Ch-Test of Trigonometrical Ratios Class 9 OP Malhotra ICSE Maths Solutions 2026-27. We Provide Step by Step Solutions / Answer of Trigonometrical Ratios OP Malhotra Maths. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics.
Trigonometrical Ratios Class 9 OP Malhotra Ch-Test ICSE Maths Solutions
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 9th |
| Chapter-19 | Trigonometrical Ratios |
| Writer | OP Malhotra |
| Ch-Test | Extra Practice Questions |
| Edition | 2026-2027 |
Ch-Test on Trigonometrical Ratios
Trigonometrical Ratios Class 9 OP Malhotra ICSE Maths Solutions 2026-27
Que-1: If A = 30° and B = 60°, then which of the following is/are correct?
I. sin A + sin B = cos A + cos B
II. tan A + tan B = cot A + cot B
Select the correct answer using the code given below:
(a) Only I
(b) Only II
(c) Both I and II
(d) Neither I nor II
Sol: A = 30° and B = 60°
I. sin A + sin B = cos A + cos B
L.H.S. = sin A + sin B = sin 30° + sin 60°
= 1/2 + 3/√2 = (1+3)/√2
R.H.S. = cos 30° + cos 60°
= 3/√2 + 1/2 = (1+3)/√2
∴ L.H.S. = R.H.S.
II. L.H.S. tan A + tan B = tan 30° + tan 60°
= 1/√3 + √3 = (1+3)/√3 = 4/√3
R.H.S. = cot A + cot B = cot 30° + cot 60°
= √3 + 1/√3 = (3+1)/√3 = 4/√3
∴ L.H.S. = R.H.S.
∴ Both I and II are correct. (c)
Que-2: What is cot 15° cot 20° cot 70° cot 75° equal to?
(a) -1 (b) 0 (c) 1 (d) 2
Sol: cot 15° cot 20° cot 70° cot 75° = cot (90° – 75°) cot (90° -70°) cot 70° cot 75°
= tan 75° tan 70° cot 70° cot 75°
= tan 75° × cot 75° tan 70° × cot 70° (∵ tan θ cot θ = 1)
= 1 × 1 = 1
Que-3: (i) If sin 3 A = cos ( A – 2°) where 3 A and
(A – 2°) are acute angles, what is the value of A?
(a) 22° (b) 23° (c) 24° (d) 25°
(ii) If sin (x – 2y) = cos (4y – x), then the value of cot 2y is:
(a) 0 (b) 1 (c) 1/√3 (d) undefined
Sol: (i) sin 3 A = cos ( A – 2°), 3 A and ( A – 2°) are acute angles
sin 3 A = cos (A – 2°) = sin {90° – A + 2°}
Comparing both sides,
3A = 90° – A + 2°
⇒ 3A + A = 92°
⇒ 4A = 92°
⇒ A = 92°/4 = 23°
∴ A = 23°
(ii) sin (x – 2y) = cos (4y – x)
sin (x – 2y) = sin (90° – 4y + x)
Comparing,
x – 2y = 90° – 4y + x
4y – 2y = 90° ⇒ 2y = 90° ⇒ y = 45°
Now, cot 2y = cot (2 × 45°) = cot 90°
not defined (d)
Que-4: (i) What is the value of cos 1° cos 2° cos 3° ………… cos 90°?
(a) 1/2 (b) 0 (c) 1 (d) 2
(ii) If tan (90°−(A/2)) = √3, then value of A is:
(a) 0 (b) 1/√2 (c) 1/2 (d) 1
Sol: (i) cos 1° cos 2° cos 3° ………. cos 90°
∵ cos 90° = 0
∴ cos 1° cos 2° cos 3° …….. cos 90° = 0
(ii) tan (90°−(A/2)) = √3 = tan 60°
Comparing, we get
90° – A/2 = 60°
⇒ A/2 = 90° – 60° = 30°
∴ A = 30° × 2 = 60°
Now cos A = cos 60° = 12 (c)
Que-5: (i) In a △ABC, ∠ABC = 90°, ∠ACB = 30°, AB = 5 cm. What is the length of AC?
(a) 10 cm (b) 5 cm (c) 5 √2 cm (d) 5 √3 cm
(ii) If ABC is a right-angled triangle at C having u units, v units and w units as the lengths of its sides opposite to the vertices A, B, C respectively, then what is tan A + tan B equal to?
(a) u²/vw (b) 1 (c) u+v (d) w²/uv
Sol: In △ABC, ∠ABC = 90°, ∠ACB = 30°, AB = 5 cm
(i) sin 30° = AB/BC
⇒ 1/2 = 5/AC
⇒ AC = 2 × 5 = 10 cm
(ii) tan A + tan B
= (u/v) + (v/u) = (u²+v²)/uv
But w² = u² + v² (Pythagoras Theorem)
∴ w² / uv (d)
Que-6: (i) If cot A =815, then what is the value of √{(1−cosA)/(1+cosA)} where A is a positive acute angle.
(a) 15 (b) 25 (c) 35 (d) 45
(ii) If A is an acute angle and sin A = √(x−1)/2x, then what is tan A equal to?
(a) √{(x-1)/(x+1)} (b) √{(x+1)/(x-1)} (c) √(x²-1) (d) √(x²+1)
Sol: (i) cot A = 8/15
∴ In right △ABC
cot A = AC/BC = 8/15
∴AC = 8, BC = 15
Then AB2 = AC2 + BC2 = 82 + 152
= 64 + 225 = 289 = (17)2
∴ AB = 17
cos A = AC/AB = 8/17
Now, √{(1-cosA)/(1+cosA)}
√[{1-(8/17)} / {1+(8/17)}]
= √[{(17-8)/17} / {(17+8)/17}]
= √{(9/17)/(25/17)}
= √{(9/17)×(17/25)}
= √(9/25) = 3/5
(ii) sin A = √(x−1)/2x = BC/AB = √(x−1)/√2x
In the △ABC,
BC = √x-1, AB = √2x
∴ AC2 = AB2 – BC2
= (√2x)² – (√(x−1))²
= 2x – (x – 1) = 2x – x + 1
= x + 1
∴ AC = √(x−1)
Now, tan A = BC/AB = √(x-1)/√(x+1)
= √(x-1)/√(x+1)
Que-7: What is the value of sin3 60° cot 30° – 2 sec2 45° + 3 cos 60° tan 45° – tan° 60°?
(a) 35/8 (b) -35/8 (c) -11/8 (d) 11/8
Sol: sin3 60° cot 30° – 2 sec2 45° + 3 cos 60° tan 45° – tan2 60°
= -35/8
Que-8: (i) If x + y = 90° and sin x : sin y = √3 : 1 then what is x : y equal to
(a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) 3 : 2
(ii) If sin x = cos y and angle x and angle y are acute, then what is the relation between x and y?
(a) x = y = π/2
(b) x + y = 3π/2
(c) x + y = π/2
(d) x + y = π/4
Sol: (i) x + y = 90°, sinx / siny = √3/1
y = 90 – x°
sinx / sin(90°−x) = √3/1
⇒ sinx / cosx = √3/1 tan x = √3
⇒ tan x = tan 60°
∴ x = 60°
and y = 90° – 60° = 30°
∴ x : y = 60° : 30° ⇒ x : y = 2 : 1 (c)
(ii) sin x = cos y = x and y are acute angle
sin x = cos y = sin (90° – y) {∵ cos θ = sin (90° – θ)}
∴ x = 90° – y x + y = 90° = π/2
Que-9: (i) What is the value of the expression {(5sin75° sin77° + 2cos13° cos15°) / (cos15° sin77°)} – {(7sin81°)/(cos9°)} ?
(a) -1 (b) 0 (c) 1 (d) 2
(ii) What is the value of the expression cos² (π/8) + 4cos² (π/4) – sec (π/3) + 5tan² (π/3) + sin² (π/8)
(a) 8 (b) 10 (c) 16 (d) 18
Sol:
= 16 + 2 – 2 = 16
Que-10: In the figure (not drawn to scale) a rocket is fired vertically upwards from its launching pad P. It first rises 20 km vertically up and then travels 80 km at 30° to the vertical. PA respresents the first state of its journey and AB the second; C is point vertically below B on the same horizontal level as P. Calculate:
(i) the height of the rocket when it is at point B.
(ii) the horizontal distance of point C from point P.
Sol: In the given figure,
A rockets is launched from P 20 km to A then 80 km to B making an angle of 30° with PA.
We have to find PC from A, draw AD || PC, Then AD = PC
Then ∠BAD = 90° – 30° = 60°
cos θ = AD/AB ⇒ cos 60° = AD/AB
⇒ 1/2 = PC/80 = PC = 80/2 = 40 Km
and sin 60° = BD/AB ⇒ √3/2 = BD/80
⇒ BD = 80√3/2 = 40√3 Km
∴ BC = BD + DC = BD + AP
= (40√3 + 20) Km
— : End of Trigonometrical Ratios Class-9 OP Malhotra Ch-Test ICSE Maths Solutions :–
Return to :– OP Malhotra S Chand Solutions for ICSE Class-9 Maths
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