Trigonometrical Ratios Class 9 OP Malhotra Exe-19D ICSE Maths Solutions Ch-19. We Provide Step by Step Solutions / Answer of OP Malhotra Maths on Problems on complementary angles. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics.

Trigonometrical Ratios Class 9 OP Malhotra Exe-19D ICSE Maths Solutions Ch-19

Board	ICSE
Subject	Maths
Class	9th
Chapter-19	Trigonometrical Ratios
Writer	OP Malhotra
Exe-19D	Problems on complementary angles
Edition	2025-2026

Problems on (90-θ) Complementary Angles

Trigonometrical Ratios Class 9 OP Malhotra Exe-19D ICSE Maths Solutions Ch-19.

Que-1: (i) sin16°/cos74°
(ii) cos25°/sin65°
(iii) tan38°/cot52°
(iv) sec62°/cosec28°

Sol: (i) sin16°/cos74° = sin16°/{cos(90°−16°)}
= sin16°/sin16° = 1 {∵ cos (90° – θ) = sin θ}

(ii) cos25°/sin65° = cos25°/{sin(90°−25°)}
= cos25°/cos25° = 1 {∵ sin (90° – θ) = cos θ}

(iii) tan38°/cot52° = tan38°/{cot(90°−38°)}
= tan38°/tan38° = 1 {∵ cot (90° – θ) = tan θ}

(iii) sec62°/cosec28° = sec62°/{cosec(90°−62°)}
= sec62°/sec62° = 1 {∵ cosec (90° – θ) = sec θ}

Que-2:  (i) sin² 67° + sin² 23°.
(ii) (sin49°/cos41°)² + (cos41°/sin49°)².
(iii) (cos²20°+cos²70°) / (sin²59°+sin²31°).
(iv) (cos70°/sin20°) + (cos59°/sin31°) – 8 sin² 30°
(v) 2 (tan53°/cot37°) – (cot80°/tan10°)
(vi) sec 50° sin 40° + cos 40° cosec 50°.
(vii) (cos80°/sin10°) + cos 59° cosec 31°.
(viii) (2sin43°/cos47°) – (cot30°/tan60°) – √2 sin 45°.
(ix) {(cos²20°+cos²70°)/(sin²20°+sin²70°)} + sin² 64° + cos 64° sin 26°.
(x) tan 35° tan 40° tan 45° tan 50° tan 55°.
(xi) sin²10° + sin2 80° + (sin15°cos75° + cos15°sin75°) / {cosθsin(90
−θ)+sinθcos(90°−θ)
(xii) (cos75°/sin15°) + (sin12°/cos78°) – (cos18°/sin72°)

Sol: (i) sin² 67° + sin² 23°
= sin² 67° + sin² (90° – 67°)
= sin² 67° + cos² 67°
= 1 {∵ sin²θ + cos²θ = 1}

(ii) (sin49°/cos41°)² + (cos41°/sin49°)²
= {sin49°/(cos(90°−49°))}² + [cos41°/{sin(90°−41°)}]²
= (sin49°/sin49°)² + (cos41°/cos41°)² = (1)² + (1)²
= 1 + 1 = 2

(iii) (cos²20°+cos²70°) / (sin²59°+sin²31°)
= [cos²(90°−70°)+cos²70°] / [sin²59°+sin²(90°−59°)²
= (sin²70°+cos²70°)/(sin²59°+cos²59°)
= 1/1 = 1

(iv) (cos70°/sin20°) + (cos59°/sin31°) – 8 sin² 30°
= [{cos(90°−20°)}/sin20°] + [{cos(90°−31°)}/sin31°] – 8 (sin 30°)²
= (sin20°/sin20°) + (sin31°/sin31°) – 8 (1/2)²
= 1 + 1 – 8 × (1/4)
= 2 – 2 = 0

(v) 2 (tan53°/cot37°) – (cot80°/tan10°)
= 2[{tan(90°−37°)}/cot37°] – [cot80°/{tan(90°−80°)}]
= 2 (cot37°/cot37°) – (cot80°/(cot80°) {∵ tan (90° – θ) = cot θ}
= 2 × 1 – 1 = 2 – 1 = 1

(vi) sec 50° sin 40° + cos 40° cosec 50°
= (sin40°/cos50°) + (cos40°/sin50)
= [sin40°/{cos(90°−40°)}] + [cos40°/{sin(90°−40°)}]
= (sin40°/sin40°) + (cos40°/cos40)
= 1 + 1 = 2

(vii) (cos80°/sin10°) + cos 59° cosec 31°.
= (cos80°)/(sin10°) + (cos59°)/(sin31°)
= [cos80°/{sin(90°−80°)}] + [cos59°/{sin(90°−59°)}]
= (cos80°/cos80°) + (cos59°/cos59°)
= 1 + 1 = 2

(viii) (2sin43°/cos47°) – (cot30°/tan60°) – √2 sin 45°

= 2 – 1 – 1 = 0

(ix) {(cos²20°+cos²70°)/(sin²20°+sin²70°)} + sin² 64° + cos 64° sin 26°.

= 1 + 1 = 2

(x) tan 35° tan 40° tan 45° tan 50° tan 55°
= tan 35° tan 40° tan 45° tan (90° – 40°) tan (90° – 35°)
= tan 35° tan 40° tan 45° cot 40° cot 35°
= tan 35° × cot 35° tan 40° × cot 40° × cot 40° × tan 45°
= 1 × 1 × 1 = 1

(xi) sin²10° + sin² 80° + (sin15°cos75° + cos15°sin75°) / {cosθsin(90°−θ)+sinθcos(90°−θ)

(xii) (cos75°/sin15°) + (sin12°/cos78°) – (cos18°/sin72°)
= [cos75°/{sin(90°−75°)}] + [sin12°/{cos(90°−12°)}] = [{cos(90°−72°)}/sin72°]
= (cos75°/cos75°) + (sin12°/sin12°) – (sin72°/sin72°)
= 1 + 1 – 1 = 1

Que-3: Express the following in terms of t-ratios of angles lying between 0° and 45°.
(i) cosec 69° + cot 69°
(ii) sin 85° + cosec 85°

Sol: Using sin (90° – θ), cosec (90° – θ) and cot (90° – θ) we get,
(i) cosec 69° + cot 69°
= cosec (90° – 21°) + cot (90° – 21°)
= sec 21° + tan 21°

(ii) sin 85° + cosec 85°
= sin (90° – 5°) + cosec (90° – 5°)
= cos 5° + sec 5°

Que-4: Prove that:
(i) {tan (90°-A)}/cosec A = cos A
(ii) [cosθ/{sin(90°−θ)}] + [sinθ/{cos(90°−θ)}] = 2
(iii) sin θ sin (90° – θ) – cos θ cos (90° – θ) = 0
(iv) sec (90° – θ) cosec (90° – θ) = sec2θ cot θ
(v) [{sin (90°-θ)}/{cosec (90°-θ)}] + [{cos (90°-θ)}/{sec (90°-θ)}] = 1
(vi) sin (60° – θ) = cos (30° – θ)
(vii) [sinθ/{sin(90°−θ)}] + [cosθ/{cos(90°−θ)}] = sec (90°–θ) cosec (90°–θ)
(viii) cos (81° + θ) = sin (9° – θ)

Sol: (i) {tan (90°-A)}/cosec A = cos A
LHS = {tan (90°-A)}/cosec A
= cot A / cosec A
= (cos A × sin A) / (sin A)
= cos A = RHS.

(ii) [cosθ/{sin(90°−θ)}] + [sinθ/{cos(90°−θ)}] = 2
LHS = [cosθ/{sin(90°−θ)}] + [sinθ/{cos(90°−θ)}]
= (cosθ/cosθ) + (sinθ/sinθ)
= 1 + 1 = 2

(iii) sin θ sin (90° – θ) – cos θ cos (90° – θ) = 0
L.H.S. = sin θ sin (90° – θ) – cos θ cos (90° – θ)
= sin θ cos θ – cos θ sin θ
= 0 = R.H.S.
(iv) sec (90° – θ) cosec (90° – θ) = sec2θ cot θ
L.H.S.
= sec (90° – θ) cosec (90° – θ)
= cosec θ sec θ
R.H.S. = sec² θ cot θ = sec² θ × (cosθ/sinθ)
= {secθ×secθ×cosθ}/sinθ = (secθ×1)/sinθ
= sec θ cosec θ = cosec θ sec θ
∴ L.H.S. = R.H.S.

(v) [{sin (90°-θ)}/{cosec (90°-θ)}] + [{cos (90°-θ)}/{sec (90°-θ)}] = 1
LHS = [{sin (90°-θ)}/{cosec (90°-θ)}] + [{cos (90°-θ)}/{sec (90°-θ)}]
= (cosθ/secθ) + (sinθ/cosecθ)
= cosθ × cosθ + sinθ × sinθ
= cos²θ + sin²θ
= 1 = RHS.
(vi) sin (60° – θ) = cos (30° – θ)
L.H.S. = sin (60° – θ) = sin {90° – (30° + θ)
= cos (30° – θ) {sin (90° – θ) = cos θ)
= R.H.S.
(vii) [sinθ/{sin(90°−θ)}] + [cosθ/{cos(90°−θ)}] = sec (90°–θ) cosec (90°–θ)
L.H.S. = [sinθ/{sin(90°−θ)}] + [cosθ/{cos(90°−θ)}]
= (sinθ/cosθ) + (cosθ/sinθ)
= (sin²θ+cos²θ)/sinθcosθ = 1/sinθ.cosθ
= cosecθ secθ
RHS = sec (90°–θ) cosec (90°–θ)
= cosecθ secθ
LHS = RHS

(viii) cos (81° + θ) = sin (9° – θ)
L.H.S. = cos (81° + θ) = cos {(90° – 9°) + θ}
= cos {90° – (9° – θ) = sin 9° – θ
= R.H.S. {cos (90° – θ) = sin θ}

Que-5: (i) Find θ, if sin (θ + 36°) = cos θ, where θ + 36° is an acute angle.
(ii) Find the value of θ, if sin 5θ = cos 4θ, where 5θ and 4θ are acute angles.
(iii) If A is an acute angle, solve sin 3 A = cos 2 A.

Sol: (i) sin (θ + 36°) = cos θ
⇒ sin (θ + 36°) = sin (90° – θ) { sin (90° – θ) = cos θ}
∴ θ + 36° = 90° – θ ⇒ θ + θ = 90° – 36°
= 2θ = 54° ⇒ θ = 54°/2 = 27°
θ = 27°
(ii) sin 5θ = cos 4 θ
⇒ sin 5θ = sin (90° – 4θ) {cos θ = sin (90° – θ)}
∴ 5θ = 90° – 4θ ⇒ 5θ + 4θ = 90°
⇒ 9θ = 90° ⇒ θ = 90°/9 = 10°
θ = 10°

(iii) sin 3A = cos 2A
sin 3A = sin (90° – 2A) {cos θ = sin (90° – θ)}
∴ 3A = 90° – 2A ⇒ 3A + 2A = 90°
⇒ 5A = 90° ⇒ A = 90°/5 = 18°
A = 18°

— : End of Trigonometrical Ratios Class 9 OP Malhotra Exe-19D ICSE Maths Solutions Ch-19.–

Return to :–  OP Malhotra S Chand Solutions for ICSE Class-9 Maths

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