 1$ ; same for cosec/cot.
$(\sin \pm \cos)^2 = 1 \pm 2\sin\cos$ ; $\sin\cos = \frac{(\sin+\cos)^2 - 1}{2}$ .
Composite fractions: divide top and bottom by $\cos\theta$ .
$\tan\theta + \cot\theta = \sec\theta\,\cosec\theta$ ; squaring then subtracting 2 gives the squared sum.

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: Conjugate pairs: sec±tan, cosec±cotvery common3 practice Q

How to spot it:

'If secθ + tanθ = m, find secθ − tanθ' (or the reverse) — or the pair used to extract individual ratios.

(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1

The conjugate is its reciprocal: $\sec\theta - \tan\theta = \frac1m$ .
To extract $\tan\theta$ : subtract the pair ( $2\tan\theta = m - \frac1m$ ).
Then a 3-4-5 style triangle gives sin/cos if asked.

Why: the product being exactly 1 makes every question in this family a two-line reciprocal computation.

Example: If $\sec\theta + \tan\theta = 5$ ( $\theta$ acute), the value of $\sec\theta - \tan\theta$ is:

$\sec\theta - \tan\theta = \frac{1}{\sec\theta + \tan\theta} = \frac15$ .

Type 2: sin²+cos² expansions and squares of sumsvery common4 practice Q

How to spot it:

Expressions like (sinθ ± cosθ)², or a product like (cosecθ − sinθ)(secθ − cosθ)(tanθ + cotθ) to simplify.

(\sin\theta \pm \cos\theta)^2 = 1 \pm 2\sin\theta\cos\theta; convert everything to sin/cos and cancel.

Expand squares: the cross term $2\sin\theta\cos\theta$ and the 1 dominate.
For composite products, convert each bracket to sin/cos: $\cosec\theta - \sin\theta = \frac{\cos^2\theta}{\sin\theta}$ etc., then cancel.
Constants fall out — most of these expressions are pure numbers (1 or 2).

Why: every term is secretly $\sin^2 + \cos^2 = 1$ wearing a costume.

Example: The value of $(\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2$ is:

$= 1 + 2\sin\theta\cos\theta + 1 - 2\sin\theta\cos\theta = 2$ . Cross terms cancel.

Type 3: Substituting tan via divide-by-cosvery common2 practice Q

How to spot it:

tanθ given; a linear fraction in sinθ and cosθ asked — coefficients on both.

Divide numerator and denominator by $\cos\theta$; the fraction becomes one in $t = \tan\theta$.

$\frac{a\sin\theta + b\cos\theta}{c\cos\theta + d\sin\theta} = \frac{a\tan\theta + b}{c + d\tan\theta}$ .
Substitute $t$ ; simplify the single fraction.
Keep fractions exact ( $t = \frac34$ , not 0.75).

Why: division by $\cos\theta$ is legal (nonzero for acute angles) and converts two unknowns into one.

Example: If $\tan\theta = \frac{3}{4}$ , the value of $\frac{4\sin\theta - \cos\theta}{4\cos\theta + \sin\theta}$ is:

Divide by $\cos\theta$ : $\frac{4 \times \frac34 - 1}{4 + \frac34} = \frac{2}{\frac{19}{4}} = \frac{8}{19}$ .

Type 4: Squared sums via squaring (tan+cot, sin+cosec)common3 practice Q

How to spot it:

'If tanθ + cotθ = k, find tan²θ + cot²θ' — or an equation like sinθ + cosecθ = 2 that forces a special value.

Square the given: $(x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2$, so subtract 2. Equations equal to 2 force $x = 1$.

Square the given sum; subtract 2 for the squared-pair answer.
AM-GM shortcut: $x + \frac1x \ge 2$ , with equality only at $x = 1$ — so "= 2" means $\tan\theta = \sin\theta = 1$ , i.e. $\theta = 45^\circ$ / $90^\circ$ respectively.
Read off the asked power exactly (squared vs fourth).

Why: the self-reciprocal structure makes both the bound and the equality case immediate.

Example: If $\tan\theta + \cot\theta = 5$ ( $\theta$ acute), the value of $\tan^2\theta + \cot^2\theta$ is:

Square: $25 = \tan^2\theta + \cot^2\theta + 2$ , so $\tan^2\theta + \cot^2\theta = 23$ .

Formulas

Pythagorean identities

\sin^2\theta+\cos^2\theta=1,\quad 1+\tan^2\theta=\sec^2\theta,\quad 1+\cot^2\theta=\text{cosec}^2\theta

Fourth powers

\sin^4\theta+\cos^4\theta=1-2\sin^2\theta\cos^2\theta

Sixth powers

\sin^6\theta+\cos^6\theta=1-3\sin^2\theta\cos^2\theta

Product-to-sum bridge

\tan\theta=\frac{\sin\theta}{\cos\theta},\quad \cot\theta=\frac{\cos\theta}{\sin\theta}

Shortcut tricks

⚡ Substitute $x=\sin^2\theta$

Everything above degree 4 collapses with $x$ and $1-x$ . Memorise the two reduced forms for 4th and 6th powers.

Example: Find the value of $2(\sin^6\theta+\cos^6\theta)-3(\sin^4\theta+\cos^4\theta)+1$ .

$2(1-3s^2c^2)-3(1-2s^2c^2)+1=2-3+1+( -6+6)s^2c^2=0$ for every $\theta$ .

⚡ Sec/tan pairs move as one

See $\sec^2$ and $\tan^2$ together → replace by 1 (or by each other via $\sec^2=1+\tan^2$ ). Same for cosec/cot.

Example: Simplify $\dfrac{\sec^2\theta-\tan^2\theta}{1+\cot^2\theta}\times\text{cosec}^2\theta$ .

Numerator $=1$ , $1+\cot^2=\text{cosec}^2$ , so the value is $1$ .

Where students lose marks

Writing $\tan^2\theta+1=\sec\theta$ (the square is dropped).
Expanding $\sin^6+\cos^6$ as $(\sin^2+\cos^2)^3$ without subtracting the cross terms.
Mixing degrees of powers: $\sin^4+\cos^4\neq1$ .

Practice sets — 17 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 8 min · wrong answers go to your mistake notebook automatically.