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MCQ Class 11 Mathematics Trigonometric Functions with Answers - Set - 2
Question 1:
The value of (sin 7x + sin 5x) /(cos 7x + cos 5x) + (sin 9x + sin 3x) / (cos 9x + cos 3x) is
(a) tan 6x
(b) 2 tan 6x
(c) 3 tan 6x
(d) 4 tan 6x
Correct Answer – (B)
Hint:
Given, (sin 7x + sin 5x) /(cos 7x + cos 5x) + (sin 9x + sin 3x) / (cos 9x + cos 3x)
⇒ [{2 × sin(7x+5x)/2 × cos(7x-5x)/2}/{2 × cos(7x+5x)/2 × cos(7x-5x)/2}] + [{2 × sin(9x+3x)/2 × cos(9x-3x)/2}/{2 × cos(9x+3x)/2 × cos(9x-3x)/2}]
⇒ [{2 × sin 6x × cosx}/{2 × cos 6x × cosx}] + [{2 × sin 6x × cosx}/{2 × cos 6x × cosx}]
⇒ (sin 6x/cos 6x) + (sin 6x/cos 6x)
⇒ tan 6x + tan 6x
⇒ 2 tan 6x
Question 2 :
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 × {-(sin 2x)/4 + (3 × cos 2x)/4}
= sin x × {-sin 2x + 3 × cos 2x}
= sin x × {-sin 2x + 3 × (1 – sin 2x)}
= sin x × {-sin 2x + 3 – 3 × sin 2x}
= sin x × {3 – 4 × sin 2x}
= 3 × sin x – 4 sin 3x
= sin 3x
So, 4 × sin x × sin(x + π/3) × sin(x + 2π/3) = sin 3x
Question 3 :
If the radius of the circumcircle of an isosceles triangle PQR is equal to PQ ( = PR), then the angle P is
(a) 2π/3
(b) π/3
(c) π/2
(d) π/6
Correct Answer – (A)
Let S be the center of the circumcircle of triangle PQR.
So, SP = SQ = SR = PQ = PR, where SP, SQ & SR are radii.
Thus SPQ & SPR are equilateral triangles.
⇒ ∠QSP = 60°;
Similarly ∠RQP = 60°
⇒ Angle at the center QSP = 120°
So, SRPQ is a rhombus, since all the four sides are equal.
Hence, its opposite angles are equal; so ∠P = ∠QSP = 120°
Question 4 :
If tan² θ = 1 – e², then the value of sec θ + tan³ θ × cosec θ is
(a) 2 – e²
(b) (2 – e²)1/2
(c) (2 – e²)²
(d) (2 – e²)3/2
Correct Answer – (D)
Given, tan² θ = 1 – e²
⇒ tan θ = √(1 – e²)
AC² = AB² + BC²
⇒ AC² = {√(1 – e²)}² + 12
⇒ AC² = 1 – e² + 1
⇒ AC² = 2 – e²
⇒ AC = √(2 – e²)
Now, sec θ = √(2 – e²)
cosec θ = √(2 – e²)/√(1 – e²)
and tan θ = √(1 – e²)
Given, sec θ + tan³ θ × cosec θ
= √(2 – e²) + {(1 – e²)3/2 × √(2 – e²)/√(1 – e²)}
= √(2 – e²) + {(1 – e²) × (1 – e²) × √(2 – e²)/√(1 – e²)}
= √(2 – e²) + (1 – e²) × √(2 – e²)
= √(2 – e²) × (1 + 1 – e²)
= √(2 – e²) × (2 – e²)
= (2 – e²)3/2
So, sec θ + tan³ θ × cosec θ = (2 – e²)3/2
Question 5 :
If the angles of a triangle be in the ratio 1 : 4 : 5, then the ratio of the greatest side to the smallest side is
(a) 4 : (√5 – 1)
(b) 5 : 4
(c) (√5 – 1) : 4
(d) none of these
Correct Answer – (A)
Hint:
Given, the angles of a triangle be in the ratio 1 : 4 : 5
⇒ x + 4x + 5x = 180
⇒ 10x = 180
⇒ x = 180/10
⇒ x = 18
So, the angle are: 18, 72, 90
Since a : b : c = sin A : sin B : sin C
⇒ a : b : c = sin 18 : sin 72 : sin 90
⇒ a : b : c = (√5 – 1)/4 : {√(10 + 2√5)}/4 : 1
⇒ a : b : c = (√5 – 1) : {√(10 + 2√5)} : 4
Now, c /a = 4/(√5 – 1)
⇒ c : a = 4 : (√5 – 1)
MCQ Class 11 Mathematics Trigonometric Functions with Answers
Question 6 :
If tan A – tan B = x and cot B – cot A = y, then the value of cot (A – B) is
(a) x + y
(b) 1/x + y
(c) x + 1/y
(d) 1/x + 1/y
Correct Answer – (D)
Given,
tan A – tan B = x ……………. 1
and cot B – cot A = y ……………. 2
From equation,
1/cot A – 1/cot B = x
⇒ (cot B – cot A)/(cot A × cot B) = x
⇒ y/(cot A × cot B) = x {from equation 2}
⇒ y = x × (cot A × cot B)
⇒ cot A × cot B = y/x
Now, cot (A – B) = (cot A × cot B + 1)/(cot B – cot A)
⇒ cot (A – B) = (y/x + 1)/y
⇒ cot (A – B) = (y/x) × (1/y) + 1/y
⇒ cot (A – B) = 1/x + 1/y
Question 7 :
If cos a + 2cos b + cos c = 2 then a, b, c are in
(a) 2b = a + c
(b) b² = a × c
(c) a = b = c
(d) None of these
Correct Answer – (A)
Given, cos A + 2 cos B + cos C = 2
⇒ cos A + cos C = 2(1 – cos B)
⇒ 2 cos((A + C)/2) × cos((A-C)/2 = 4 sin² (B/2)
⇒ 2 sin(B/2)cos((A-C)/2) = 4sin² (B/2)
⇒ cos((A-C)/2) = 2sin (B/2)
⇒ cos((A-C)/2) = 2cos((A+C)/2)
⇒ cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2)
⇒ 2sin(A/2)sin(C/2) = sin(B/2)
⇒ 2{√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac
⇒ 2(s – b) = b
⇒ a + b + c – 2b = b
⇒ a + c – b = b
⇒ a + c = 2b
Question 8 :
The value of cos 20 + 2sin² 55 – √2 sin65 is
(a) 0
(b) 1
(c) -1
(d) None of these
Correct Answer – (B)
Given, cos 20 + 2sin² 55 – √2 sin65
= cos 20 + 1 – cos 110 – √2 sin65 {since cos 2x = 1 – 2sin² x}
= 1 + cos 20 – cos 110 – √2 sin65
= 1 – 2 × sin {(20 + 110)/2 × sin{(20 – 110)/2} – √2 sin65 {Apply cos C – cos D formula}
= 1 – 2 × sin 65 × sin (-45) – √2 sin65
= 1 + 2 × sin 65 × sin 45 – √2 sin65
= 1 + (2 × sin 65)/√2 – √2 sin65
= 1 + √2 ( sin 65 – √2 sin 65
= 1
So, cos 20 + 2sin² 55 – √2 sin65 = 1
Question 9 :
The general solution of √3 cos x – sin x = 1 is
(a) x = n × π + (-1)n × (π/6)
(b) x = π/3 – n × π + (-1)n × (π/6)
(c) x = π/3 + n × π + (-1)n × (π/6)
(d) x = π/3 – n × π + (π/6)
Correct Answer – (C)
√3 cos x-sin x=1
⇒ (√3/2)cos x – (1/2)sin x = 1/2
⇒ sin 60 × cos x – cos 60 × sin x = 1/2
⇒ sin (x – 60) = 1/2
⇒ sin (x – π/3) = sin 30
⇒ sin (x – π/3) = sinπ/6
⇒ x – π/3 = n × π + (-1)n × (π/6) {where n ∈ Z}
⇒ x = π/3 + n × π + (-1)n × (π/6)
Question 10 :
The value of tan 20 × tan 40 × tan 80 is
(a) tan 30
(b) tan 60
(c) 2 tan 30
(d) 2 tan 60
Correct Answer – (B)
Given, tan 20 × tan 40 × tan 80
= tan 40 × tan 80 × tan 20
= [{sin 40 × sin 80}/{cos 40 × cos 80}] × (sin 20/cos 20)
= [{2 * sin 40 × sin 80}/{2 × cos 40 × cos 80}] × (sin 20/cos 20)
= [{cos 40 – cos 120}/{cos 120 + cos 40}] × (sin 20/cos 20)
= [{cos 40 – cos (90 + 30)}/{cos (90 + 30) + cos 40}] × (sin 20/cos 20)
= [{cos 40 + sin30}/{-sin30 + cos 40}] × (sin 20/cos 20)
= [{(2 × cos 40 + 1)/2}/{(-1 + cos 40)/2}] × (sin 20/cos 20)
= [{2 × cos 40 + 1}/{-1 + cos 40}] × (sin 20/cos 20)
= [{2 × cos 40 × sin 20 + sin 20}/{-cos 20 + cos 40 × cos 20}]
= (sin 60 – sin 20 + sin 20)/(-cos 20 + cos 60 + cos 20)
= sin 60/cos 60
= tan 60
So, tan 20 × tan 40 × tan 80 = tan 60
- NCERT Solutions Class 11 Mathematics Relations and Functions : Exercise 2.1
- NCERT Solutions Class 11 Mathematics Relations and Functions : Exercise 2.2
- NCERT Solutions Class 11 Mathematics Relations and Functions : Exercise 2.3
- NCERT Solutions Class 11 Mathematics Relations and Functions : Exercise 2 Misc
- NCERT Solutions Class 11 Mathematics Textbook download
