Multiple Choice Questions on Trigonometrical Functions Class 11 OP Malhotra Exe-4F ISC Maths Solutions Ch-4. In this article you would learn to solve objective problem on Trigonometrical Function. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.

Trigonometrical Functions Class 11 OP Malhotra Multiple Choice Questions ISC Maths Solutions Ch-4

Board	ISC
Publications	S Chand
Subject	Maths
Class	11th
Chapter-4	Trigonometrical Functions
Writer	OP Malhotra
Exe-4(F)	Multiple Choice Questions.

Multiple Choice Questions on Trigonometrical Functions

OP Malhotra ISC Class 11 Maths Solutions

Que-1: If sin θ + cosec θ = 2, then sin² θ + cosec² θ is equal to

(a) 1    (b) 4    (c) 2    (d) None of these

Sol: (c) 2
Given: sinθ + cosecθ = 2
But cosecθ = 1/sinθ
So, sinθ + 1/sinθ = 2
Multiply both sides by sinθ:
sin²θ + 1 = 2 sinθ
sin²θ − 2sinθ + 1 = 0
(sinθ − 1)² = 0
⇒ sinθ = 1
Therefore, cosecθ = 1
Hence,
sin²θ + cosec²θ = 1² + 1² = 2

Que-2: sin² 17.5° + sin² 72.5° is equal to

(a) cos² 90°    (b) tan² 45°    (c) cos² 30°    (d) sin² 45°

Sol: (b) tan² 45°
sin² 72.5° = cos² 17.5°   (since sin(90° − θ) = cos θ)
So,
sin² 17.5° + sin² 72.5°
= sin² 17.5° + cos² 17.5°
= 1
Now check options:
cos² 90° = 0
tan² 45° = 1 ✔
cos² 30° = 3/4
sin² 45° = 1/2

Que-3: If θ lies in the first quadrant and 5 tan θ = 4, then
(5 sin θ − 3 cos θ) / (sin θ + 2 cos θ) is equal to

(a) 5/14    (b) 3/14    (c) 1/14    (d) 0

Sol: (a) 5/14
5 tanθ = 4 ⇒ tanθ = 4/5
Take a right triangle:
Opposite = 4, Adjacent = 5 ⇒ Hypotenuse = √(4²+5²) = √41
sinθ = 4/√41,    cosθ = 5/√41
Substitute:
(5 sinθ − 3 cosθ)/(sinθ + 2 cosθ)
= [5(4/√41) − 3(5/√41)] / [(4/√41) + 2(5/√41)]
= (20 − 15)/√41 ÷ (4 + 10)/√41
= 5/√41 ÷ 14/√41
= 5/14

Que-4: The value of 2 cot²(π/6) + 4 tan²(π/6) − 3 cosec(π/6) is

(a) 2    (b) 4    (c) 4/3    (d) 3/4

Sol: (c) 4/3
We know standard values:
tan(π/6) = 1/√3,
cot(π/6) = √3,
cosec(π/6) = 2
⇒ cot²(π/6) = 3
⇒ tan²(π/6) = 1/3
Substitute in expression:
2(3) + 4(1/3) − 3(2)
= 6 + 4/3 − 6
= 4/3

Que-5: Which one of the following is not correct?

(a) |sin x| ≤ 1
(b) −1 ≤ cos x ≤ 1
(c) |sec x| < 1
(d) cosec x ≥ 1 or cosec x ≤ −1

Sol: (c) |sec x| < 1
We know standard trigonometric ranges:
|sin x| ≤ 1 ✔️
−1 ≤ cos x ≤ 1 ✔️
cosec x ≥ 1 or ≤ −1 ✔️
But sec x = 1/cos x, so |sec x| ≥ 1 (not less than 1)
❌ Therefore, option (c) |sec x| < 1 is incorrect.

Que-6: cos⁴ θ − sin⁴ θ is equal to

(a) 1 − 2 sin²(θ/2)
(b) 1 + 2 sin²(θ/2)
(c) 2 cos² θ − 1
(d) 1 + 2 cos² θ

Sol: (c) 2 cos² θ − 1
cos⁴θ − sin⁴θ = (cos²θ − sin²θ)(cos²θ + sin²θ)
= (cos²θ − sin²θ)(1)
= cos 2θ
= 2cos²θ − 1

Que-7: sin 200° + cos 200° is

(a) negative    (b) zero    (c) positive    (d) either zero or positive

Sol: (a) negative
200° lies in the 3rd quadrant.
In 3rd quadrant: sinθ is negative and cosθ is negative.
Therefore, sin 200° + cos 200° = negative + negative = negative.

Que-8: If A = 130° and x = sin A + cos A, then

(a) x < 0    (b) x = 0    (c) x > 0    (d) x ≥ 0

Sol: (c) x > 0
A = 130° lies in II quadrant
⇒ sin A is positive and cos A is negative
sin 130° = sin 50° (positive)
cos 130° = −cos 50° (negative)
So, x = sin A + cos A = sin50° − cos50°
Since sin50° > cos50°, therefore x > 0

Que-9: If tan θ = −4/3, then sin θ is

(a) −4/5 or 4/5
(b) 4/5 but not −4/5
(c) −4/5 or 4/5
(d) None of these

Sol: (c) −4/5 or 4/5
tan θ = −4/3 = opposite / adjacent
Using Pythagoras: hypotenuse = √(4² + 3²) = 5
So, sin θ = opposite / hypotenuse = 4/5
Since tan θ is negative, θ lies in either 2nd or 4th quadrant.
Thus sin θ can be positive or negative.
∴ sin θ = ± 4/5

Que-10: The expression tan² α + cot² α is

(a) ≥ −2    (b) ≥ 2    (c) ≤ 2    (d) None of these

Sol: (b) ≥ 2
tan²α + cot²α ≥ 2 (by AM ≥ GM)

Que-11: If x ∈ [0, π/2], y ∈ [0, π/2] and sin x + cos y = 2, then the value of x + y is equal to

(a) 2π    (b) π    (c) π/4    (d) π/2    (e) 0

Sol: (d) π/2
Since sin x ≤ 1 and cos y ≤ 1 for all x, y ∈ [0, π/2]
Given: sin x + cos y = 2
This is possible only when:
sin x = 1 and cos y = 1
⇒ x = π/2 and y = 0
Therefore,
x + y = π/2 + 0 = π/2

Que-12: If A lies in the second quadrant and 3 tan A + 4 = 0, then the value of 2 cot A − 5 cos A + sin A is

(a) −53/10    (b) 23/10    (c) 37/10    (d) 7/10

Sol: (b) 23/10
Given: 3tanA + 4 = 0
⇒ tanA = −4/3
Since A is in 2nd quadrant ⇒ sinA (+), cosA (−)
Take triangle ratio:
sinA = 4/5,    cosA = −3/5
cotA = 1/tanA = −3/4
Now substitute:
2cotA − 5cosA + sinA
= 2(−3/4) − 5(−3/5) + (4/5)
= (−3/2) + 3 + 4/5
= 3/2 + 4/5
= (15 + 8) / 10
= 23/10

Que-13: The expression (tan A)/(1 − cot A) + (cot A)/(1 − tan A) can be written as

(a) sin A cos A + 1    (b) sec A cosec A + 1    (c) tan A + cot A    (d) sec A + cosec A

Sol: (b) sec A cosec A + 1
(tanA)/(1 − cotA) + (cotA)/(1 − tanA)
Convert into sin and cos:
= (sinA/cosA)/(1 − cosA/sinA) + (cosA/sinA)/(1 − sinA/cosA)
= (sin²A / cosA(sinA − cosA)) + (cos²A / sinA(cosA − sinA))
= [sin³A − cos³A] / [sinA cosA (sinA − cosA)]
= [(sinA − cosA)(sin²A + sinAcosA + cos²A)] / [sinA cosA (sinA − cosA)]
Cancel (sinA − cosA):
= (sin²A + cos²A + sinAcosA) / (sinA cosA)
= (1 + sinAcosA) / (sinA cosA)
= 1/(sinA cosA) + 1
= secA cosecA + 1

Que-14: If (cos A)/3 = (cos B)/4 = 1/5, −π/2 < A < 0 and −π/2 < B < 0, then the value of 2 sin A + 4 sin B is

(a) 4    (b) −2    (c) −4    (d) 0

Sol: (c) -4
⇒ cosA = 3/5 and cosB = 4/5
Since A and B lie in (−π/2, 0), sinA and sinB are negative.
sinA = −√(1 − cos²A) = −√(1 − 9/25) = −4/5
sinB = −√(1 − cos²B) = −√(1 − 16/25) = −3/5
2 sinA + 4 sinB = 2(−4/5) + 4(−3/5)
= −8/5 − 12/5 = −20/5 = −4

Que-15: The expression cos(10π/13) + cos(8π/13) + cos(3π/13) + cos(5π/13) is equal to

(a) −1    (b) 0    (c) 1    (d) None of these

Sol: (b) 0
Use identity: cos(π − x) = −cos x
cos(10π/13) = cos(π − 3π/13) = −cos(3π/13)
cos(8π/13) = cos(π − 5π/13) = −cos(5π/13)
Therefore,
cos(10π/13) + cos(8π/13) + cos(3π/13) + cos(5π/13)
= [−cos(3π/13) + −cos(5π/13) + cos(3π/13) + cos(5π/13)]
= 0

Que-16: Which one of the following is possible?

(a) sin θ = (a² + b²)/(a² − b²), (a ≠ b)
(b) sec θ = 4/5
(c) tan θ = 45
(d) cos θ = 7/3

Sol: (c) tan θ = 45

Que-17: If cos θ = −√3/2 and sin α = −3/5, where θ does not lie in 3rd quadrant and α lies in the 3rd quadrant, then (2 tan α + √3 tan θ) / (cos² θ + cos α) is equal to

(a) 7/22    (b) 5/22    (c) 9/22    (d) 22/5

Sol: (b) 5/22
cosθ = −√3/2 and θ not in 3rd quadrant ⇒ θ in 2nd quadrant
⇒ sinθ = 1/2
⇒ tanθ = sinθ/cosθ = (1/2)/(−√3/2) = −1/√3
sinα = −3/5 and α in 3rd quadrant
⇒ cosα = −4/5
⇒ tanα = sinα/cosα = (−3/5)/(−4/5) = 3/4
Now,
2tanα + √3tanθ = 2(3/4) + √3(−1/√3) = 3/2 − 1 = 1/2
cos²θ + cosα = (−√3/2)² + (−4/5) = 3/4 − 4/5 = (15 − 16)/20 = −1/20
Required value = (1/2) ÷ (−1/20) = (1/2) × (−20) = −10

Que-18: If for real values of x, cos θ = x + 1/x, then

(a) θ is an acute angle
(b) θ is right angle
(c) θ is an obtuse angle
(d) No value of θ is possible

Sol: (d) No value of θ is possible
For any real x, we know that:
x + 1/x ≥ 2   or   x + 1/x ≤ −2
But the range of cos θ is:
−1 ≤ cos θ ≤ 1
Since x + 1/x can never lie between −1 and 1 for real x,
it cannot be equal to cos θ.
Therefore, no real value of θ is possible.

Que-19: The cotangent of the angles π/3, π/4 and π/6 are in

(a) A.P.    (b) G.P.    (c) H.P.    (d) None of these

Sol: (b) G.P.
cot(π/3) = 1/√3
cot(π/4) = 1
cot(π/6) = √3
So the numbers are:
1/√3 , 1 , √3
Check G.P. ratio:
1 ÷ (1/√3) = √3
√3 ÷ 1 = √3
Common ratio same ⇒ numbers are in G.P.

Que-20: The value of tan 1° tan 2° tan 3° … tan 89° is equal to

(a) −1    (b) 1    (c) π/2    (d) 2

Sol: (b) 1
We use the identity:
tan(90° − θ) = cot θ = 1 / tan θ
So we pair terms:
tan 1° × tan 89° = 1
tan 2° × tan 88° = 1
tan 3° × tan 87° = 1
… and so on
Each pair = 1, and there are 44 such pairs, and the middle term:
tan 45° = 1
Therefore,
tan 1° × tan 2° × … × tan 89° = 1

–: End Trigonometrical Functions Class 11 OP Malhotra Exe-4F ISC Maths Ch-4 Latest editions :–

Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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