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CDS Mathematics · Trigonometric Ratios and Identities

Complementary Angles

The sine of an angle is the cosine of its complement, and likewise tangent and cotangent, secant and cosecant — so angles that add to 90° cancel, pair off, or collapse to 1.

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Why this matters

Twenty-one PYQs, nearly all MODERATE and nearly all one idea: spot the pairs of angles that add to 90° and replace one ratio of each pair by its co-ratio. The long products and sums that look fearsome — tan 1° tan 2° … tan 89° — are the easiest marks on the page once the pairing is seen.

Concept 1 of 3: The co-function rule: sin(90° − θ) = cos θ

In a right triangle the two acute angles add to 90∘90^\circ, and the side opposite one of them is the side adjacent to the other. So the sine of one angle is literally the same fraction as the cosine of the other. The 'co' in cosine means 'of the complement'.

Definition

For every angle θ\theta:

  • sin⁡(90∘−θ)=cos⁡θ\sin(90^\circ - \theta) = \cos\theta and cos⁡(90∘−θ)=sin⁡θ\cos(90^\circ - \theta) = \sin\theta;
  • tan⁡(90∘−θ)=cot⁡θ\tan(90^\circ - \theta) = \cot\theta and cot⁡(90∘−θ)=tan⁡θ\cot(90^\circ - \theta) = \tan\theta;
  • sec⁡(90∘−θ)=cosec⁡θ\sec(90^\circ - \theta) = \operatorname{cosec}\theta and cosec⁡(90∘−θ)=sec⁡θ\operatorname{cosec}(90^\circ - \theta) = \sec\theta.

So a quotient like sin⁡19∘cos⁡71∘\dfrac{\sin 19^\circ}{\cos 71^\circ} is 11, and sec⁡256∘−cot⁡234∘\sec^2 56^\circ - \cot^2 34^\circ is sec⁡256∘−tan⁡256∘=1\sec^2 56^\circ - \tan^2 56^\circ = 1.

Co-function rule

sin⁡(90∘−θ)=cos⁡θ,tan⁡(90∘−θ)=cot⁡θ,sec⁡(90∘−θ)=cosec⁡θ\sin(90^\circ-\theta) = \cos\theta, \quad \tan(90^\circ-\theta) = \cot\theta, \quad \sec(90^\circ-\theta) = \operatorname{cosec}\theta
θ90° − θopposite θadjacent to 90° − θadjacent to θopposite 90° − θhypotenuseABC

sin θ = opposite ÷ hypotenuse = cos (90° − θ): the same side, named from the other angle.

Worked example

Evaluate cos⁡58∘sin⁡32∘+tan⁡15∘cot⁡75∘−sin⁡240∘−sin⁡250∘\dfrac{\cos 58^\circ}{\sin 32^\circ} + \dfrac{\tan 15^\circ}{\cot 75^\circ} - \sin^2 40^\circ - \sin^2 50^\circ.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2019 · CDS (I) 2019 — Elementary Mathematics · Q35Easy

Example 1 · Trigonometric Ratios and Identities · Complementary and Allied Angles

What is the value of sin⁡19∘cos⁡71∘+cos⁡73∘sin⁡17∘\dfrac{\sin 19^\circ}{\cos 71^\circ} + \dfrac{\cos 73^\circ}{\sin 17^\circ} ?

Check the angles really add to 90°

cos⁡61∘\cos 61^\circ and sin⁡29∘\sin 29^\circ are equal; cos⁡61∘\cos 61^\circ and sin⁡31∘\sin 31^\circ are not. Statement questions plant a pair that misses by two degrees. Add the angles before you cancel.
Drill 9 more on the co-function rule: sin(90° − θ) = cos θ

Concept 2 of 3: Pairing the terms of a long product or sum

A product tan⁡1∘tan⁡2∘⋯tan⁡89∘\tan 1^\circ \tan 2^\circ \cdots \tan 89^\circ is 89 factors that nobody can evaluate one by one — and does not need to. Pair 1∘1^\circ with 89∘89^\circ, 2∘2^\circ with 88∘88^\circ: each pair is a tangent times its own cotangent, which is 11. The same pairing turns a sum of sin⁡2\sin^2 terms into a count of pairs.

Definition

  • Products: tan⁡θ⋅tan⁡(90∘−θ)=tan⁡θcot⁡θ=1\tan\theta \cdot \tan(90^\circ - \theta) = \tan\theta\cot\theta = 1, and the same for cot⁡\cot. The middle term, if any, is tan⁡45∘=1\tan 45^\circ = 1.
  • Sums of squares: sin⁡2θ+sin⁡2(90∘−θ)=1\sin^2\theta + \sin^2(90^\circ - \theta) = 1, and the same for cos⁡2\cos^2. Count the pairs, then add any unpaired term (sin⁡245∘=12\sin^2 45^\circ = \dfrac12, sin⁡290∘=1\sin^2 90^\circ = 1, sin⁡20∘=0\sin^2 0^\circ = 0).
  • Mixed: sin⁡Acos⁡(90∘−A)≠1\sin A\cos(90^\circ - A) \ne 1 — it is sin⁡2A\sin^2 A. Pair a ratio with its co-ratio of the complement, not with itself.

Pairs that collapse

tan⁡θ tan⁡(90∘−θ)=1,sin⁡2θ+sin⁡2(90∘−θ)=1\tan\theta\,\tan(90^\circ-\theta) = 1, \qquad \sin^2\theta + \sin^2(90^\circ-\theta) = 1

Worked example

Find sin⁡25∘+sin⁡210∘+sin⁡215∘+⋯+sin⁡285∘\sin^2 5^\circ + \sin^2 10^\circ + \sin^2 15^\circ + \cdots + \sin^2 85^\circ.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2016 · CDS (II) 2016 — Elementary Mathematics · Q85Moderate

Example 2 · Trigonometric Ratios and Identities · Complementary and Allied Angles

What is the value of tan⁡1∘tan⁡2∘tan⁡3∘tan⁡4∘…tan⁡89∘\tan 1^\circ \tan 2^\circ \tan 3^\circ \tan 4^\circ \ldots \tan 89^\circ ?

Count the terms before pairing

With an odd number of terms one is left unpaired, and it is not always 45∘45^\circ: in sin⁡26∘+⋯+sin⁡290∘\sin^2 6^\circ + \cdots + \sin^2 90^\circ the stray term is sin⁡290∘=1\sin^2 90^\circ = 1. Write the first and last angles and the step, count, then pair.
Drill 4 more on pairing the terms of a long product or sum

Concept 3 of 3: tan A = cot B, and angles of a triangle

If the tangent of one acute angle equals the cotangent of another, the two angles are complements. The same happens when a product like tan⁡3xtan⁡6x\tan 3x \tan 6x equals 11. In a triangle the three angles add to 180∘180^\circ, so half of B+CB + C is the complement of half of AA.

Definition

  • For acute angles, tan⁡A=cot⁡B\tan A = \cot B (or sin⁡A=cos⁡B\sin A = \cos B) means A+B=90∘A + B = 90^\circ.
  • tan⁡P⋅tan⁡Q=1\tan P \cdot \tan Q = 1 means tan⁡P=cot⁡Q\tan P = \cot Q, so P+Q=90∘P + Q = 90^\circ (then check the stated range).
  • In a triangle, B+C2=90∘−A2\dfrac{B + C}{2} = 90^\circ - \dfrac{A}{2}, so sin⁡B+C2=cos⁡A2\sin\dfrac{B+C}{2} = \cos\dfrac{A}{2} and tan⁡B+C2=cot⁡A2\tan\dfrac{B+C}{2} = \cot\dfrac{A}{2}.

Complementary condition

tan⁡A=cot⁡B  ⇒  A+B=90∘(A,B acute)\tan A = \cot B \;\Rightarrow\; A + B = 90^\circ \quad (A, B \text{ acute})

Worked example

If sin⁡(2A+10∘)=cos⁡(A+20∘)\sin(2A + 10^\circ) = \cos(A + 20^\circ) with both angles acute, find AA.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2024 · CDS (I) 2024 — Elementary Mathematics · Q51Easy

Example 3 · Trigonometric Ratios and Identities · Complementary and Allied Angles

If tan⁡(3A)=cot⁡(A−22∘)\tan(3A) = \cot(A - 22^\circ), where 3A is an acute angle, then what is the value of A ?

The rule needs acute angles — check the range

tan⁡Ptan⁡Q=1\tan P \tan Q = 1 in general gives P+Q=90∘+180∘kP + Q = 90^\circ + 180^\circ k. A question with a range like 0≤x<30∘0 \le x < 30^\circ is there to make you pick the one value that fits, and to reject the edge value that the range excludes.
Drill 5 more on tan a = cot b, and angles of a triangle

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • The co-function rule: sin(90° − θ) = cos θ

    Co-function rule

    sin⁡(90∘−θ)=cos⁡θ,tan⁡(90∘−θ)=cot⁡θ,sec⁡(90∘−θ)=cosec⁡θ\sin(90^\circ-\theta) = \cos\theta, \quad \tan(90^\circ-\theta) = \cot\theta, \quad \sec(90^\circ-\theta) = \operatorname{cosec}\theta
  • Pairing the terms of a long product or sum

    Pairs that collapse

    tan⁡θ tan⁡(90∘−θ)=1,sin⁡2θ+sin⁡2(90∘−θ)=1\tan\theta\,\tan(90^\circ-\theta) = 1, \qquad \sin^2\theta + \sin^2(90^\circ-\theta) = 1
  • tan A = cot B, and angles of a triangle

    Complementary condition

    tan⁡A=cot⁡B  ⇒  A+B=90∘(A,B acute)\tan A = \cot B \;\Rightarrow\; A + B = 90^\circ \quad (A, B \text{ acute})

Watch out for (3)

Test yourself on Trigonometric Ratios and Identities

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