ExamShortcut
🔍Log in

Trigonometry

🔒 Log in to track
high importance~3 Q in Tier 119 formulas⚡ 11 shortcuts5 subtopics
OverviewLearn (5)Formulas (19)Examples (5)Practice (72)
Subtopic 4 of 5·← Complementary angles·Maximum and minimum values →

Value-putting and given-ratio questions

🔒 Log in to track

Two CGL workhorses:

  1. Given one ratio, draw the triangle. If sin⁡θ=513\sin\theta=\frac{5}{13}, take opposite =5=5, hypotenuse =13=13, adjacent =132−52=12=\sqrt{13^2-5^2}=12 (a 5-12-13 triplet!) and read off every other ratio.
  2. Value-putting an identity. If only sin⁡θ+cosecθ\sin\theta+\text{cosec}\theta or a quadratic condition like sin⁡θ+sin⁡2θ=1\sin\theta+\sin^2\theta=1 is given, convert the asked expression into the given pieces — or test a convenient θ\theta when the expression is identically constant.
📖 Notes🧩 4 question types⚡ Formulas🎯 15 practice Q

Detailed notes

Putting values into expressions

Two flavours: (a) the angle is a STANDARD one (30∘,45∘,60∘30^\circ, 45^\circ, 60^\circ) — read the table and compute; (b) an angle satisfying a condition ("tan⁡θ=2\tan\theta = 2") — reduce the expression to a single ratio, then substitute. Both end in one small fraction; the work is entirely in the reduction. 2tan⁡245∘+cos⁡230∘−sin⁡260∘=2+34−34=22\tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ = 2 + \tfrac34 - \tfrac34 = 2

The sin+cos condition web

Given sin⁡θ+cos⁡θ=m\sin\theta + \cos\theta = m:

  • sin⁡θcos⁡θ=m2−12\sin\theta\cos\theta = \frac{m^2 - 1}{2} (square the given).
  • sin⁡θ−cos⁡θ=2−m2\sin\theta - \cos\theta = \sqrt{2 - m^2} (from the difference squared).
  • sin⁡3θ+cos⁡3θ=m3−3sin⁡θcos⁡θ⋅m\sin^3\theta + \cos^3\theta = m^3 - 3\sin\theta\cos\theta \cdot m via x3+y3=(x+y)(x2−xy+y2)x^3 + y^3 = (x+y)(x^2 - xy + y^2). Special anchors: m=2m = \sqrt2 gives sin⁡θcos⁡θ=12\sin\theta\cos\theta = \frac12 (θ=45∘\theta = 45^\circ); m=1m = 1 gives sin⁡θcos⁡θ=0\sin\theta\cos\theta = 0 (endpoints).

cot/tan triplet substitution

cot⁡θ=2120\cot\theta = \frac{21}{20} means a 20-21-29 triangle: cos⁡θ=2129\cos\theta = \frac{21}{29} directly. Any cot/tan given as a fraction of smallish integers is a triangle waiting to be drawn — faster than identities and immune to sign slips.

Reducing linear fractions in sin and cos

asin⁡θ+bcos⁡θccos⁡θ+dsin⁡θ→atan⁡θ+bc+dtan⁡θ\frac{a\sin\theta + b\cos\theta}{c\cos\theta + d\sin\theta} \to \frac{a\tan\theta + b}{c + d\tan\theta} — divide by cos⁡θ\cos\theta, substitute tt, simplify. Works the same with cot⁡\cot if the expression is cot-heavy.

The cosec/cot condition twin

Given cosec⁡θ+cot⁡θ=k\cosec\theta + \cot\theta = k: its conjugate cosec⁡θ−cot⁡θ=1k\cosec\theta - \cot\theta = \frac{1}{k} (product 1), so cosec⁡θ=k+1k2\cosec\theta = \frac{k + \frac1k}{2} and cot⁡θ=k−1k2\cot\theta = \frac{k - \frac1k}{2}. Since cosec⁡2−cot⁡2=1\cosec^2 - \cot^2 = 1, this family is the sec±tan engine wearing a different coat — and it finishes with sin⁡θ=2kk2+1\sin\theta = \frac{2k}{k^2 + 1} directly. Example: cosec⁡θ+cot⁡θ=3\cosec\theta + \cot\theta = 3 gives cosec⁡θ=3+132=53\cosec\theta = \frac{3 + \frac13}{2} = \frac53, so sin⁡θ=35\sin\theta = \frac35 — a 3-4-5 triangle in disguise.

Reading the options like a setter

Options in this family are built from predictable slips: substituting the reciprocal of the given tan, forgetting the constant +b+b term, dropping a minus sign when the denominator is negative, or reporting tan⁡2\tan^2 when tan⁡\tan was asked. After computing, name the slip each wrong option encodes — if you cannot, re-check your own arithmetic. And keep every substitution exact: tan⁡θ=43\tan\theta = \frac43 stays a fraction; a decimal 1.33 is how sign and size errors sneak in.

Quick revision

  • Table entries at 0/30/45/60/900/30/45/60/90; reduce everything to them.
  • sin⁡cos⁡=(sin⁡+cos⁡)2−12\sin\cos = \frac{(\sin+\cos)^2 - 1}{2}; (sin⁡−cos⁡)2=2−(sin⁡+cos⁡)2(\sin-\cos)^2 = 2 - (\sin+\cos)^2.
  • cot given as pq\frac{p}{q} → triangle (q,p,p2+q2)(q, p, \sqrt{p^2+q^2}).
  • Divide linear fractions by cos⁡θ\cos\theta; one substitution finishes them.

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 angles in an expressionvery common2 practice Q
How to spot it:

An arithmetic expression in sin/cos/tan of 0/30/45/60/90 asked as a single value.

Substitutetablevalues;simplifywithexactfractions.Substitute table values; simplify with exact fractions.
  1. Replace each ratio with its table entry (reciprocal functions included).
  2. Compute with fractions (14,34,12\frac14, \frac34, \frac12) — no decimals.
  3. Cancel before adding where possible.

Why: it is pure table recall plus one tidy computation — the marks are in avoiding the sin⁡260\sin^2 60 vs sin⁡60\sin 60 slips.

Example: The value of 2tan⁡245∘+cos⁡230∘−sin⁡260∘2\tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ is:

2(1)+34−34=22(1) + \frac34 - \frac34 = 2. The last two terms cancel exactly.

Type 2: sin+cos given → products and differencesvery common4 practice Q
How to spot it:

'sinθ + cosθ = m' given; sinθcosθ, sinθ − cosθ, or a cubic combo asked.

$\sin\cos = \frac{m^2-1}{2}$; $(\sin-\cos)^2 = 2 - m^2$; cube sums via $x^3+y^3 = (x+y)(x^2-xy+y^2)$.
  1. Square the given: m2=1+2sin⁡θcos⁡θm^2 = 1 + 2\sin\theta\cos\theta → the product.
  2. Difference: square root of 2−m22 - m^2 (positive since sin > cos or vice versa is possible — check signs only if options demand).
  3. Cubes: factor as (sin⁡+cos⁡)(1−sin⁡cos⁡)(\sin + \cos)(1 - \sin\cos) and substitute both knowns.

Why: every quantity in this family is a symmetric function of sin⁡\sin and cos⁡\cos — the pair (m,sin⁡cos⁡)(m, \sin\cos) determines all of them.

Example: If sin⁡θ+cos⁡θ=2\sin\theta + \cos\theta = \sqrt{2}, the value of sin⁡θcos⁡θ\sin\theta\cos\theta is:

2=1+2sin⁡θcos⁡θ2 = 1 + 2\sin\theta\cos\theta → sin⁡θcos⁡θ=12\sin\theta\cos\theta = \frac12 (indeed θ=45∘\theta = 45^\circ).

Type 3: tan/cot condition into a linear fractionvery common3 practice Q
How to spot it:

'If 5 tanθ = 4' or tanθ = p/q; a fraction like (5sinθ − 3cosθ)/(5sinθ + 3cosθ) asked.

Divide by $\cos\theta$; the fraction becomes $\frac{5t - 3}{5t + 3}$ in $t = \tan\theta$.
  1. Convert the condition to t=tan⁡θt = \tan\theta first.
  2. Divide numerator and denominator by cos⁡θ\cos\theta.
  3. Substitute and simplify the single fraction exactly.

Why: the coefficients of the question are the SAME numbers as in the fraction — substitution must be mechanical, not improvised.

Example: If 5tan⁡θ=45\tan\theta = 4, the value of 5sin⁡θ−3cos⁡θ5sin⁡θ+3cos⁡θ\frac{5\sin\theta - 3\cos\theta}{5\sin\theta + 3\cos\theta} is:

=5×45−35×45+3=17= \frac{5 \times \frac45 - 3}{5 \times \frac45 + 3} = \frac{1}{7}. (Divide top and bottom by cos⁡θ\cos\theta first.)

Type 4: cot/tan fraction → triangle → ratiocommon2 practice Q
How to spot it:

cotθ or tanθ given as a fraction like 21/20; a DIFFERENT ratio (cos, sin, sec) asked.

Draw the triangle with legs $q, p$; hyp $= \sqrt{p^2+q^2}$; read the asked ratio.
  1. cot⁡θ=2120\cot\theta = \frac{21}{20} → adj 21, opp 20 → hyp 29.
  2. Read the requested ratio from the same triangle.
  3. Triplet families (3-4-5, 5-12-13, 8-15-17, 20-21-29) cover the exams.

Why: the condition fully determines the shape — one drawing replaces three identity manipulations.

Example: If cot⁡θ=2120\cot\theta = \frac{21}{20} (θ\theta acute), the value of cos⁡θ\cos\theta is:

Triangle 20-21-29: cos⁡θ=adjhyp=2129\cos\theta = \frac{\text{adj}}{\text{hyp}} = \frac{21}{29}.

Formulas

From sin to the rest
sin⁡θ=ph⇒cos⁡θ=h2−p2h, tan⁡θ=ph2−p2\sin\theta=\frac{p}{h}\Rightarrow \cos\theta=\frac{\sqrt{h^2-p^2}}{h},\ \tan\theta=\frac{p}{\sqrt{h^2-p^2}}
Standard triplets
(3,4,5), (5,12,13), (7,24,25), (8,15,17), (9,40,41)(3,4,5),\ (5,12,13),\ (7,24,25),\ (8,15,17),\ (9,40,41)
Sum-product of a ratio and reciprocal
x+1x=k⇒x2+1x2=k2−2x+\frac1x=k\Rightarrow x^2+\frac{1}{x^2}=k^2-2

Shortcut tricks

⚡ Draw the triplet triangle

Ratio given → sides in hand in 5 seconds. Check the triplet table before any algebra.

Example: If cos⁡θ=725\cos\theta=\dfrac{7}{25}, find tan⁡θ\tan\theta.

Adjacent 7, hypotenuse 25 → opposite 625−49=24\sqrt{625-49}=24; tan⁡θ=247\tan\theta=\frac{24}{7}.

⚡ Reduce the asked expression

Rewrite the target in terms of the given relation. For x+1xx+\frac1x chains, square and subtract 2.

Example: If sin⁡θ+cosecθ=2\sin\theta+\text{cosec}\theta=2, find sin⁡2θ+cosec2θ\sin^2\theta+\text{cosec}^2\theta.

(x+1x)2=x2+1x2+2⇒4−2=2(x+\frac1x)^2=x^2+\frac1{x^2}+2\Rightarrow 4-2=2. (Here x=1x=1 in fact.)

Where students lose marks

  • Triplets applied with hypotenuse as a leg (132−122\sqrt{13^2-12^2} vs 132+122\sqrt{13^2+12^2}).

  • Forgetting that a given ratio fixes θ\theta up to the quadrant — CGL keeps θ\theta acute, so take the positive root.

  • Value-putting with θ=45∘\theta=45^\circ when the expression is not constant — verify a second angle before trusting it.

Practice sets — 15 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.