CBSE Class 10 Mathematics Applications Of Trigonometry Worksheet Set D

Question. The value of sin 60° ⋅ cos 30° + sin 30° ⋅ cos 60° is
(a) 0
(b) 1
(c) 2
(d) 8
Answer : B

Question. The value of (1 + tan2 θ)(1 – sin θ)(1 + sin θ) =
(a) 0
(b) 1
(c) 2
(d) None of these
Answer : B

Question. Value of cos 0° ⋅ cos 30° ⋅ cos 45° ⋅ cos 60° ⋅ cos 90° is
(a) 0
(b) 1
(c) 2
(d) 9
Answer : A

Question. The value of (sin² θ = 1/1+ tan² θ) =
(a) 0
(b) 1
(c) 2
(d) 5
Answer : B

Question. 2 tan² 45° + cos² 30° – sin² 60° equals
(a) 1
(b) 2
(c) 5
(d) 6
Answer : B

Question. If 15 cot A = 8, then the value of cosec A is
(a) 15/12
(b) 13/15
(c) 4/15
(d) 17/15
Answer : D

Question. If tan(A + B) = √3 and tan(A – B) = 1/√3, A > B, then the value of A is
(a) A = 30°
(b) A = 60°
(c) A = 90°
(d) A = 45°
Answer : D

Question. The value of sin 60° cos 30° + sin 30° cos 60° is
(a) 1
(b) 2
(c) 11
(d) 0
Answer : A

Question. The value of cos30° + sin 60°/1 + cos60° + sin30° is
(a) √3/2
(b) 2/√3
(c) 1/√2
(d) 0
Answer : A

Question. ABC is a triangle right angled at C and AC = v3 BC. Then ∠ABC =
(a) 30°
(b) 60°
(c) 90°
(d) 0°
Answer : B

Question. Evaluate: 4 sin² 60° + 3 tan² 30° – 8 sin 45° cos 45°
(a) 0
(b) 1
(c) 2
(d) 5
Answer : A

Question. If sin θ = x and sec q = y, then the value of cot θ is
(a) xy
(b) 2xy
(c) 1/xy
(d) x + y
Answer : C

Question. If (1 + cos A)(1 – cos A) = 3/4, the value of sec A is
(a) 2
(b) –2
(c) ±2
(d) 0
Answer : C

Question. Evaluate: sin30° + tan45° - cosec60°/sec30° + cos 60° + cot45°
(a) 3√3+2/3√3+2
(b) 3√3-4/3√3+4
(c) 3√3+8/3√3-9
(d) None of these
Answer : B
 

Assertion-Reason Type Questions

In the following questions, a statement of assertion (A) is followed by a statement reason (R). Choose
the correct choice as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question. Assertion (A): In a right-angled triangle, if tan θ = 3/4, the greatest side of the triangle is 5 units.
Reason (R): (Greatest side)² = (Hypotenuse)2 = (Perpendicular)2 + (Base)².
Answer : A

Question. Assertion (A): In a right-angled triangle, if cos θ = 1/2 and sin θ = 3/2, then tan θ = 3 .
Reason (R): sinθ/cosθ
Answer : A

ONE MARKERS

1) If θ = 45⁰, the value of cosec² θ is a) 1/√2 b)1 c) 1/2 d) 2

2) sin (60+ θ ) – cos (30 – θ ) is equal to a)2 cos θ b) 2 sin θ c) 0 d) 1

3) If sin A + sin² A=1, then the value of cos²A + cos 4A is

a) 2 b) 1 c) – 2 d) 0

4) The value of [(sec A + tan A) (1 – sin A)] is equal to

a) tan² A b) sin² A c) cos A d) sin A.

5) In fig .if D is mid-point of BC, the value of tan x⁰/ tan y⁰ is A

∟CAD= x⁰, ∟CAB =y⁰

C D B

a) 1/3 b) 1 c) 2 d) 1 / 2

6) If x = 3 sec² θ – 1, y = tan² θ – 2 then x – 3y is equal to

a) 3 b) 4 c) 8 d) 5

7) In the given fig. ∟ACB =90⁰,∟BDC = 90⁰,CD = 4cm, BD = 3 cm, AC =12 cm.cos A – sinA Is equal to a) 5/12 b) 5/13 c) 7/12 d) 7/13. A C

8) The value of cos (90 - θ θ ) cos θ – 1 is DDDD tan θ B

a) − sin² θ b) − cosec² θ c) − cos² θ d) − cot θ

9) If tan θ + 1/ tan θ = 2, then the value of tan² θ + 1/ tan² θ is

a) 3 b) 4 c) 2 d) − 4

10) If sec 4A = cosec (A −20⁰), where 4A is an acute angle, then the value of A is

a) 21⁰ b) 22⁰ c) 23⁰ d) 24⁰.

TWO MARKERS

1) If 7 sin² θ + 3 cos² θ = 4, show that tan θ = 1 / √3.

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