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MCQ Class 11 Mathematics Trigonometric Functions with Answers - Set - 3
Question 1:
In any triangle ABC, if cos A/a = cos B/b = cos C/c and the side a = 2, then the area of the triangle is
(a) √3
(b) √3/4
(c) √3/2
(d) 1/√3
Correct Answer – (A)
Given cos A/a = cos B/b = cos C/c
= cos A/k × sin A = cos B/k × sin B = cos C/k × sin C {since sin A/a = sin B/b = sin C/c = k}
= cot A = cot B = cot C
⇒ A = B = C = 60
So, triangle ABS is equilateral.
Now area of the triangle = (√3/4) × a² = (√3/4) × 2² = (√3/4) × 4 = √3
Question 2 :
If a × cos x + b × cos x = c, then the value of (a × sin x – b × cos x)² is
(a) a² + b² + c²
(b) a² – b² – c²
(c) a² – b² + c²
(d) a² + b² – c²
Correct Answer – (D)
We have
(a × cos x + b × sin x)² + (a × sin x – b × cos x)² = a² + b²
⇒ c² + (a × sin x – b × cos x)² = a² + b²
⇒ (a × sin x – b × cos x)² = a² + b² – c²
Question 3 :
The value of cos 420° is
(a) 0
(b) 1
(c) 1/2
(d) √3/2
Correct Answer – (C)
cos 420° = cos(360° + 60° ) = cos 60° = 1/2
Question 4 :
If tan x = (cos 9 + sin 9)/(cos 9 – sin 9), then x =
(a) 45
(b) 54
(c) 36
(d) None of these
Correct Answer – (B)
Given, tan x = (cos 9 + sin 9)/(cos 9 – sin 9)
⇒ tan x = {cos 9(1 + sin 9/cos 9)}/{cos 9(1 – sin 9/cos 9)}
⇒ tan x = (1 + tan 9)}/(1 – tan 9)
⇒ tan x = (tan 45 + tan 9)}/(1 – tan 45 × tan 9) {since tan 45 = 1}
⇒ tan x = tan(45 + 9) {Apply tan(A + B) formula}
⇒ tan x = tan(54)
⇒ x = 54
Question 5 :
The value of tan A/2 – cot A/2 + 2cot A is
(a) 0
(b) 1
(c) -1
(d) None of these
Correct Answer – (A)
Answer: (a) 0
Given, tan A/2 – cot A/2 + 2cot A
= {sin(A/2)/cos(A/2)} – {cos(A/2)/sin(A/2)} + 2cotA
= {sin² (A/2) – cos² (A/2)}/{cos(A/2) × sin(A/2)} + 2cotA
= -{cos² (A/2) – sin² (A/2)}/{cos(A/2) × sin(A/2)} + 2cotA
= -{cosA}/{cos(A/2) × sin(A/2)} + 2cotA (since cos² A – sin² A = cos²A )
= -{cos(2A/2)}/{cos(A/2) × sin(A/2)} + 2cotA
= -{2 × cosA}/{2 × cos(A/2) × sin(A/2)} + 2cotA
= -{2cosA}/{sin(2A/2)} + 2cotA
= {-(2cosA)/(sinA)} + 2cotA (since sin2A = 2 × sinA × cosA)
= -2cotA + 2cotA
= 0
MCQ Class 11 Mathematics Trigonometric Functions with Answers
Question 6 :
When the length of the shadow of a pole is equal to the height of the pole, then the elevation of source of light is
(a) 30°
(b) 60°
(c) 75°
(d) 45°
Correct Answer – (D)
Let AB is the length of the pole and BC is the shadow of the pole.
Given AB = BC
Now from triangle ABC,
tan θ = AB/BC
⇒ tan θ = 1
⇒ θ = 45°
So, the elevation of source of light is 45°
Question 7 :
If in a triangle ABC, tan A + tan B + tan C = 6 then the value of cot A × cot B × cot C is
(a) 1/2
(b) 1/3
(c) 1/4
(d) 1/6
Correct Answer – (D)
Given tanA + tanB + tanC = 6
Now tan(A + B + C) = {(tanA + tanB + tanC) – tanA × tanB × tanC}/{1 – (tanA × tanB + tanB × tanC + tanA × tanC)}
We know that,
A + B + C = π
⇒ tan(A + B + C) = tan π
⇒ tan(A + B + C) = 0
Now
0 = {(tanA + tanB + tanC) – tanA × tanB × tanC}/{1 – (tanA × tanB + tanB × tanC + tanA × tanC)}
⇒ tanA + tanB + tanC – tanA × tanB × tanC = 0
⇒ tanA + tanB + tanC = tanA × tanB × tanC
⇒ tanA × tanB × tanC = 6
⇒ (1/cotA) × (1/cotB) × (1/cotC) = 6
⇒ 1/(cot A × cot B × cot C) = 6
⇒ cot A × cot B × cot C = 1/6
Question 8 :
In a triangle ABC, sin A – cos B = cos C, then angle B is
(a) π/2
(b) π/3
(c) π/4
(d) π/6
Correct Answer – (A)
Given, sin A – cos B = sin C
⇒ sin A = cos B + sin C
⇒ 2 × sin (A/2) × cos (A/2) = 2 × cos {(B + C)/2} × cos {(B – C)/2}
⇒ 2 × sin (A/2) × cos (A/2) = 2 × cos {π/2 – A/2} × cos {(B – C)/2}
⇒ 2 × sin (A/2) × cos (A/2) = 2 × sin (A/2) × cos {(B – C)/2}
⇒ cos (A/2) = cos {(B – C)/2}
⇒ A/2 = (B – C)/2
⇒ A = B – C
⇒ B = A + C
⇒ B = π – B {Since A + B + C = π}
⇒ 2B = π
⇒ B = π/2
Question 9 :
The value of 4 × sin x × sin(x + π/3) × sin(x + 2π/3) is
(a) sin x
(b) sin 2x
(c) sin 3x
(d) sin 4x
Correct Answer – (C)
Given, 4 × sin x × sin(x + π/3) × sin(x + 2π/3)
= 4 × sin x × {sin x × cos π/3 + cos x × sin π/3} × {sin x × cos 2π/3 + cos x × sin 2π/3}
= 4 × sin x × {(sin x)/2 + (√3 × cos x)/2} × {-(sin x)/2 + (√3 × cos x)/2}
= 4 × sin x × {-(sin² x)/4 + (3 × cos² x)/4}
= sin x × {-sin² x + 3 × cos² x}
= sin x × {-sin² x + 3 × (1 – sin² x)}
= sin x × {-sin² x + 3 – 3 × sin² x}
= sin x × {3 – 4 × sin² x}
= 3 × sin x – 4sin³ x
= sin 3x
So, 4 × sin x × sin(x + π/3) × sin(x + 2π/3) = sin 3x
Question 10 :
The value of sin 15 + cos 15 is
(a) 1
(b) 1/2
(c) √3/2
(d) √3
Correct Answer – (C)
Given, sin 15 + cos 15
= sin 15 + cos(90 – 15)
= sin 15 + sin 15
= 2 × sin 45 × cos 30
= 2 × (1/√2) × (√3/2)
= √3/2
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