Exercise 11.3 — Complementary Angles

Trigonometric ratios of complementary angles.

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Exercise 11.3 — Trigonometric Ratios of Complementary Angles

Exercise 11.3 continues Chapter 11, Trigonometry, in Class 10 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) by introducing one of the most powerful shortcuts in the entire chapter: the complementary angle relationships. These six identities let you instantly rewrite any ratio of angle (90° − θ) in terms of θ, which turns many seemingly complicated expressions into simple cancellations.

This exercise has six question types: direct evaluation of expressions using the complementary identities; "show that" proofs where both sides reduce to the same value; solving equations involving 2A and (A − 18°) or two acute angles A and B; a triangle-angle proof connecting half-angles; and finally, expressing an angle-sum expression in terms of angles below 45°.

Complementary Angles Evaluating Expressions Proving Identities Solving for Unknown Angles
💡 Foundation fact: Two angles are complementary if they add up to exactly 90°. In a right triangle, the two acute angles are always complementary to each other — and this single geometric fact is what generates all six identities used throughout this exercise.

Deriving the Complementary Angle Identities

Consider a right triangle ABC with ∠B = 90°. The two acute angles, ∠A and ∠C, must add up to 90° (since all three angles of a triangle total 180° and one of them is already 90°). If we call ∠A = θ, then ∠C must equal 90° − θ.

A B C θ 90°−θ

Right triangle ABC with ∠B = 90°, ∠A = θ, and ∠C = 90° − θ (complementary angles).

From this triangle, the six basic ratios for angle θ (at vertex A) are written as usual:

sin θ = BC/AC,   cos θ = AB/AC,   tan θ = BC/AB
cosec θ = AC/BC,   sec θ = AC/AB,   cot θ = AB/BC

Now look at the same triangle from the point of view of angle C, which equals (90° − θ). Because the side opposite to C is AB and the side adjacent to C is BC, the ratios for (90° − θ) come out as the AB and BC versions of the ratios above — but matched to a different name than before:

sin(90° − θ) = AB/AC = cos θ cos(90° − θ) = BC/AC = sin θ tan(90° − θ) = AB/BC = cot θ cosec(90° − θ) = AC/AB = sec θ sec(90° − θ) = AC/BC = cosec θ cot(90° − θ) = BC/AB = tan θ

These six results are the complementary angle identities — the single most useful toolkit in this exercise:

IdentityIn Words
sin(90° − θ) = cos θsin of the complement equals cos of the angle
cos(90° − θ) = sin θcos of the complement equals sin of the angle
tan(90° − θ) = cot θtan of the complement equals cot of the angle
cosec(90° − θ) = sec θcosec of the complement equals sec of the angle
sec(90° − θ) = cosec θsec of the complement equals cosec of the angle
cot(90° − θ) = tan θcot of the complement equals tan of the angle
💡 Memory trick: Notice the ratios pair up neatly into three "swap" pairs: sin ↔ cos, tan ↔ cot, and sec ↔ cosec. Whenever you see (90° − θ) inside a trig function, just swap that function for its pair and drop the (90° − θ), replacing it with θ.
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Question 1 — Evaluate Using Complementary Angle Identities

This question presents five expressions where the two angles inside always add up to 90°. The strategy every time is identical: rewrite one of the two angles as (90° minus the other), apply the matching identity, and watch the expression collapse to a simple number.

Part (i)
tan 36° ÷ cot 54°
tan 36° = tan(90° − 54°) = cot 54° [∵ tan(90°−θ) = cotθ] So the expression becomes: cot 54° / cot 54° = 1 (numerator and denominator are identical)
Answer: tan 36° / cot 54° = 1
Part (ii)
cos 12° − sin 78°
cos 12° = cos(90° − 78°) = sin 78° [∵ cos(90°−θ) = sinθ] So the expression becomes: sin 78° − sin 78° = 0 (the two terms cancel exactly)
Answer: cos 12° − sin 78° = 0
Part (iii)
cosec 31° − sec 59°
cosec 31° = cosec(90° − 59°) = sec 59° [∵ cosec(90°−θ) = secθ] So the expression becomes: sec 59° − sec 59° = 0 (the two terms cancel exactly)
Answer: cosec 31° − sec 59° = 0
Part (iv)
sin 15° · sec 75°
sin 15° = sin(90° − 75°) = cos 75° [∵ sin(90°−θ) = cosθ] So the expression becomes: cos 75° · sec 75° = cos 75° × (1/cos 75°) [∵ secθ = 1/cosθ] = 1 (cos 75° cancels with its reciprocal)
Answer: sin 15° · sec 75° = 1
Part (v)
tan 26° · tan 64°
tan 26° = tan(90° − 64°) = cot 64° [∵ tan(90°−θ) = cotθ] So the expression becomes: cot 64° · tan 64° = cot 64° × (1/cot 64°) [∵ tanθ = 1/cotθ] = 1 (cot 64° cancels with its reciprocal)
Answer: tan 26° · tan 64° = 1
📌 Pattern across Question 1: Every single part of this question evaluates to either 0 or 1. This is not a coincidence — whenever two angles inside an expression add up to exactly 90°, converting one into the other's complementary form always produces a matching pair that either cancels by subtraction (giving 0) or by reciprocal multiplication (giving 1).

Question 2 — Show That the Following Identities Hold

These two "show that" problems use the same complementary-angle substitution technique as Question 1, but ask you to prove the full equality rather than just compute a final number — meaning you must display both the left-hand side (LHS) and right-hand side (RHS) explicitly to demonstrate they match.

Part (i)
Show that tan 48° · tan 16° · tan 42° · tan 74° = 1
LHS = tan 48° · tan 16° · tan 42° · tan 74° tan 48° = tan(90° − 42°) = cot 42° tan 16° = tan(90° − 74°) = cot 74° LHS = cot 42° · cot 74° · tan 42° · tan 74° = (cot 42° · tan 42°) · (cot 74° · tan 74°) = 1 × 1 = 1 = RHS
Hence proved: tan 48° · tan 16° · tan 42° · tan 74° = 1, since each cot–tan pair of complementary angles multiplies to exactly 1.
Part (ii)
Show that cos 36° cos 54° − sin 36° sin 54° = 0
LHS = cos 36° cos 54° − sin 36° sin 54° cos 36° = cos(90° − 54°) = sin 54° cos 54° = cos(90° − 36°) = sin 36° LHS = sin 54° · sin 36° − sin 36° · sin 54° = 0 (identical terms subtract to zero) = RHS
Hence proved: cos 36° cos 54° − sin 36° sin 54° = 0, because 36° and 54° are complementary angles.
💡 Exam strategy: When a "show that" question contains two angles whose sum is 90° (like 36° and 54°, or 16°/74° and 42°/48°), it is almost always solvable purely through complementary angle substitution — no calculator values are ever needed.

Question 3 — Solving tan 2A = cot(A − 18°)

Here the complementary identity is used in reverse: instead of simplifying a known expression, it is used to convert two different trig functions (tan and cot) into the same function so that the angles themselves can be equated.

Worked Solution
Given: tan 2A = cot(A − 18°), where 2A is an acute angle. Find A.
tan 2A = cot(90° − 2A) [∵ cot(90°−θ) = tanθ, used here as tanθ = cot(90°−θ)] So: cot(90° − 2A) = cot(A − 18°) Equating the angles: 90° − 2A = A − 18° 90° + 18° = A + 2A 108° = 3A A = 108° / 3 = 36°
Answer: The value of A is 36°.

Question 4 — Proving A + B = 90° When tan A = cot B

Worked Solution
Given: tan A = cot B, where A and B are acute angles. Prove A + B = 90°.
tan A = cot B cot(90° − A) = cot B [∵ cot(90°−θ) = tanθ, so tanA = cot(90°−A)] Since the cot function is the same on both sides, the angles must be equal: 90° − A = B ⟹ A + B = 90° ✓ Hence proved
Hence proved: If tan A = cot B for acute angles A and B, then A and B must be complementary, i.e. A + B = 90°.
📌 Key technique in Q3 and Q4: Whenever an equation mixes two different trig functions that belong to the same complementary pair (tan/cot, sin/cos, sec/cosec), convert one side using the (90° − θ) identity so both sides use the same function — then simply equate the angles.

Question 5 — Proving tan((A+B)/2) = cot(C/2) in a Triangle

This question connects the angle sum property of a triangle (the three interior angles always total 180°) with the complementary angle identities, producing a relationship between half-angles that is frequently used in advanced trigonometry problems.

Worked Solution
Given: A, B, C are interior angles of triangle ABC. Show that tan((A+B)/2) = cot(C/2)
Since A, B, C are angles of a triangle: A + B + C = 180° ⟹ A + B = 180° − C Dividing both sides by 2: (A + B)/2 = (180° − C)/2 = 90° − C/2 Taking tan on both sides: tan((A+B)/2) = tan(90° − C/2) tan((A+B)/2) = cot(C/2) [∵ tan(90°−θ) = cotθ]
Hence proved: tan((A+B)/2) = cot(C/2) for any triangle ABC.
💡 Why this works: Half of (A + B) and half of C are themselves complementary angles whenever A + B + C = 180°, because (A+B)/2 + C/2 = 90°. This is exactly the same complementary relationship used throughout this exercise, just applied to half-angles of a triangle instead of standalone angles.

Question 6 — Expressing sin 75° + cos 65° Using Angles Below 45°

This question asks for an expression to be rewritten entirely using angles between 0° and 45°. Since both 75° and 65° are above 45°, each one needs to be converted into its complementary form so that the angle inside each function drops below the 45° threshold.

Worked Solution
Express sin 75° + cos 65° in terms of angles between 0° and 45°
sin 75° = sin(90° − 15°) = cos 15° [∵ sin(90°−θ) = cosθ] cos 65° = cos(90° − 25°) = sin 25° [∵ cos(90°−θ) = sinθ] So: sin 75° + cos 65° = cos 15° + sin 25°
Answer: sin 75° + cos 65° = cos 15° + sin 25°, where both 15° and 25° lie between 0° and 45° as required.
75° becomes
cos 15°
65° becomes
sin 25°

Common Mistakes to Avoid

  • Swapping the wrong pair: Remember the three fixed pairs — sin↔cos, tan↔cot, sec↔cosec. Mixing up which function pairs with which (e.g. confusing tan↔sec) leads to an incorrect identity.
  • Forgetting to check the angle sum first: Before applying any complementary identity, always confirm the two angles in the question genuinely add up to 90° — the technique only works for true complementary pairs.
  • Sign errors when rearranging equations: In Question 3, carefully expand 90° − 2A = A − 18° step by step; rushing this rearrangement is the most common source of wrong answers.
  • Stopping at the substitution instead of finishing the proof: In "show that" questions (Question 2), always explicitly write out the final LHS = RHS line — a half-finished substitution is not considered a complete proof in board exams.
  • Confusing reciprocal identities with complementary identities: tanθ = 1/cotθ is a reciprocal identity (same angle), while tan(90°−θ) = cotθ is a complementary identity (different angle) — both are used together in parts (iv) and (v) of Question 1, so keep them distinct.
  • Not simplifying half-angle expressions fully: In Question 5, make sure to divide both sides of A + B = 180° − C by 2 correctly before applying the tan ratio.
High-risk exam trap: Questions 3 and 4 are popular because they disguise a simple complementary-angle substitution inside an algebraic equation. Students who try to solve them using triangle properties or calculator values (instead of the cot(90°−θ)=tanθ trick) often get stuck — always look first for a hidden complementary pair.

Quick Reference — All Answers at a Glance

QuestionTopicKey Result
Q1(i)tan 36° ÷ cot 54°1
Q1(ii)cos 12° − sin 78°0
Q1(iii)cosec 31° − sec 59°0
Q1(iv)sin 15° · sec 75°1
Q1(v)tan 26° · tan 64°1
Q2(i)tan48°tan16°tan42°tan74°= 1, proved
Q2(ii)cos36cos54 − sin36sin54= 0, proved
Q3tan2A = cot(A−18°)A = 36°
Q4tanA = cotBA+B=90°, proved
Q5Triangle half-angle identitytan((A+B)/2)=cot(C/2), proved
Q6sin75° + cos65°= cos15° + sin25°

What This Exercise Prepares You For

Exercise 11.3 builds the algebraic flexibility needed for the rest of the Trigonometry chapter. The complementary angle identities practised here are used constantly when proving the fundamental trigonometric identities and when simplifying mixed-angle expressions in later exercises. The standard-angle values from Exercise 11.2 combine directly with these complementary rules — for instance, sin 30° = cos 60° is simply the complementary identity in action at a specific angle.

The angle-sum reasoning used in Question 5 also reinforces the triangle angle-sum property from Triangles, while the equation-solving technique in Questions 3 and 4 is a direct preview of the algebraic manipulation skills needed in Trigonometric Identities, the next major topic in this chapter.

📐 Board Exam Tip (CBSE, Telangana & AP): Complementary angle questions are extremely common as 2–3 mark "evaluate" or "show that" problems. The fastest way to spot them is to add the two angles in the question — if they sum to exactly 90°, a complementary identity will almost always provide the shortest path to the answer.
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