NDA Trigonometric Ratios and Identities previous year questions with solutions

Answer: (c) 0

Method. reduce each angle using the period 180° of tan and cot

tan 1125° = tan(1125° - 1080°) = tan 45° = 1.
cot 405° = cot(405° - 360°) = cot 45° = 1.
tan 765° = tan(765° - 720°) = tan 45° = 1.
cot 675° = cot(675° - 720°) = cot(-45°) = -1.
Value = 1 · 1 + 1 · (-1) = 0.

Answer: (b) x − y + xy + 1 = 0

Method. write x and y in terms of t = tan(θ2)

y = 1 + cos θsin θ = 2cos²(θ/⁠2)2sin(θ/⁠2)cos(θ/⁠2) = cot(θ2) = 1t.
x = 1 − sin θcos θ = (1 + t²) − 2t1 − t² = (1 − t)²(1 − t)(1 + t) = 1 − t1 + t.
Put t = 1y: x = y − 1y + 1.
xy + x = y − 1 ⇒ x − y + xy + 1 = 0.

Answer: (b) 0

Method. evaluate the greatest-integer coefficients first

π² ≈ 9.87, so [π²] = 9 and [-π²] = -10.
f(x) = sin 9x + cos(-10x) = sin 9x + cos 10x.
f(π2) = sin 9π2 + cos 5π = 1 + (-1) = 0.

Answer: (d) 1⁠√2⁠

Method. f(x) = sin 9x + cos 10x from the previous item

f(π4) = sin 9π4 + cos 10π4.
sin 9π4 = sin(2π + π4) = 1⁠√2;⁠ cos 5π2 = cos(2π + π2) = 0.
f(π4) = 1⁠√2.⁠

Answer: (c) 2 + ⁠√3⁠ cm

Method. inradius of a regular n-gon r = a2 cot πn

Each side subtends 2π12 = π6 at the centre; the apothem bisects it, so r = a2 cot π12.
cot 15° = 2 + ⁠√3,⁠ a = 1 cm: r = 2 + ⁠√32⁠ cm.
Diameter = 2r = 2 + ⁠√3⁠ cm.

Answer: (b) -2⁠√3⁠

Method. odd function and period 2π

cosec(-x) = -cosec x, so the value is -cosec 73π3.
73π3 = 24π + π3, and 24π is a multiple of 2π, so cosec 73π3 = cosec π3 = 2⁠√3.⁠
Value = -2⁠√3.⁠

Answer: (b) -343600

Method. fourth-quadrant signs from a 7-24-25 triangle

cos x = 2425; sin²x = 1 − 576625 = 49625.
In the fourth quadrant sin x < 0: sin x = -725, tan x = -724.
tan x + sin x = -724 − 725 = -7(25 + 24)600 = -343600.

Answer: (a) -14

Method. periodicity 2π and sin(-x) = -sin x

sin(2nπ + 5π6) = sin 5π6 = 12.
sin(2nπ − 5π6) = -sin 5π6 = -12.
Product = -14.

Answer: (c) 513 or -513

Method. sign of sin θ in the quadrants where tan θ < 0

tan θ < 0, so θ lies in the second or the fourth quadrant.
With |tan θ| = 512, the reference triangle has sides 5, 12, 13, so |sin θ| = 513.
Second quadrant: sin θ = 513; fourth quadrant: sin θ = -513. Both are possible.

Answer: (b) 34

Method. reduce to cos⁴x + sin⁴x = 1 − 12sin²2x

cos 7π8 = -cos π8 and cos 5π8 = -sin π8.
So the sum = cos⁴π8 + sin⁴π8 = 1 − 2sin²π8cos²π8 = 1 − 12sin²π4.
= 1 − 12 · 12 = 34.

Answer: (b) 0

Method. pair terms using sin(360° - x) = -sin x

The terms are sin k·10° for k = 0, 1, …, 36.
sin 0° = sin 180° = sin 360° = 0.
Every other term sin x (0° < x < 180°) pairs with sin(360° - x) = -sin x, and the pair sums to 0.
Hence the whole sum is 0.

Answer: (d) -1225

Method. signs of sin and cos in the second quadrant

tan x = -34 with x in the second quadrant: sin x > 0, cos x < 0, and the reference triangle is 3-4-5.
sin x = 35, cos x = -45.
sin x · cos x = 35 × (-45) = -1225.

Answer: (c) -4

Method. reduce 210° and 300° to reference angles

7π6 = 210° = 180° + 30°: sin 210° = -12, so cosec 7π6 = -2.
5π3 = 300° = 360° - 60°: cos 300° = 12, so sec 5π3 = 2.
Product = (-2)(2) = -4.

Answer: (c) 1

Method. pair complementary angles: tan θ tan (90° - θ) = 1

The angles are 31°, 33°, …, 59° (15 terms); pair 31° with 59°, 33° with 57°, …, 43° with 47°, leaving 45° in the middle.
Each pair gives tan θ cot θ = 1, and tan 45° = 1.
So the product is 1.

Answer: (b) 0

Method. cos 66° = sin 24° (complementary angles)

cos 66° = cos (90° - 24°) = sin 24°.
So the second factor sin 24° - cos 66° = 0, and the product is 0.

Answer: (b) 2

Method. write in sin and cos, then (s + c)² - 1 = 2sc

Let s = sin θ, c = cos θ.
1 + cot θ − cosec θ = s + c − 1s; 1 + tan θ + sec θ = c + s + 1c.
Product = (s + c)² - 1sc = 1 + 2sc − 1sc = 2.

Answer: (a) 0

Method. sec²/⁠cosec² = tan², and simplify the fraction with t = tan θ

1 + tan² θ1 + cot² θ = sec² θcosec² θ = tan² θ.
With t = tan θ: 1 − t1 − 1t = t(1 − t)t − 1 = -t, so its square is t² = tan² θ.
Difference: tan² θ − tan² θ = 0.

Answer: (b) 3π4

Method. interior angle of a regular n-gon = (n − 2)πn

For n = 8: interior angle = (8 − 2)π8 = 6π8 = 3π4 (135°). The side length plays no part.

Answer: (a) c − ba − c

Method. divide by cos² x and use sec² x = 1 + tan² x

Divide a sin² x + b cos² x = c by cos² x: a tan² x + b = c sec² x = c(1 + tan² x).
So tan² x (a − c) = c − b, i.e. tan² x = c − ba − c.

Answer: (c) tan² y

Method. divide by cos² y and use sec² y = 1 + tan² y

Divide b sin² y + a cos² y = d by cos² y: b tan² y + a = d(1 + tan² y).
So tan² y (b − d) = d − a, i.e. d − ab − d = tan² y.

Answer: (b) (a − d)(c − a)(b − c)(d − b)

Method. p/⁠q = tan y / tan x, then substitute the two previous results

p tan x = q tan y gives p²q² = tan² ytan² x.
From the two previous items: tan² y = d − ab − d, tan² x = c − ba − c.
So p²q² = (d − a)(a − c)(b − d)(c − b).
Rewrite (d − a)(a − c) = (a − d)(c − a) and (b − d)(c − b) = (b − c)(d − b) (two sign changes each):
p²q² = (a − d)(c − a)(b − c)(d − b), which is option (b).

Answer: (a) t₁t₃

Method. factor sin² θ cos² θ out of each difference

Write s = sin θ, c = cos θ.
t₃ - t₅ = s³(1 − s²) + c³(1 − c²) = s³c² + c³s² = s²c²(s + c) = s²c² t₁.
t₅ - t₇ = s⁵(1 − s²) + c⁵(1 − c²) = s⁵c² + c⁵s² = s²c²(s³ + c³) = s²c² t₃.
So t₃ - t₅t₅ - t₇ = t₁t₃.

Answer: (b) sin 2θ

Method. expand (sin θ + cos θ)²

t₁² = (sin θ + cos θ)² = sin² θ + cos² θ + 2 sin θ cos θ = 1 + sin 2θ.
t₂ = sin² θ + cos² θ = 1.
So t₁² - t₂ = sin 2θ.

Answer: (c) 116

Method. substitute sin 45° = cos 45° = 1⁠√2⁠

t₁₀ = (1⁠√2)¹⁰⁠ + (1⁠√2)¹⁰⁠ = 2 · 12⁵ = 232 = 116.

Answer: (b) 10

Method. right triangle with hypotenuse 29, opposite side 21

sin θ = 2129, so the adjacent side is ⁠√(29²⁠ - 21²) = ⁠√(841⁠ −⁠ 441) = 20, and cos θ = 2029 (θ acute).
sec θ = 2920, tan θ = 2120.
4 sec θ + 4 tan θ = 4 × 29 + 2120 = 4 × 5020 = 10.

Answer: (c) 2sin(θ2)

Method. isosceles triangle with two radii: chord = 2r sin(θ2)

The two radii (length 1) and the chord form an isosceles triangle with apex angle θ.
The perpendicular from the centre bisects the chord and the angle, so half the chord = 1 × sin(θ2).
Chord = 2 sin(θ2).

Answer: (b) 0

Method. a zero factor: cos 90° = 0

The angles run from 46° to 135° in steps of 1°, so 90° is among them.
cos 90° = 0, so the whole product is 0.

Answer: (a) 0

Method. expand: the first two terms give (1 + tan² α)(1 + tan² β)

(1 + tan α tan β)² + (tan α − tan β)² = 1 + tan² α tan² β + tan² α + tan² β (the 2 tan α tan β terms cancel).
This is (1 + tan² α)(1 + tan² β) = sec² α sec² β.
Expression = sec² α sec² β − sec² α sec² β = 0.

Answer: (b) p = q

Method. cosec² θ − cot² θ = 1

(cosec θ − cot θ)(cosec θ + cot θ) = cosec² θ − cot² θ = 1.
So q = 1cosec θ + cot θ = cosec θ − cot θ = p.
Hence p = q.

Answer: (c) ⁠√2⁠ sin θ

Method. solve for sin θ in terms of cos θ, then rationalise

sin θ = (⁠√2⁠ − 1) cos θ, so cos θ = sin θ⁠√2⁠ − 1 = (⁠√2⁠ + 1) sin θ.
cos θ − sin θ = (⁠√2⁠ + 1) sin θ − sin θ = ⁠√2⁠ sin θ.

Answer: (a) 48421 cm

Method. chord equal to the radius subtends 60°; arc = rθ

Radius r = 22 cm and the chord is 22 cm, so the chord and the two radii form an equilateral triangle.
The chord subtends 60° = π3 at the centre.
Minor arc = rθ = 22 × π3 = 223 × 227 = 48421 cm.

Answer: (c) Third

Method. signs of sin and tan in the four quadrants

sin θ < 0: θ is in the third or fourth quadrant.
tan θ > 0: θ is in the first or third quadrant.
Both hold only in the third quadrant (e.g. θ = 210°: sin 210° = -12, tan 210° = 1⁠√3).⁠

Answer: (a) -43

Method. product of roots of 4x² - 3 = 0

cos α cos β = product of roots = -34 (the roots are ±⁠√32,⁠ i.e. α = π6, β = 5π6).
sec α sec β = 1cos α cos β = -43.

Answer: (b) 1 and 2

Method. B + C = π − A, so (B + C)/⁠2 = π/⁠2 − A/⁠2

(1) sin(π2 − A2) = cos A2: correct.
(2) tan(π2 − A2) = cot A2: correct.
(3) sin(π − A) = sin A, not cos A: wrong.
(4) tan(π − A) = -tan A, not -cot A: wrong.
So 1 and 2.

Answer: (b) 12 only

Method. let t = sin θ − cos θ, so sin θ cos θ = (1 − t²)/⁠2; reject values with |t| > ⁠√2⁠

sec θ − cosec θ = sin θ − cos θsin θ cos θ = 43.
With t = sin θ − cos θ, (sin θ − cos θ)² = 1 − 2 sin θ cos θ gives sin θ cos θ = 1 − t²2.
2t1 − t² = 43 ⇒ 6t = 4 − 4t² ⇒ 2t² + 3t − 2 = 0 ⇒ (2t − 1)(t + 2) = 0.
t = -2 is impossible, since sin θ − cos θ = ⁠√2⁠ sin(θ − π4) lies in [-⁠√2,⁠ ⁠√2].⁠
t = 12 gives sin θ cos θ = 38, which is attainable. So 12 only.

Answer: (b) p + q1 − pq

Method. A = (A2 + B2) + (A2 − B2) and tan of a sum

A = (A2 + B2) + (A2 − B2).
tan A = p + q1 − pq (defined since pq ≠ 1).

Answer: (a) p − q1 + pq

Method. B = (A2 + B2) − (A2 − B2) and tan of a difference

B = (A2 + B2) − (A2 − B2).
tan B = p − q1 + pq (defined since pq ≠ -1).

Answer: (b) 2

Method. tan(A + B) = 1 when A + B = π4

A + B = π4 ⇒ tan A + tan B1 − tan A tan B = 1 ⇒ tan A + tan B + tan A tan B = 1.
(1 + tan A)(1 + tan B) = 1 + (tan A + tan B + tan A tan B) = 1 + 1 = 2.

Answer: (d) 2

Method. cot(A + B) = 1 when A + B = π4

cot(A + B) = cot A cot B − 1cot A + cot B = 1 ⇒ cot A cot B − cot A − cot B = 1.
(cot A − 1)(cot B − 1) = (cot A cot B − cot A − cot B) + 1 = 1 + 1 = 2.

Answer: (a) 2

Method. ratio of tangents as sin(α + β)/⁠sin(α − β)

From m tan β = n tan α: mn = tan αtan β.
m + nm − n = tan α + tan βtan α − tan β = sin α cos β + cos α sin βsin α cos β − cos α sin β = sin(α + β)sin(α − β).
α + β = π2 and α − β = π6, so the value is 11/⁠2 = 2.

Shortcut. α = π3, β = π6: m/⁠√3⁠ = n⁠√3⁠ ⇒ m = 3n ⇒ 4n2n = 2.

Answer: (c) 1

Method. ⁠√3⁠ cos θ − sin θ = 2 cos(θ + 30°); sin 2θ

⁠√3⁠ p − q4 = 14(⁠√3sin⁠ 20° − 1cos 20°) = ⁠√3⁠ cos 20° - sin 20°4 sin 20° cos 20°.
Numerator = 2 cos 50°; denominator = 2 sin 40° = 2 cos 50°.
Value = 1.

Answer: (b) 1

Method. 1p² + 1q² with complementary angles

p² + q²p² q² = 1q² + 1p² = sin² 70° + sin² 20° = cos² 20° + sin² 20° = 1.

Answer: (d) cot 62°

Method. 1 − tan A1 + tan A = tan(45° - A)

Divide numerator and denominator by cos 17°: 1 − tan 17°1 + tan 17° = tan(45° - 17°) = tan 28°.
tan 28° = cot 62°.

Answer: (d) 2 cos 2θ

Method. componendo-dividendo; tan A + tan Btan A − tan B = sin(A + B)sin(A − B)

pq = tan(θ + 120°)tan(θ − 30°). With A = θ + 120°, B = θ − 30°:
p + qp − q = tan A + tan Btan A − tan B = sin A cos B + cos A sin Bsin A cos B − cos A sin B = sin(A + B)sin(A − B).
A + B = 2θ + 90°, A − B = 150°: sin(2θ + 90°)sin 150° = cos 2θ12 = 2 cos 2θ.

Answer: (d) 4

Method. a cos θ − b sin θ as R cos(θ + φ); double angle

⁠√3⁠ cos 10° - sin 10° = 2(cos 30° cos 10° - sin 30° sin 10°) = 2 cos 40°.
sin 25° cos 25° = 12 sin 50° = 12 cos 40°.
Ratio = 2 cos 40°12 cos 40° = 4.

Answer: (b) 12

Method. tan(α + β) = 1 gives (1 + tan α)(1 + tan β) = 2

tan(α + β) = tan 5π4 = 1 ⟹ tan α + tan β = 1 − tan α tan β.
(1 + tan α)(1 + tan β) = 1 + (tan α + tan β) + tan α tan β = 1 + 1 = 2.
f(α) f(β) = 1(1 + tan α)(1 + tan β) = 12.

Answer: (b) 1385

Method. sum and product of roots; cos 2θ = 1 − tan² θ1 + tan² θ

tan α + tan β = 6, tan α tan β = 8.
tan(α + β) = 61 − 8 = -67.
cos 2(α + β) = 1 − 36491 + 3649 = 1385.

Answer: (c) 2

Method. tan(A − B) with tan 65° tan 25° = 1

40° = 65° - 25°, and tan 65° tan 25° = tan 65° cot 65° = 1.
tan 40° = tan 65° - tan 25°1 + tan 65° tan 25° = tan 65° - tan 25°2 ⟹ tan 65° - tan 25° = 2 tan 40°.
Expression = (tan 65° - tan 25° - 2 tan 40°) + 2 tan 45° = 0 + 2 = 2.

Answer: (c) 1⁠√2⁠

Method. tan(α + 2β) from tan α and tan 2β, then fix the quadrant

sin β = 1⁠√10,⁠ β acute: cos β = 3⁠√10,⁠ tan β = 13.
tan 2β = 2/⁠31 − 1/⁠9 = 34.
tan(α + 2β) = 1/⁠7 + 3/⁠41 − 3/⁠28 = 25/⁠2825/⁠28 = 1.
tan α < 1 and tan β < 1 put α, β in (0, π4), so 0 < α + 2β < 3π4; hence α + 2β = π4.
cos(α + 2β) = 1⁠√2.⁠

Answer: (b) 14

Method. tan 165° = -tan 15° = -(2 - ⁠√3)⁠

tan 165° = tan(180° - 15°) = -tan 15° = -(2 - ⁠√3),⁠ so cot 165° = -(2 + ⁠√3).⁠
tan²165° + cot²165° = (2 - ⁠√3)²⁠ + (2 + ⁠√3)²⁠ = 2(4 + 3) = 14.

Answer: (b) f(α + β)

Method. f(α) = tan α, then tan(α + β)

⁠√(sec²α⁠ −⁠ 1) = ⁠√(tan²α)⁠ = |tan α|; with α, β (and α + β) acute as the question intends, f(α) = tan α.
f(α) + f(β)1 − f(α)f(β) = tan α + tan β1 − tan α tan β = tan(α + β) = f(α + β).

Answer: (b) 14

Method. exact values tan 15° = 2 - ⁠√3,⁠ cot 15° = 2 + ⁠√3⁠

tan 15° = tan(45° - 30°) = 1 − 1/⁠√31⁠ + 1/⁠√3⁠ = ⁠√3⁠ − 1⁠√3⁠ + 1 = 2 - ⁠√3,⁠ so cot 15° = 2 + ⁠√3.⁠
cot²15° + tan²15° = (2 + ⁠√3)²⁠ + (2 - ⁠√3)²⁠ = 2(4 + 3) = 14.

Shortcut. (cot + tan)² - 2 = (2/⁠sin 30°)² - 2 = 16 − 2 = 14.

Answer: (c) 34

Method. tan of a difference, then double-angle formula for tan

tan α = 12 and β = π4 − α.
tan β = 1 − tan α1 + tan α = 1/⁠23/⁠2 = 13.
tan 2β = 2 tan β1 − tan²β = 2/⁠38/⁠9 = 34.

Answer: (a) sin2θ

Method. sin²A − sin²B = sin(A + B) sin(A − B)

With A = π4 + θ, B = π4 − θ: A + B = π2, A − B = 2θ.
sin²A − sin²B = sin(A + B) sin(A − B) = sin π2 · sin 2θ = sin 2θ.

Answer: (b) π4

Method. compound-angle formula for tan(α + β)

Since tan α tan β ≠ 1, divide by 1 − tan α tan β: tan α + tan β1 − tan α tan β = 1.
The left side is tan(α + β), so tan(α + β) = 1 and α + β = nπ + π4.
Among the options, α + β = π4.

Answer: (b) 1

Method. expand and recognise tan(θ + 9θ)

Let a = tan θ, b = tan 9θ. (1 + a)(1 + b) = 2 gives a + b + ab = 1, i.e. a + b = 1 − ab.
ab ≠ 1 (else a + b = 0 and ab = 1, impossible for real a, b, since then a² = -1).
tan 10θ = a + b1 − ab = 1.

Answer: (d) 3 : 1

Method. solve for A + B and A − B, then compare tangents

sin (A + B) = 1 with 0 < A + B < π gives A + B = π2.
sin (A − B) = 12 with -π2 < A − B < π2 gives A − B = π6.
So A = π3, B = π6: tan A = ⁠√3,⁠ tan B = 1⁠√3.⁠
tan A: tan B = ⁠√3⁠ : 1⁠√3⁠ = 3 : 1.

Answer: (c) 1

Method. cot (A + B) formula with 6x = 2x + 4x

cot 6x = cot (2x + 4x) = cot 2x cot 4x − 1cot 2x + cot 4x.
Cross-multiply: cot 6x cot 2x + cot 6x cot 4x = cot 2x cot 4x − 1.
Rearrange: cot 2x cot 4x − cot 4x cot 6x − cot 6x cot 2x = 1.

Answer: (b) 7

Method. tan (45° + θ) = 1 + tan θ1 − tan θ

3 cos θ = 4 sin θ gives tan θ = 34.
tan (45° + θ) = 1 + 341 − 34 = 7/⁠41/⁠4 = 7.

Answer: (b) 28°

Method. tan(45° − A) in the form (1 − tan A)/⁠(1 + tan A)

Divide numerator and denominator by cos 17°: tan θ = 1 − tan 17°1 + tan 17°.
Since tan 45° = 1, this is tan 45° - tan 17°1 + tan 45° tan 17° = tan (45° - 17°) = tan 28°.
So θ = 28°.

Answer: (d) π2

Method. show cos (A + 2B) = 0 using the two given relations

From 3 sin 2A = 2 sin 2B: sin 2B = 32 sin 2A = 3 sin A cos A.
cos (A + 2B) = cos A cos 2B − sin A sin 2B
= cos A · 3 sin² A − sin A · 3 sin A cos A = 0.
A and B are acute, so 0 < A + 2B < 3π2; the only angle there with cosine 0 is π2.
So A + 2B = π2.

Answer: (d) ⁠√3√2⁠

Method. square the sum: (sin 15° + cos 15°)² = 1 + sin 30°

sin α + cos β = sin 15° + cos 15°.
(sin 15° + cos 15°)² = 1 + 2 sin 15° cos 15° = 1 + sin 30° = 32.
The sum is positive, so it equals ⁠√(32)⁠ = ⁠√3√2.⁠

Shortcut. sin 15° = ⁠√3⁠ − 12⁠√2,⁠ cos 15° = ⁠√3⁠ + 12⁠√2;⁠ their sum is 2⁠√32√2⁠ = ⁠√3√2.⁠

Answer: (d) ⁠√3√2⁠

Method. reduce 105° to 15° with sin (90° + A) = cos A, cos (90° + A) = -sin A

7α = 7β = 105°.
sin 105° = sin (90° + 15°) = cos 15°; cos 105° = cos (90° + 15°) = -sin 15°.
So sin 7α − cos 7β = cos 15° + sin 15° = ⁠√3√2⁠ (as in the previous item).

No correct option. This question is defective as printed in UPSC's paper.

Method. expand sin 16° and cos 16° as compound angles 15° + 1°

sin 16° + cos 16° = (sin 15° cos 1° + cos 15° sin 1°) + (cos 15° cos 1° - sin 15° sin 1°)
= (sin 15° + cos 15°) cos 1° + (cos 15° - sin 15°) sin 1°.
sin 15° + cos 15° = ⁠√3√2⁠ and cos 15° - sin 15° = 1⁠√2⁠ (since (cos 15° - sin 15°)² = 1 − sin 30° = 12).
So the value is 1⁠√2(√3⁠ cos 1° + sin 1°) ≈ 1.2369.
The options evaluate to (a) 1.7492, (b) 1.7231, (c) 1.2122, (d) 0.8746; none equals it.

The correct value is 1⁠√2(√3⁠ cos 1° + sin 1°), but option (c) prints a minus sign, 1⁠√2(√3⁠ cos 1° - sin 1°), so no option is right; with the sign read as + the intended answer is (c).

Answer: (b) 13

Method. cot (α + β) = cot α cot β − 1cot α + cot β with sum and product of roots

Roots of x² - 3x + 2 = 0: cot α + cot β = 3, cot α cot β = 2.
cot (α + β) = cot α cot β − 1cot α + cot β = 2 − 13 = 13.

Answer: (a) 1x + 1y

Method. express tan A tan B through x and y, then tan(A − B)

y = 1tan B − 1tan A = tan A − tan Btan A tan B = xtan A tan B, so tan A tan B = xy.
tan(A − B) = tan A − tan B1 + tan A tan B = x1 + x/⁠y = xyx + y.
cot(A − B) = x + yxy = 1x + 1y.

Answer: (a) 0

Method. sin(α + β) + sin(α − β) = 2 sin α cos β

sin(α + β) + sin(α − β) = (sin α cos β + cos α sin β) + (sin α cos β − cos α sin β) = 2 sin α cos β.
Expression = 2 sin α cos β − 2 sin α cos β = 0.

Answer: (c) 17

Method. compound-angle formula for tan(A − B)

tan A = 12, tan B = 13.
tan(A − B) = 1/⁠2 − 1/⁠31 + 1/⁠6 = 1/⁠67/⁠6 = 17.

Answer: (b) 1

Method. if A + B + C = 90°, then tan A tan B + tan B tan C + tan C tan A = 1

25° + 15° + 50° = 90°, so 25° + 15° = 90° - 50° and tan(25° + 15°) = cot 50°.
tan 25° + tan 15°1 − tan 25° tan 15° = 1tan 50°.
tan 50°(tan 25° + tan 15°) = 1 − tan 25° tan 15°.
tan 25° tan 15° + tan 15° tan 50° + tan 25° tan 50° = 1.

Answer: (a) -2

Method. reduce every angle to an acute one, then compound-angle formulas

cos 236° = cos(180° + 56°) = -cos 56°, sin 124° = sin(180° - 56°) = sin 56°, and sin 34° = cos 56°.
Numerator = -cos 56° cos 56° - sin 56° sin 56° = -(cos² 56° + sin² 56°) = -1.
cos 178° = -cos 2°, sin 208° = -sin 28°, and cos 88° = sin 2°.
Denominator = cos 28° sin 2° + cos 2° sin 28° = sin 30° = 12.
Value = -11/⁠2 = -2.

Answer: (c) cos 9° + sin 9°cos 9° - sin 9°

Method. tan(45° + A) = 1 + tan A1 − tan A

tan 54° = tan(45° + 9°) = 1 + tan 9°1 − tan 9°.
Multiply numerator and denominator by cos 9°: tan 54° = cos 9° + sin 9°cos 9° - sin 9°.
Option (a) is the negative of this, (b) is -tan 36°, and (d) is tan 36°; only (c) equals tan 54°.

Answer: (a) ab + ⁠√(1⁠ −⁠ a²) ⁠√(1⁠ −⁠ b²)

Method. α − β = (θ − β) − (θ − α), then cos(A − B)

α − β = (θ − β) − (θ − α).
cos(α − β) = cos(θ − β) cos(θ − α) + sin(θ − β) sin(θ − α) = ab + sin(θ − α) sin(θ − β).
Taking the sines as the non-negative roots, sin(θ − α) = ⁠√(1⁠ −⁠ a²) and sin(θ − β) = ⁠√(1⁠ −⁠ b²).
cos(α − β) = ab + ⁠√(1⁠ −⁠ a²) ⁠√(1⁠ −⁠ b²).

Answer: (a) a² + b²

Method. square the relation cos(α − β) − ab = sin(θ − α) sin(θ − β)

Let C = cos(α − β). From the previous item, C − ab = sin(θ − α) sin(θ − β).
Squaring: C² - 2abC + a²b² = (1 − a²)(1 − b²) = 1 − a² - b² + a²b².
So C² - 2abC = 1 − a² - b², i.e. (1 − C²) + 2abC = a² + b².
sin²(α − β) + 2ab cos(α − β) = a² + b² (whatever the signs of the sines).

Answer: (c) π4

Method. compound-angle formula for tan(A + B)

tan(θ + φ) = tan θ + tan φ1 − tan θ tan φ = 1/⁠2 + 1/⁠31 − 1/⁠6 = 5/⁠65/⁠6 = 1.
θ and φ are taken as the acute angles with these tangents, so 0 < θ + φ < π, and tan(θ + φ) = 1 gives θ + φ = π4.

Answer: (b) 4

Method. tan x + cot x = 2sin 2x

tan x + cot x = sin² x + cos² xsin x cos x = 1sin x cos x = 2sin 2x.
With x = 75°: tan 75° + cot 75° = 2sin 150° = 21/⁠2 = 4.

Answer: (a) q − p = 1

Method. 19° + 26° = 45°, with sum and product of roots

tan 19° + tan 26° = -p and tan 19° tan 26° = q.
tan 45° = tan 19° + tan 26°1 − tan 19° tan 26°, so 1 = -p1 − q.
1 − q = -p, i.e. q − p = 1.

Answer: (c) π/⁠4

Method. convert to tangents, then tan(x + y)

sin x = 1⁠√5,⁠ x acute ⇒ cos x = 2⁠√5,⁠ tan x = 12.
sin y = 1⁠√10,⁠ y acute ⇒ cos y = 3⁠√10,⁠ tan y = 13.
tan(x + y) = 1/⁠2 + 1/⁠31 − 1/⁠6 = 5/⁠65/⁠6 = 1.
0 < x + y < π, so x + y = π4.

Answer: (c) 1/⁠√2⁠

Method. sin θ + cos θ = ⁠√2⁠ sin(θ + 45°)

sin 105° + cos 105° = ⁠√2⁠ sin(105° + 45°) = ⁠√2⁠ sin 150° = ⁠√2⁠ × 12 = 1⁠√2.⁠

No correct option. This question is defective as printed in UPSC's paper.

Method. angle sum of a triangle; GP condition tan² x = tan(x − y) tan(x + y)

Angle sum: x + (x − y) + (x + y) = π ⇒ x = π3.
GP condition: tan² x = tan(x − y) tan(x + y) = tan² x − tan² y1 − tan² x tan² y.
With tan² x = 3: 3(1 − 3 tan² y) = 3 − tan² y ⇒ 8 tan² y = 0 ⇒ y = 0.
So the GP condition forces all three angles to be π3, the equilateral triangle the question excludes.

Defective as printed: the angle sum gives x = π/⁠3, and the GP condition then forces y = 0, which makes the triangle equilateral, the case the question excludes. The angle sum alone gives x = π/⁠3, option (b), while the GP condition alone gives tan² x = 1, x = π/⁠4, option (a); the two cannot both hold, so no option is defensible.

Answer: (a) ab

Method. componendo-dividendo, then sum and difference of sines

sin(x + y) + sin(x − y)sin(x + y) − sin(x − y) = (a + b) + (a − b)(a + b) − (a − b) = ab.
Left side = 2 sin x cos y2 cos x sin y = tan xtan y.
So tan xtan y = ab.

Answer: (c) c = a + b

Method. P/⁠2 + Q/⁠2 = π/⁠4, so tan(P/⁠2 + Q/⁠2) = 1; use sum and product of roots

∠R = π2 ⇒ P + Q = π2 ⇒ P2 + Q2 = π4.
Sum of roots = -ba, product = ca.
tan(P2 + Q2) = -b/⁠a1 − c/⁠a = -ba − c = 1.
-b = a − c ⇒ c = a + b.

Answer: (a) 4

Method. combine into one fraction; numerator becomes 2 sin(60° - 20°)

⁠√3⁠ cosec 20° - sec 20° = ⁠√3⁠ cos 20° - sin 20°sin 20° cos 20°.
Numerator = 2(sin 60° cos 20° - cos 60° sin 20°) = 2 sin 40°.
Denominator = 12 sin 40°.
Value = 4.

Answer: (d) (p − q) sin αp + q

Method. componendo-dividendo on sin A cos B: cos A sin B

A + B = α, A − B = x.
tan Atan B = sin A cos Bcos A sin B = pq.
Componendo-dividendo: sin A cos B − cos A sin Bsin A cos B + cos A sin B = p − qp + q, i.e. sin(A − B)sin(A + B) = p − qp + q.
sin x = (p − q) sin αp + q.

Answer: (b) 1 − cb

Method. sum and product of roots with the compound-angle formula for cot(α + β)

cot α + cot β = -b and cot α cot β = c.
cot(α + β) = cot α cot β − 1cot α + cot β = c − 1-b = 1 − cb.

Answer: (d) 4

Method. combine into one fraction and write the numerator as 2 sin(30° - θ) form

1sin 10° − ⁠√3cos⁠ 10° = cos 10° - ⁠√3⁠ sin 10°sin 10° cos 10°.
Numerator = 2(12 cos 10° - ⁠√32⁠ sin 10°) = 2 sin(30° - 10°) = 2 sin 20°.
Denominator = 12 sin 20°.
Value = 2 sin 20°12 sin 20° = 4.

Answer: (b) 0

Method. write θ = (θ + α) − α and θ + 2α = (θ + α) + α, then expand

Let u = θ + α. Then sin(u − α) = 3 sin(u + α).
sin u cos α − cos u sin α = 3 sin u cos α + 3 cos u sin α
⇒ -2 sin u cos α = 4 cos u sin α.
Divide by cos u cos α: tan u = -2 tan α.
So tan(θ + α) + 2 tan α = 0.

Answer: (a) -3

Method. 2α = (α + β) + (α − β), then tan(A + B)

tan 2α = tan[(α + β) + (α − β)] = 2 + 11 − 2 × 1 = 3-1 = -3.

Answer: (c) 1

Method. triple-angle formula sin 3θ = 3sin θ − 4sin³ θ

6sin(π18) − 8sin³(π18) = 2[3sin(π18) − 4sin³(π18)]
= 2sin(3π18) = 2sin(π6) = 2 · 12 = 1.

Answer: (a) 59

Method. product-to-sum, then cos 2θ = 2cos² θ − 1

sin(θ2) sin(3θ2) = 12[cos θ − cos 2θ].
cos 2θ = 2(19) − 1 = -79.
Value = 12[13 + 79] = 12 · 109 = 59.

Answer: (b) -1725

Method. fix sin θ and cos θ by quadrant, then double-angle formulas

cot θ = -43; in the fourth quadrant cos θ > 0, sin θ < 0.
So cos θ = 45, sin θ = -35.
sin 2θ = 2(-35)(45) = -2425; cos 2θ = 1625 − 925 = 725.
sin 2θ + cos 2θ = -1725.

Answer: (c) 2cos 3θ

Method. cube of x + 1x and the triple-angle formula for cos 3θ

x³ + 1x³ = (x + 1x)³ - 3(x + 1x)
= 8cos³ θ − 6cos θ
= 2(4cos³ θ − 3cos θ)
= 2cos 3θ.

Shortcut. Take x = cos θ + i sin θ; then x³ + 1x³ = 2cos 3θ by De Moivre.

Answer: (c) tan α · tan 2α

Method. tan A − tan B = sin(A − B)cos A cos B, cot B − cot A = sin(A − B)sin A sin B

p = tan 2α − tan α = sin(2α − α)cos 2α cos α = sin αcos 2α cos α.
q = cot α − cot 2α = sin(2α − α)sin α sin 2α = sin αsin α sin 2α.
pq = sin α sin 2αcos α cos 2α = tan α · tan 2α.

Answer: (d) 2 cosec 4α

Method. combine the forms of p and q from the previous item

p + q = sin α [1cos α cos 2α + 1sin α sin 2α] = sin α · sin α sin 2α + cos α cos 2αsin α cos α sin 2α cos 2α.
Numerator: sin α sin 2α + cos α cos 2α = cos(2α − α) = cos α.
p + q = sin α cos αsin α cos α sin 2α cos 2α = 1sin 2α cos 2α = 2sin 4α = 2 cosec 4α.

Answer: (c) p/⁠(p + 2q)

Method. p/⁠q = tan α tan 2α; solve for tan² α

Let u = tan² α. tan α tan 2α = 2u1 − u, so pq = 2u1 − u.
p(1 − u) = 2qu ⟹ p = u(p + 2q) ⟹ tan² α = pp + 2q.

Answer: (c) 3/⁠4

Method. half-angle substitution t = tan(α/⁠2)

With t = tan α2: 4t1 + t² + 1 − t²1 + t² = 2 ⟹ 4t + 1 − t² = 2 + 2t² ⟹ 3t² - 4t + 1 = 0.
t = 1 (α = 90°, excluded) or t = 13.
tan α = 2t1 − t² = 2/⁠38/⁠9 = 34.

Shortcut. Check: sin α = 3/⁠5, cos α = 4/⁠5 gives 2(3/⁠5) + 4/⁠5 = 2.

Answer: (b) 11/⁠5

Method. double-angle values from tan α = 3/⁠4

tan α = 34, α acute: sin α = 35, cos α = 45.
sin 2α = 2425, cos 2α = 16 − 925 = 725.
2 sin 2α + cos 2α = 48 + 725 = 115.

Answer: (c) Only two

Method. tan 22.5° = ⁠√2⁠ − 1 from the half-angle formula

tan 22.5° = 1 − cos 45°sin 45° = ⁠√2⁠ − 1: irrational.
cot 22.5° = 1⁠√2⁠ − 1 = ⁠√2⁠ + 1: irrational.
tan 22.5° - cot 22.5° = (⁠√2⁠ − 1) − (⁠√2⁠ + 1) = -2: rational.
So only two are irrational.

Answer: (a) -⁠√(5⁠ - ⁠√5)⁠2

Method. square the expression; sin 18° = ⁠√5⁠ − 14

(sin 9° - cos 9°)² = 1 − 2 sin 9° cos 9° = 1 − sin 18° = 1 − ⁠√5⁠ − 14 = 5 - ⁠√54.⁠
sin 9° < cos 9° (since 9° < 45°), so the expression is negative.
sin 9° - cos 9° = -⁠√(5⁠ - ⁠√5)⁠2.

Answer: (d) -83

Method. triple-angle formula for tan 3x

Let t = tan² x. tan 3x = tan x (3 − t)1 − 3t, so tan 3xtan x = 3 − t1 − 3t.
p = 13 − 3 − t1 − 3t = (1 − 3t) − 3(3 − t)3(1 − 3t) = -83(1 − 3t).
q = 1 − 3t, so pq = -83.

Answer: (a) No value

Method. triple-angle formula for tan 3x; domain of p

From the previous item, wherever p is defined, p = -83(1 − 3 tan² x), so 1p = -3(1 − 3 tan² x)8.
This would vanish only if tan² x = 13, i.e. x = π6 or 5π6 in (0, π).
But at these x, 3x = π2 or 5π2, so tan 3x is undefined and p itself does not exist there.
Hence 1p is never zero: no value of x.

1/⁠p could vanish only where p is infinite. Simplifying p to −83(1 − 3 tan² x) suggests two such values, x = π/⁠6 and 5π/⁠6, option (c); but at both points tan 3x is undefined, so p itself does not exist there. Everywhere else p is finite and nonzero, so 1/⁠p is never zero.

Answer: (a) (cosα − sinα)²

Method. sin²β = sin α cos α, then cos 2β = 1 − 2sin²β

GM: sin β = ⁠√(sin⁠ α cos α), so sin²β = sin α cos α.
cos 2β = 1 − 2sin²β = 1 − 2sin α cos α = cos²α + sin²α − 2sin α cos α = (cos α − sin α)².

Answer: (b) 5 + sin2α3 − sin2α

Method. sec 2γ = 1 + tan²γ1 − tan²γ

AM: tan γ = sin α + cos α2, so tan²γ = 1 + sin 2α4.
sec 2γ = 1cos 2γ = 1 + tan²γ1 − tan²γ = 4 + 1 + sin 2α4 − 1 − sin 2α = 5 + sin 2α3 − sin 2α.

Answer: (a) -sin3θ

Method. triple-angle formula sin 3θ = 3sin θ − 4sin³θ

f(sin θ) = sin θ(4sin²θ − 3) = 4sin³θ − 3sin θ.
= -(3sin θ − 4sin³θ) = -sin 3θ.

Answer: (b) ⁠√2⁠ + 1

Method. half-angle formula tan x2 = 1 − cos xsin x

3π8 = 67.5° = 135°2; cos 135° = -1⁠√2,⁠ sin 135° = 1⁠√2.⁠
tan 67.5° = 1 − cos 135°sin 135° = 1 + 1/⁠√21/⁠√2⁠ = ⁠√2⁠ + 1.

Shortcut. tan 67.5° = cot 22.5° = 1⁠√2⁠ − 1 = ⁠√2⁠ + 1.

Answer: (c) 12

Method. exact values of cos 36° and cos 72°

cos 36° = ⁠√5⁠ + 14, cos 72° = sin 18° = ⁠√5⁠ − 14.
cos 36° - cos 72° = (⁠√5⁠ + 1) − (⁠√5⁠ − 1)4 = 24 = 12.

Answer: (c) 1

Method. write both sides in t = tan θ

Let t = tan θ; for -π4 < θ < π4, -1 < t < 1, so 1 + t ≠ 0.
tan(45° + θ) = 1 + t1 − t; 1 + sin 2θ = 1 + 2t1 + t² = (1 + t)²1 + t².
Equating and cancelling 1 + t: 1 + t² = (1 + t)(1 − t) = 1 − t², so t² = 0, t = 0.
Then θ = 0 and cos 2θ = 1.

Answer: (c) ⁠√5⁠

Method. sin 2θ = cos 3θ with θ acute gives 2θ + 3θ = 90°

sin 2θ = cos 3θ = sin(90° - 3θ). With θ acute the general solutions are θ = 18° + 72°k or θ = -90° + 360°k; the only acute one is θ = 18°.
1 + 4 sin 18° = 1 + 4 · ⁠√5⁠ − 14 = ⁠√5.⁠

Answer: (c) 0

Method. find x from sin x = ⁠√32,⁠ then evaluate tan 3x

sin² x = 34; on [0, π], sin x ≥ 0, so sin x = ⁠√32⁠ and x = π3 or 2π3.
Then 3x = π or 2π, and tan 3x = 0 in both cases.

Answer: (a) 2425

Method. cos 2A = 1 − tan² A1 + tan² A

tan² A = 149.
cos 2A = 1 − 1491 + 149 = 4850 = 2425.

Answer: (b) K − 33K − 1

Method. triple-angle formula for tan 3A, then solve for tan² A

Let t = tan A. tan 3A = 3t − t³1 − 3t², so K = tan 3Atan A = 3 − t²1 − 3t².
K − 3Kt² = 3 − t², so t²(1 − 3K) = 3 − K.
tan² A = 3 − K1 − 3K = K − 33K − 1.

Answer: (a) 13 and 3

Method. tan² A ≥ 0 forces K − 33K − 1 ≥ 0

From the previous item, tan² A = K − 33K − 1, which must be ≥ 0 for real tan A.
The fraction is ≥ 0 only when K > 3 or K < 13 (K = 3 would need tan A = 0, excluded); it is negative for 13 < K < 3.
So K cannot lie between 13 and 3.

Answer: (a) 0

Method. triple-angle formulas for sin 3x and cos 3x

sin 3x = 3 sin x − 4 sin³ x, so sin 3x + 4 sin³ x − 3 sin x = 0.
cos 3x = 4 cos³ x − 3 cos x, so cos 3x + 3 cos x − 4 cos³ x = 0.
The whole expression is the sum of these two groups: 0 + 0 = 0.

Answer: (d) 2cot A

Method. half-angle: cos² - sin² = cos A, sin cos = 12 sin A

cot(A2) − tan(A2) = cos²(A/⁠2) − sin²(A/⁠2)sin(A/⁠2) cos(A/⁠2) = cos A12 sin A = 2 cot A.

Answer: (b) cot(A2)

Method. 1 + cos A = 2cos²(A2), sin A = 2 sin(A2) cos(A2)

cot A + cosec A = cos A + 1sin A = 2cos²(A/⁠2)2 sin(A/⁠2) cos(A/⁠2) = cos(A/⁠2)sin(A/⁠2) = cot(A2).

Answer: (c) π4

Method. rotation of axes: the XY term must vanish

p² + q² = X² + Y² (cross terms cancel).
pq = (X cos θ − Y sin θ)(X sin θ + Y cos θ) = (X² - Y²) sin θ cos θ + XY(cos² θ − sin² θ) = 12(X² - Y²) sin 2θ + XY cos 2θ.
The right side AX² + BY² has no XY term, so cos 2θ = 0.
With 0 ≤ θ ≤ π2: 2θ = π2, θ = π4.

Answer: (b) 3

Method. rotation of axes with θ = π4

With θ = π4: sin 2θ = 1, cos 2θ = 0, so pq = 12(X² - Y²).
p² + 4pq + q² = (X² + Y²) + 2(X² - Y²) = 3X² - Y².
Comparing with AX² + BY²: A = 3.

Answer: (a) -1

Method. rotation of axes with θ = π4

With θ = π4: pq = 12(X² - Y²).
p² + 4pq + q² = (X² + Y²) + 2(X² - Y²) = 3X² - Y².
Comparing with AX² + BY²: B = -1.

Answer: (c) p²(2 − p²)

Method. square sin α + cos α to get sin 2α, then cos² 2α = 1 − sin² 2α

(sin α + cos α)² = 1 + sin 2α = p², so sin 2α = p² - 1.
cos² 2α = 1 − (p² - 1)² = 1 − p⁴ + 2p² - 1 = 2p² - p⁴ = p²(2 − p²).

Answer: (b) 516

Method. product-to-sum: 2 sin A sin B = cos(A − B) − cos(A + B)

sin(A2) sin(3A2) = 12[cos A − cos 2A].
cos 2A = 2cos² A − 1 = 2 × 916 − 1 = 18.
Value = 12(34 − 18) = 12 × 58 = 516.