Trigonometry

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

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Subtopic 1 of 5·Fundamental identities →

Ratios and standard values

For an acute angle $\theta$ in a right triangle: $\sin\theta=\frac{\text{opp}}{\text{hyp}}$ , $\cos\theta=\frac{\text{adj}}{\text{hyp}}$ , $\tan\theta=\frac{\text{opp}}{\text{adj}}$ , and $\text{cosec}=\frac1{\sin}$ , $\sec=\frac1{\cos}$ , $\cot=\frac1{\tan}$ .

Standard value table (memorise cold — most CGL questions are one lookup away):

$\theta$	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$
$\sin$	$0$	$\frac12$	$\frac{1}{\sqrt2}$	$\frac{\sqrt3}{2}$	$1$
$\cos$	$1$	$\frac{\sqrt3}{2}$	$\frac{1}{\sqrt2}$	$\frac12$	$0$
$\tan$	$0$	$\frac{1}{\sqrt3}$	$1$	$\sqrt3$	undef.

📖 Notes 🧩 4 question types ⚡ Formulas 🎯 12 practice Q

Detailed notes

The standard-value table (build it, don't memorise it)

$\sin: \frac{\sqrt0}{2}, \frac{\sqrt1}{2}, \frac{\sqrt2}{2}, \frac{\sqrt3}{2}, \frac{\sqrt4}{2} \text{ at } 0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$ cos is the same list read backwards; tan = sin/cos gives $0, \frac{1}{\sqrt3}, 1, \sqrt3, \infty$ . Then reciprocals give cosec, sec, cot. The $\frac{\sqrt{k}}{2}$ construction means the table can never be forgotten under pressure.

One ratio → the whole triangle

Any single ratio pins the triangle's SHAPE: draw a right triangle with sides in that ratio, get the third side by Pythagoras, read off every other ratio.

$\sin\theta = \frac{3}{5}$ → sides 3-4-5 → $\tan\theta = \frac34$ .
$\cos\theta = \frac{12}{13}$ → 5-12-13 → $\sin\theta = \frac{5}{13}$ , $\tan\theta = \frac{5}{12}$ .
$\tan\theta = \frac{1}{\sqrt7}$ → hyp $\sqrt{1+7} = 2\sqrt2$ → $\cosec\theta = 2\sqrt2$ . Keep ratios as fractions to the end; rationalise only the final answer if options are surd-free.

Two sides → every ratio

Given two sides of the right triangle: third side first (Pythagoras), then name the angle carefully — the ratio asked is relative to a SPECIFIC acute angle: opposite/hyp for its sine, adjacent/hyp for its cosine. "The smaller angle" sits opposite the shorter leg. Triangles 5-12-13, 7-24-25, 8-15-17, 9-40-41 and their multiples supply almost all exam numbers.

Combining ratios safely

When an expression mixes functions ( $\sec 60^\circ + \cosec 30^\circ$ , $\cot 30^\circ + \tan 60^\circ$ ), reduce each term to the $\sin$ / $\cos$ table values first, then add — do not invent product rules. The angle-difference formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$ occasionally appears disguised as a pure 60/30 evaluation; recognising it turns a messy fraction into $\tan 30^\circ$ .

The reciprocal rows (never leave home without them)

Flip the main table for the rest: $\cosec 30^\circ = 2$ , $\cosec 45^\circ = \sqrt2$ , $\cosec 60^\circ = \frac{2}{\sqrt3}$ ; $\sec 60^\circ = 2$ , $\sec 45^\circ = \sqrt2$ , $\sec 30^\circ = \frac{2}{\sqrt3}$ (sec runs opposite to cosec); $\cot 30^\circ = \sqrt3$ , $\cot 45^\circ = 1$ , $\cot 60^\circ = \frac{1}{\sqrt3}$ — cot is tan read backwards. Expressions like $\cosec^2 30^\circ + \sec^2 45^\circ$ are then pure table work: $4 + 2 = 6$ . Spot the reciprocal twins that cancel: $\cosec 30^\circ - \sec 60^\circ = 2 - 2 = 0$ and $\tan 30^\circ \tan 60^\circ = 1$ .

Quick revision

$\sin$ column $\frac{\sqrt{0,1,2,3,4}}{2}$ ; cos reversed; tan $= \frac{\sin}{\cos}$ .
One ratio → 3-4-5 / 5-12-13 / 7-24-25 triangle → all ratios.
Smaller acute angle ⟷ shorter opposite leg.
$\tan(60^\circ - 30^\circ)$ identity rescues $\frac{\tan A - \tan B}{1+\tan A\tan B}$ expressions.

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: Standard values at 0/30/45/60/90very common3 practice Q

How to spot it:

A sum/difference of ratios at special angles asked directly.

Read each term off the $\frac{\sqrt{k}}{2}$ table; add last.

Convert every function to its table value (sec/cot too — reciprocals of the cos/tan rows).
Keep surds as surds; add only matching radicals.
Rationalise only at the very end if options demand it.

Why: each term is a fixed constant — the question only tests retrieval plus one line of arithmetic.

Example: The value of $\sin 30^\circ + \cos 60^\circ$ is:

$\frac12 + \frac12 = 1$ . Both are the table's middle entries.

Type 2: One given ratio → the other ratiosvery common3 practice Q

How to spot it:

'If sinθ = 3/5, find tanθ' — one ratio given (sometimes in a surd form), others asked.

Construct the 3-4-5 (or matching) triangle; read the asked ratio off it.

Write opp/adj from the given ratio; complete the triangle with Pythagoras.
Read the asked ratio; rationalise surds at the end.
Sign checks are unnecessary for acute exam angles — everything positive.

Why: one ratio fixes the shape, so every other ratio is a re-reading of the same triangle.

Example: If $\sin\theta = \frac{3}{5}$ ( $\theta$ acute), the value of $\tan\theta$ is:

Triangle 3-4-5: $\tan\theta = \frac{\text{opp}}{\text{adj}} = \frac34$ .

Type 3: Two sides of the triangle → ratioscommon2 practice Q

How to spot it:

Two sides of a right triangle given (or hyp + one side); a named ratio of one acute angle asked.

Third side by Pythagoras, then $\sin = \frac{\text{opp}}{\text{hyp}}$, $\cos = \frac{\text{adj}}{\text{hyp}}$ w.r.t. the NAMED angle.

Complete the triangle (5-12-13, 7-24-25, 8-15-17 multiples).
Identify the angle: 'opposite the shorter leg' = smaller angle.
Opposite/adjacent w.r.t. that angle — the top exam trap is flipping them.

Why: the triangle is fully determined; only the naming of the angle can go wrong.

Example: In a triangle right-angled at B, AB = 5 cm and BC = 12 cm. The value of sin A is:

AC $= 13$ ; $\sin A = \frac{\text{opp}}{\text{hyp}} = \frac{BC}{AC} = \frac{12}{13}$ . Note the opposite side to A is BC, not AB.

Type 4: Mixed-function combinationscommon2 practice Q

How to spot it:

Expressions mixing sec, cosec, cot at special angles — or a tan-difference formula hidden inside a fraction.

Reciprocal table entries; $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A\tan B}$ rescues the disguised ones.

Replace sec/cosec/cot by reciprocals of table values.
If the expression matches $\frac{\tan A - \tan B}{1 + \tan A \tan B}$ , rewrite as $\tan(A - B)$ and read the table once.
Compare with the options in surd form first; rationalise only if needed.

Why: the fraction pattern is invisible until you look for it — that's the entire difficulty of these questions.

Example: The value of $\frac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ}$ is:

$= \tan(60^\circ - 30^\circ) = \tan 30^\circ = \frac{1}{\sqrt3}$ .

Formulas

Reciprocal pairs

\text{cosec}\theta=\frac{1}{\sin\theta},\quad \sec\theta=\frac{1}{\cos\theta},\quad \cot\theta=\frac{1}{\tan\theta}

sin column

\sin0^\circ=0,\ \sin30^\circ=\frac12,\ \sin45^\circ=\frac{1}{\sqrt2},\ \sin60^\circ=\frac{\sqrt3}{2},\ \sin90^\circ=1

cos column

\cos\theta=\sin(90^\circ-\theta)

tan column

\tan0^\circ=0,\ \tan30^\circ=\frac{1}{\sqrt3},\ \tan45^\circ=1,\ \tan60^\circ=\sqrt3

Shortcut tricks

⚡ The 1-2-3 memory trick for sin

Write $\frac{\sqrt0}{2},\frac{\sqrt1}{2},\frac{\sqrt2}{2},\frac{\sqrt3}{2},\frac{\sqrt4}{2}$ for $0^\circ,30^\circ,45^\circ,60^\circ,90^\circ$ — the sin row falls out; the cos row is the same read backwards. Divide the two rows to get tan.

Example: Find the value of $\sin30^\circ+\cos60^\circ+\tan45^\circ$ .

$\frac12+\frac12+1=2$ .

⚡ Reciprocal first, table second

When an expression has cosec/sec/cot, flip each to sin/cos/tan before touching the table.

Example: Find the value of $\text{cosec}30^\circ-\sec60^\circ$ .

$2-2=0$ — both flip to $\frac{1}{1/2}=2$ .

Where students lose marks

$\sin45^\circ$ written as $\frac{1}{2}$ (confusing the 45 and 30 entries).
$\tan90^\circ$ treated as $0$ or $1$ — it is undefined.
Reciprocal flipped the wrong way ( $\sec\theta$ taken as $\frac{1}{\sin\theta}$ ).

Practice sets — 12 questions

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

Topic test · 12 questions

Suggested time 7 min · wrong answers go to your mistake notebook automatically.