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Class 11 Maths
Trigonometric Functions
Quiz 1
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The value of cos² x + cos² y – 2cos x × cos y × cos (x + y) is
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sin (x + y)
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sin² (x + y)
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sin³ (x + y)
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sin
4
(x + y)
Explanation
sin² (x + y)
Hint:
cos² x + cos² y – 2cos x × cos y × cos(x + y)
{since cos(x + y) = cos x × cos y – sin x × sin y }
= cos² x + cos² y – 2cos x × cos y × (cos x × cos y – sin x × sin y)
= cos² x + cos² y – 2cos² x × cos² y + 2cos x × cos y × sin x × sin y
= cos² x + cos² y – cos² x × cos² y – cos² x × cos² y + 2cos x × cos y × sin x × sin y
= (cos² x – cos² x × cos² y) + (cos² y – cos² x × cos² y) + 2cos x × cos y × sin x × sin y
= cos² x(1- cos² y) + cos² y(1 – cos² x) + 2cos x × cos y × sin x × sin y
= sin² y × cos² x + sin² x × cos² y + 2cos x × cos y × sin x × sin y (since sin² x + cos² x = 1 )
= sin² x × cos² y + sin² y × cos² x + 2cos x × cos y × sin x × sin y
= (sin x × cos y)² + (sin y × cos x)² + 2cos x × cos y × sin x × sin y
= (sin x × cos y + sin y × cos x)²
= {sin (x + y)}²
= sin² (x + y)
If cos a + 2cos b + cos c = 2 then a, b, c are in
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a² + b² + c²
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a² – b² – c²
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a² – b² + c²
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a² + b² – c²
Explanation
a² + b² – c²
Hint:
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
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
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4 : (√5 – 1)
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5 : 4
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(√5 – 1) : 4
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none of these
Explanation
4 : (√5 – 1)
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)
If a×cos x + b × cos x = c, then the value of (a × sin x – b²cos x)² is
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a² + b² + c²
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a² – b² – c²
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a² – b² + c²
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a² + b² – c²
Explanation
a² + b² – c²
Hint:
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
The value of cos 5π is
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0
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1
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-1
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None of these
Explanation
-1
Hint:
Given, cos 5π = cos (π + 4π) = cos π = -1
In a triangle ABC, cosec A (sin B cos C + cos B sin C) equals
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none of these
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c/a
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1
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a/c
Explanation
1
Hint:
Given cosec A (sin B cos C + cos B sin C)
= cosec A × sin(B+C)
= cosec A × sin(180 – A)
= cosec A × sin A
= cosec A × 1/cosec A
= 1
The value of cos 180° is
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0
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1
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-1
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infinite
Explanation
-1
Hint:
180 is a standard degree generally we all know their values but if we want to go theoretically then
cos(90 + x) = – sin(x)
So, cos 180 = cos(90 + 90)
= -sin 90
= -1 {sin 90 = 1}
So, cos 180 = -1
The perimeter of a triangle ABC is 6 times the arithmetic mean of the sines of its angles. If the side b is 2, then the angle B is
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30°
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90°
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60°
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120°
Explanation
90°
Hint:
Let the lengths of the sides if ∆ABC be a, b and c
Perimeter of the triangle = 2s = a + b + c = 6(sinA + sinB + sinC)/3
⇒ (sinA + sinB + sinC) = ( a + b + c)/2
⇒ (sinA + sinB + sinC)/( a + b + c) = 1/2
From sin formula,Using
sinA/a = sinB/b = sinC/c = (sinA + sinB + sinC)/(a + b + c) = 1/2
Now, sinB/b = 1/2
Given b = 2
So, sinB/2 = 1/2
⇒ sinB = 1
⇒ B = π/2
If 3 × tan(x – 15) = tan(x + 15), then the value of x is
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30
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45
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60
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90
Explanation
45
Hint:
Given, 3×tan(x – 15) = tan(x + 15)
⇒ tan(x + 15)/tan(x – 15) = 3/1
⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = (3 + 1)/(3 – 1)
⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 4/2
⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 2
⇒ sin(x + 15 + x – 15)/sin(x + 15 – x + 15) = 2
⇒ sin 2x/sin 30 = 2
⇒ sin 2x/(1/2) = 2
⇒ 2 × sin 2x = 2
⇒ sin 2x = 1
⇒ sin 2x = sin 90
⇒ 2x = 90
⇒ x = 45
If the sides of a triangle are 13, 7, 8 the greatest angle of the triangle is
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π/3
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π/2
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2π/3
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3π/2
Explanation
2π/3
Hint:
Given, the sides of a triangle are 13, 7, 8
Since greatest side has greatest angle,
Now Cos A = (b² + c² – a²)/2bc
⇒ Cos A = (7² + 8² – 13²)/(2×7×8)
⇒ Cos A = (49 + 64 – 169)/(2×7×8)
⇒ Cos A = (113 – 169)/(2×7×8)
⇒ Cos A = -56/(2×56)
⇒ Cos A = -1/2
⇒ Cos A = Cos 2π/3
⇒ A = 2π/3
So, the greatest angle is
= 2π/3
The value of tan 20 × tan 40 × tan 80 is
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tan 30
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tan 60
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2 tan 30
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2 tan 60
Explanation
tan 60
Hint:
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
The general solution of √3 cos x – sin x = 1 is
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x = n × π + (-1)n × (π/6)
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x = π/3 – n × π + (-1)n × (π/6)
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x = π/3 + n × π + (-1)n × (π/6)
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x = π/3 – n × π + (π/6)
Explanation
x = π/3 + n × π + (-1)n × (π/6)
Hint:
√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)
The value of cos 20 + 2sin² 55 – √2 sin65 is
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0
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1
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-1
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None of these
Explanation
1
Hint:
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
If the radius of the circumcircle of an isosceles triangle PQR is equal to PQ ( = PR), then the angle P is
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2π/3
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π/3
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π/2
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π/6
Explanation
2π/3
Hint:
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°
The value of 4 × sin x × sin(x + π/3) × sin(x + 2π/3) is
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sin x
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sin 2x
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sin 3x
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sin 4x
Explanation
sin 3x
Hint:
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
If tan A – tan B = x and cot B – cot A = y, then the value of cot (A – B) is
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x + y
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1/x + y
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x + 1/y
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1/x + 1/y
Explanation
1/x + 1/y
Hint:
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
The greatest value of sin x cos x is
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2
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1/ 2
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1
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$\sqrt {2}$
If sin A and cos A are the roots of the equation $ax^2 – bx + c = 0$, then a, b and c satisfy the relation.
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$a^2 + c^2 + 2ab =0$
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$a^2 – b^2 – 2ac = 0$
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$a^2 + b^2 + 2ac = 0$
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$a^2 – b^2 + 2ac = 0$
The value of tan 3A – tan 2A – tan A is equal to
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– tan 3A tan 2A tan A
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tan A tan 2A – tan 2A tan 3A – tan 3A tan A
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None of these
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tan 3A tan 2A tan A
The minimum value of 3 cosx + 4 sinx + 9 is
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5
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9
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4
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2
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