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SSC CGL Maths · Trigonometric Ratios and Identities

Complementary and Allied Angles

Angles that add to 90° swap sin for cos, tan for cot and sec for cosec; any larger or negative angle reduces to an acute one with a sign fixed by its quadrant.

All Trigonometric Ratios and Identities notes

Why this matters

61 questions in the bank, the largest page in the chapter, with 49 EASY and 18 from 2025. Most look long (a product of five tangents, a sum of eight squares) and are really one pairing step that collapses to 1 or 0. The 15 on quadrants and large angles carry most of this page's MODERATE questions, and the marks there go on signs.

Concept 1 of 3: The co-function rule for complementary angles

The two acute angles of a right triangle add to 90°, and the side opposite one of them is adjacent to the other. So the sine of one angle is the cosine of the other, and the same swap links tan with cot and sec with cosec.

Definition

  • 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) = \text{cosec}\,\theta and cosec(90∘−θ)=sec⁡θ\text{cosec}(90^\circ - \theta) = \sec\theta.
  • So a pair of angles adding to 90° gives: tan⁡Atan⁡(90∘−A)=1\tan A \tan(90^\circ - A) = 1, sin⁡2A+sin⁡2(90∘−A)=1\sin^2 A + \sin^2(90^\circ - A) = 1, and sin⁡Acos⁡(90∘−A)=1\frac{\sin A}{\cos(90^\circ - A)} = 1.
  • Fast route: in a long product or sum, add the angles in pairs. Each pair that makes 90° collapses to 1 or cancels, and tan⁡45∘=1\tan 45^\circ = 1 is usually the middle term.
  • Limit: only angles adding to exactly 90° pair this way. 20° and 60° do not.

Co-function rule

sin⁡(90∘−θ)=cos⁡θtan⁡(90∘−θ)=cot⁡θsec⁡(90∘−θ)=cosec θ\sin(90^\circ - \theta) = \cos\theta \qquad \tan(90^\circ - \theta) = \cot\theta \qquad \sec(90^\circ - \theta) = \text{cosec}\,\theta

Worked example

Evaluate sin⁡36∘cos⁡54∘+sec⁡28∘cosec 62∘−tan⁡15∘tan⁡75∘\frac{\sin 36^\circ}{\cos 54^\circ} + \frac{\sec 28^\circ}{\text{cosec}\,62^\circ} - \tan 15^\circ \tan 75^\circ.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 21 Sep 2025, Shift 1 · Q71Easy

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

If sin⁡A=45\sin A = \frac{4}{5} and A is acute, find cot⁡(90∘−A)\cot(90^\circ - A).

Pairing angles that add to 180°

The co-function rule needs a sum of exactly 90°. tan⁡50∘tan⁡130∘\tan 50^\circ \tan 130^\circ is not 1: those angles add to 180°, and 130° is handled by the allied-angle rule, not by swapping tan for cot.

Pairing a ratio with itself

sin⁡20∘sin⁡70∘\sin 20^\circ \sin 70^\circ is not 1. Swap one of them first: sin⁡70∘=cos⁡20∘\sin 70^\circ = \cos 20^\circ, so the product is sin⁡20∘cos⁡20∘\sin 20^\circ \cos 20^\circ, and it is tan times cot, or a square plus a square, that gives 1.
Drill 27 more on the co-function rule for complementary angles

Concept 2 of 3: Angles that add up to 90°

If the sine of one acute angle equals the cosine of another, the two must be partners that make 90°. Turn the equation of ratios into an equation of angles, and solve that.

Definition

  • sin⁡A=cos⁡B\sin A = \cos B, tan⁡A=cot⁡B\tan A = \cot B or sec⁡A=cosec B\sec A = \text{cosec}\,B, with A and B acute: A+B=90∘A + B = 90^\circ.
  • tan⁡Atan⁡B=1\tan A \tan B = 1 is the same statement, since it says tan⁡A=cot⁡B\tan A = \cot B.
  • With brackets, add the brackets: sin⁡(2x+10∘)=cos⁡(x+20∘)\sin(2x + 10^\circ) = \cos(x + 20^\circ) gives 3x+30∘=90∘3x + 30^\circ = 90^\circ.
  • In a triangle right-angled at C, A+B=90∘A + B = 90^\circ, so cos⁡B=sin⁡A\cos B = \sin A and tan⁡B=cot⁡A\tan B = \cot A with no working.
  • Limit: the rule needs both angles acute. sin⁡120∘=cos⁡30∘\sin 120^\circ = \cos 30^\circ, yet the angles add to 150°. SSC writes 'acute' in the question for this reason.

Partners that make 90°

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

Worked example

If sec⁡4x=cosec(x+15∘)\sec 4x = \text{cosec}(x + 15^\circ), with both angles acute, find x.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 14 Sep 2025, Shift 3 · Q70Easy

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

If tan⁡A=cot⁡(2A−30∘)\tan A = \cot(2A - 30^\circ), then what is the value of A?

Setting the angles equal

sin⁡2x=cos⁡3x\sin 2x = \cos 3x does not mean 2x=3x2x = 3x. Different ratios on the two sides mean the angles ADD to 90°; only the same ratio on both sides means the angles are equal.

Using a non-standard answer as a sign of error

sin⁡x=cos⁡(2x−10∘)\sin x = \cos(2x - 10^\circ) gives x=33.33∘x = 33.33^\circ, which is not a table angle. That is fine: this rule never needs the table, only the sum of 90°.
Drill 17 more on angles that add up to 90°

Concept 3 of 3: Allied angles, quadrant signs and large angles

Every angle beyond 90° behaves like an acute angle wearing a sign. Find the acute angle it is tied to, take that value from the table, and let the quadrant decide plus or minus.

Definition

  • Signs (ASTC): from 0° to 90° All ratios are positive; 90° to 180° only Sine (and cosec); 180° to 270° only Tan (and cot); 270° to 360° only Cos (and sec).
  • 180∘±θ180^\circ \pm \theta and 360∘−θ360^\circ - \theta keep the same ratio: sin⁡150∘=sin⁡30∘\sin 150^\circ = \sin 30^\circ, cos⁡240∘=−cos⁡60∘\cos 240^\circ = -\cos 60^\circ.
  • 90∘+θ90^\circ + \theta and 270∘±θ270^\circ \pm \theta switch to the co-ratio: cos⁡(90∘+θ)=−sin⁡θ\cos(90^\circ + \theta) = -\sin\theta.
  • Large angles: take away whole turns of 360°: cos⁡750∘=cos⁡30∘\cos 750^\circ = \cos 30^\circ. tan and cot repeat every 180°, so you may take away 180° for them.
  • Negative angles: sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta, tan⁡(−θ)=−tan⁡θ\tan(-\theta) = -\tan\theta, cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta.
  • A ratio given in a quadrant: draw the acute triangle for the sizes, then give each ratio the quadrant's sign.

Allied angles

sin⁡(180∘−θ)=sin⁡θcos⁡(180∘−θ)=−cos⁡θtan⁡(180∘+θ)=tan⁡θcos⁡(360∘−θ)=cos⁡θsin⁡(−θ)=−sin⁡θ\sin(180^\circ - \theta) = \sin\theta \qquad \cos(180^\circ - \theta) = -\cos\theta \qquad \tan(180^\circ + \theta) = \tan\theta \qquad \cos(360^\circ - \theta) = \cos\theta \qquad \sin(-\theta) = -\sin\theta

Worked example

Find sin⁡495∘+cos⁡(−300∘)\sin 495^\circ + \cos(-300^\circ).
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 22 Sep 2025, Shift 1 · Q67Moderate

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

If sin⁡A=513\sin A = \frac{5}{13} and A lies in 2nd quadrant, then find the value of sin⁡A×sec⁡A\sin A \times \sec A.

Dropping the sign in the second quadrant

Between 90° and 180° only sin and cosec are positive. If sin⁡x=35\sin x = \frac{3}{5} there, then cos⁡x=−45\cos x = -\frac{4}{5} and tan⁡x=−34\tan x = -\frac{3}{4}; the positive versions are always among the options.

Taking away 180° from a sine

Only tan and cot repeat every 180°. For sin and cos, sin⁡(θ+180∘)=−sin⁡θ\sin(\theta + 180^\circ) = -\sin\theta, so remove whole turns of 360° instead.

Putting a minus on cos of a negative angle

cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta: cos ignores the sign of the angle. sin and tan are the ones that change sign.
Drill 14 more on allied angles, quadrant signs and large angles

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 for complementary angles

    Co-function rule

    sin⁡(90∘−θ)=cos⁡θtan⁡(90∘−θ)=cot⁡θsec⁡(90∘−θ)=cosec θ\sin(90^\circ - \theta) = \cos\theta \qquad \tan(90^\circ - \theta) = \cot\theta \qquad \sec(90^\circ - \theta) = \text{cosec}\,\theta
  • Angles that add up to 90°

    Partners that make 90°

    sin⁡A=cos⁡B or tan⁡A=cot⁡B (A,B acute)⇒A+B=90∘\sin A = \cos B \ \text{or} \ \tan A = \cot B \ (A, B \text{ acute}) \Rightarrow A + B = 90^\circ
  • Allied angles, quadrant signs and large angles

    Allied angles

    sin⁡(180∘−θ)=sin⁡θcos⁡(180∘−θ)=−cos⁡θtan⁡(180∘+θ)=tan⁡θcos⁡(360∘−θ)=cos⁡θsin⁡(−θ)=−sin⁡θ\sin(180^\circ - \theta) = \sin\theta \qquad \cos(180^\circ - \theta) = -\cos\theta \qquad \tan(180^\circ + \theta) = \tan\theta \qquad \cos(360^\circ - \theta) = \cos\theta \qquad \sin(-\theta) = -\sin\theta

Watch out for (7)

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