Trigonometry — Identities, Equations & Triangles — Practice

Question 1

If tanθ = 5/12 (θ acute), the value of secθ − cosecθ is:
(a) -91/60  (b) 221/60  (c) 7/13  (d) 91/60

Accepted Answer

5-12-13 is a Pythagorean triple, so sinθ=5/13, cosθ=12/13, secθ=13/12, cosecθ=13/5.

secθ − cosecθ = 13/12 − 13/5 = 65/60 − 156/60 = −91/60.

Answer: (a) -91/60. Trap (b) 221/60 comes from adding instead of subtracting (secθ+cosecθ = 65/60+156/60). Trap (c) 7/13 comes from forgetting the reciprocal step and using cosθ−sinθ = 12/13−5/13 directly in place of secθ−cosecθ. Trap (d) 91/60 is the correct magnitude with the subtraction order reversed (cosecθ−secθ).

Question 2

The value of (secθ − tanθ)(secθ + tanθ), for any θ where both are defined, is:
(a) 1  (b) 0  (c) 2secθtanθ  (d) Cannot be determined without knowing θ

Accepted Answer

(secθ−tanθ)(secθ+tanθ) is a difference of squares: sec²θ − tan²θ. From 1+tan²θ=sec²θ, this equals exactly 1 — a constant, true for every valid θ.

Answer: (a) 1. Trap (b) 0 wrongly assumes the two bracketed terms are equal and cancel. Trap (c) 2secθtanθ mistakenly expands the product as if it were (secθ+tanθ)² rather than a difference of squares. Trap (d) fails to recognise sec²θ−tan²θ=1 as a genuine identity that holds for ALL θ, not an expression that varies.

Question 3

If secθ + tanθ = 3, then secθ equals:
(a) 5/3  (b) 3/2  (c) 4/3  (d) 10/3

Accepted Answer

Since sec²θ−tan²θ=1, (secθ−tanθ)(secθ+tanθ)=1, so secθ−tanθ = 1/3.

Adding the two equations: 2secθ = 3 + 1/3 = 10/3, so secθ = 5/3.

Answer: (a) 5/3. Trap (b) 3/2 naively splits the given value 3 equally between secθ and tanθ without using the identity at all. Trap (c) 4/3 comes from subtracting instead of adding the two equations (2tanθ = 3−1/3 = 8/3, tanθ=4/3) and reporting tanθ's value where secθ was asked. Trap (d) 10/3 is the correct numerator 3+1/3 with the final ÷2 step forgotten.

Question 4

The value of cos15° is:
(a) (√6+√2)/4  (b) (√6−√2)/4  (c) (√6+√2)/2  (d) (√2−√6)/4

Accepted Answer

cos15° = cos(45°−30°) = cos45°cos30° + sin45°sin30° = (√2/2)(√3/2) + (√2/2)(1/2) = √6/4 + √2/4 = (√6+√2)/4.

Answer: (a) (√6+√2)/4. Trap (b) (√6−√2)/4 comes from using the cos(A+B) minus-sign formula instead of cos(A−B) — this is actually cos75°, not cos15°. Trap (c) (√6+√2)/2 drops a factor of 2 in the denominator (forgetting (1/2)×(1/2)=1/4, not 1/2). Trap (d) (√2−√6)/4 flips the subtraction order, giving a negative value — a red flag that should be caught by noting cos15° must be close to 1 and positive.

Question 5

If cosθ = −3/5 and θ lies in the second quadrant, the value of sin2θ is:
(a) -24/25  (b) 24/25  (c) -7/25  (d) 7/25

Accepted Answer

In QII, sine is positive: sinθ = √(1−9/25) = 4/5.

sin2θ = 2sinθcosθ = 2(4/5)(−3/5) = −24/25.

Answer: (a) -24/25. Trap (b) 24/25 comes from wrongly taking sinθ as negative in QII (a quadrant-sign slip, mixing QII up with QIII), giving 2(−4/5)(−3/5)=+24/25. Trap (c) -7/25 is cos2θ = 2cos²θ−1 = 18/25−1, computed instead of sin2θ — the wrong double-angle formula entirely. Trap (d) 7/25 is that same cos2θ confusion with the subtraction order reversed (sin²θ−cos²θ instead of cos²θ−sin²θ).

Question 6

If sinθ = 1/3, the value of sin3θ is:
(a) 23/27  (b) 31/27  (c) -23/27  (d) -1/3

Accepted Answer

sin3θ = 3sinθ − 4sin³θ = 3(1/3) − 4(1/27) = 1 − 4/27 = 23/27.

Answer: (a) 23/27. Trap (b) 31/27 flips the sign inside the formula (3sinθ+4sin³θ = 1+4/27), confusing sin3A's sign pattern with cos3A's. Trap (c) -23/27 uses cos3A's structure (4sin³θ−3sinθ) with sine values, giving the correct magnitude but the wrong sign. Trap (d) -1/3 forgets to cube sinθ, using sinθ itself instead of sin³θ in the second term: 3(1/3)−4(1/3)=−1/3.

Question 7

The number of solutions of 2cos²θ − 3cosθ + 1 = 0 in [0, 2π) is:
(a) 3  (b) 2  (c) 4  (d) 6

Accepted Answer

Let x=cosθ: 2x²−3x+1=0 ⇒ (2x−1)(x−1)=0 ⇒ x=1/2 or x=1.

cosθ=1/2 gives θ=π/3, 5π/3 (2 solutions in [0,2π)). cosθ=1 gives θ=0 only — θ=2π is excluded since the interval is half-open. Total = 3.

Answer: (a) 3. Trap (b) 2 drops the θ=0 root entirely, e.g. by treating the linear factor (x−1) as not worth counting. Trap (c) 4 double-counts the endpoint by including both θ=0 and θ=2π, forgetting the interval is [0,2π) not [0,2π]. Trap (d) 6 comes from mistakenly using a period of π instead of 2π (as if solving for 2θ), doubling every root's count.

Question 8

The general solution of tan²θ = 3 is:
(a) θ = nπ ± π/3  (b) θ = nπ + π/3  (c) θ = 2nπ ± π/3  (d) θ = nπ + (−1)ⁿπ/3

Accepted Answer

tan²θ=3 ⇒ tanθ=±√3. tanθ=√3 gives θ=nπ+π/3; tanθ=−√3 gives θ=nπ−π/3. Combined, these form the single family θ=nπ±π/3, matching the standard rule that tan²θ=tan²α collapses to θ=nπ±α.

Answer: (a) θ = nπ ± π/3. Trap (b) keeps only the tanθ=+√3 branch, dropping half the solutions. Trap (c) borrows cosine's 2nπ±α period, which doesn't apply to a tangent-based equation. Trap (d) borrows sine's nπ+(−1)ⁿα form, another formula-family mix-up.

Question 9

The value of tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3) is:
(a) π  (b) 3π/4  (c) π/2  (d) 0

Accepted Answer

For tan⁻¹2+tan⁻¹3: xy=6>1 with both positive, so the addition formula needs the +π correction: tan⁻¹[(2+3)/(1−6)] + π = tan⁻¹(−1) + π = −π/4+π = 3π/4.

Adding tan⁻¹1 = π/4: total = π/4 + 3π/4 = π.

Answer: (a) π. Trap (b) 3π/4 stops after computing tan⁻¹2+tan⁻¹3 and forgets to add tan⁻¹1. Trap (c) π/2 misapplies the complementary identity tan⁻¹x+tan⁻¹(1/x)=π/2 to values that aren't reciprocals. Trap (d) 0 uses the plain addition formula on tan⁻¹2+tan⁻¹3 WITHOUT the +π correction (xy=6>1 ignored), getting tan⁻¹(−1)=−π/4, which then cancels tan⁻¹1=π/4 to give 0 — exactly the xy>1 trap this chapter flags.

Question 10

The principal-value range of cos⁻¹x is:
(a) [0, π]  (b) [−π/2, π/2]  (c) (−π/2, π/2)  (d) [−π, π]

Accepted Answer

By definition, the principal-value branch of cos⁻¹x is [0, π].

Answer: (a) [0, π]. Trap (b) [−π/2,π/2] is sin⁻¹x's range, the single most common inverse-trig confusion. Trap (c) (−π/2,π/2), an open interval, is tan⁻¹x's range. Trap (d) [−π,π] doubles the correct interval, as if the full circle were used instead of the restricted principal branch.

Question 11

In triangle ABC, A=60°, b=6 cm, c=4 cm. The length of side a is:
(a) 2√7 cm  (b) 2√10 cm  (c) 2√19 cm  (d) 2√13 cm

Accepted Answer

a² = b²+c²−2bc·cosA = 36+16−2(6)(4)(1/2) = 52−24 = 28, so a=2√7 cm.

Answer: (a) 2√7 cm. Trap (b) 2√10 cm forgets the factor of 2 in the cosine rule, computing 52−(6)(4)(1/2)=40. Trap (c) 2√19 cm adds instead of subtracts (52+24=76). Trap (d) 2√13 cm drops the cosine term entirely, treating the triangle as right-angled via plain Pythagoras (36+16=52).

Question 12

From a point on level ground, the angle of elevation of the top of a tower is 30°. After walking 60 m towards the tower, the angle of elevation becomes 45°. The height of the tower is:
(a) 30(√3+1) m  (b) 60 m  (c) 30(3+√3) m  (d) 30(√3−1) m

Accepted Answer

Let x be the distance from the nearer point to the foot. h=x·tan45°=x, and h=(x+60)·tan30°=(x+60)/√3. Setting equal: x√3=x+60 ⇒ x(√3−1)=60 ⇒ x=60(√3+1)/[(√3−1)(√3+1)] = 60(√3+1)/2 = 30(√3+1) ≈ 81.96 m.

Answer: (a) 30(√3+1) m. Trap (b) 60 m uses tan30°=1/2, confusing it with sin30°=1/2. Trap (c) 30(3+√3) m ≈141.96 m comes from using h=x/(tanβ−tanα), omitting the tanα·tanβ product that belongs in the numerator of the two-position formula. Trap (d) 30(√3−1) m ≈21.96 m rationalises 60/(√3−1) with the wrong sign in the numerator, i.e. multiplying by (√3−1) instead of the conjugate (√3+1).

Practice — Trigonometry — Identities, Equations & Triangles

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