Sunday, 21 February 2021

CBSE Class 11 Maths - MCQ and Online Tests - Unit 3 - Trigonometric Functions

CBSE Class 11 Maths  – MCQ and Online Tests – Unit 3 – Trigonometric Functions

Every year CBSE schools conducts Annual Assessment exams for 6,7,8,9,11th standards. These exams are very competitive to all the students. So our website provides online tests for all the 6,7,8,9,11th standard’s subjects. These tests are also very effective and useful for those who preparing for any competitive exams like Olympiad etc. It can boost their preparation level and confidence level by attempting these chapter wise online tests.

These online tests are based on latest CBSE syllabus. While attempting these our students can identify the weak lessons and continuously practice those lessons for attaining high marks. It also helps to revise the NCERT textbooks thoroughly.


 

CBSE Class 11 Maths  – MCQ and Online Tests – Unit 3 – Trigonometric Functions

Question 1.
If tanē ? = 1 – eē, then the value of sec ? + tanģ ? Ũ cosec ? is
(a) 2 – eē
(b) (2 – eē)1/2
(c) (2 – eē)ē
(d) (2 – eē)3/2

Answer

Answer: (d) (2 – eē)3/2
Hint:
Given, tanē ? = 1 – eē
? tan ? = v(1 – eē)
MCQ Questions for Class 11 Maths Chapter 3 Trigonometric Functions with Answers 1
From the figure and Pythagorus theorem,
ACē = ABē + BCē
? ACē = {v(1 – eē)}ē + 12
? ACē = 1 – eē + 1
? ACē = 2 – eē
? AC = v(2 – eē)
Now, sec ? = v(2 – eē)
cosec ? = v(2 – eē)/v(1 – eē)
and tan ? = v(1 – eē)
Given, sec ? + tanģ ? Ũ cosec ?
= v(2 – eē) + {(1 – eē)3/2 Ũ v(2 – eē)/v(1 – eē)}
= v(2 – eē) + {(1 – eē) Ũ (1 – eē) Ũ v(2 – eē)/v(1 – eē)}
= v(2 – eē) + (1 – eē) Ũ v(2 – eē)
= v(2 – eē) Ũ (1 + 1 – eē)
= v(2 – eē) Ũ (2 – eē)
= (2 – eē)3/2
So, sec ? + tanģ ? Ũ cosec ? = (2 – eē)3/2

Question 2.
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
(a) 30°
(b) 90°
(c) 60°
(d) 120°

Answer

Answer: (b) 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 = p/2


Question 3:
If 3 Ũ tan(x – 15) = tan(x + 15), then the value of x is
(a) 30
(b) 45
(c) 60
(d) 90

Answer

Answer: (b) 45
Hint:
Given, 3Ũtan(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 Ũ sin 2x = 2
? sin 2x = 1
? sin 2x = sin 90
? 2x = 90
? x = 45


Question 4.
If cos a + 2cos b + cos c = 2 then a, b, c are in
(a) 2b = a + c
(b) bē = a Ũ c
(c) a = b = c
(d) None of these

Answer

Answer: (a) 2b = a + c
Hint:
Given, cos A + 2 cos B + cos C = 2
? cos A + cos C = 2(1 – cos B)
? 2 cos((A + C)/2) Ũ cos((A-C)/2 = 4 sinē(B/2)
? 2 sin(B/2)cos((A-C)/2) = 4sinē (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{v(s-b)(s-c)vbc} Ũ {v(s-a)(s-b)vab} = v(s-a)(s-c)vac
? 2(s – b) = b
? a + b + c – 2b = b
? a + c – b = b
? a + c = 2b


Question 5.
The value of cos 5p is
(a) 0
(b) 1
(c) -1
(d) None of these

Answer

Answer: (c) -1
Hint:
Given, cos 5p = cos (p + 4p) = cos p = -1


Question 6.
In a triangle ABC, cosec A (sin B cos C + cos B sin C) equals
(a) none of these
(b) c/a
(c) 1
(d) a/c

Answer

Answer: (c) 1
Hint:
Given cosec A (sin B cos C + cos B sin C)
= cosec A Ũ sin(B+C)
= cosec A Ũ sin(180 – A)
= cosec A Ũ sin A
= cosec A Ũ 1/cosec A
= 1


Question 7.
The value of cosē x + cosē y – 2cos x Ũ cos y Ũ cos (x + y) is
(a) sin (x + y)
(b) sinē (x + y)
(c) sinģ (x + y)
(d) sin4 (x + y)

Answer

Answer: (b) sinē (x + y)
Hint:
cosē x + cosē y – 2cos x Ũ cos y Ũ cos(x + y)
{since cos(x + y) = cos x Ũ cos y – sin x Ũ sin y }
= cosē x + cosē y – 2cos x Ũ cos y Ũ (cos x Ũ cos y – sin x Ũ sin y)
= cosē x + cosē y – 2cosē x Ũ cosē y + 2cos x Ũ cos y Ũ sin x Ũ sin y
= cosē x + cosē y – cosē x Ũ cosē y – cosē x Ũ cosē y + 2cos x Ũ cos y Ũ sin x Ũ sin y
= (cosē x – cosē x Ũ cosē y) + (cosē y – cosē x Ũ cosē y) + 2cos x Ũ cos y Ũ sin x Ũ sin y
= cosē x(1- cosē y) + cosē y(1 – cosē x) + 2cos x Ũ cos y Ũ sin x Ũ sin y
= sinē y Ũ cosē x + sinē x Ũ cosē y + 2cos x Ũ cos y Ũ sin x Ũ sin y (since sinē x + cosē x = 1 )
= sinē x Ũ cosē y + sinē y Ũ cosē x + 2cos x Ũ cos y Ũ sin x Ũ sin y
= (sin x Ũ cos y)ē + (sin y Ũ cos x)ē + 2cos x Ũ cos y Ũ sin x Ũ sin y
= (sin x Ũ cos y + sin y Ũ cos x)ē
= {sin (x + y)}ē
= sinē (x + y)


Question 8.
If aŨcos x + b Ũ cos x = c, then the value of (a Ũ sin x – bēcos x)ē is
(a) aē + bē + cē
(b) aē – bē – cē
(c) aē – bē + cē
(d) aē + bē – cē

Answer

Answer: (d) aē + bē – cē
Hint:
We have
(aŨcos x + b Ũ sin x)ē + (a Ũ sin x – b Ũ cos x)ē = aē + bē
? cē + (a Ũ sin x – b Ũ cos x)ē = aē + bē
? (a Ũ sin x – b Ũ cos x)ē = aē + bē – cē


Question 9.
If the sides of a triangle are 13, 7, 8 the greatest angle of the triangle is
(a) p/3
(b) p/2
(c) 2p/3
(d) 3p/2

Answer

Answer: (c) 2p/3
Hint:
Given, the sides of a triangle are 13, 7, 8
Since greatest side has greatest angle,
Now Cos A = (bē + cē – aē)/2bc
? Cos A = (7ē + 8ē – 13ē)/(2Ũ7Ũ8)
? Cos A = (49 + 64 – 169)/(2Ũ7Ũ8)
? Cos A = (113 – 169)/(2Ũ7Ũ8)
? Cos A = -56/(2Ũ56)
? Cos A = -1/2
? Cos A = Cos 2p/3
? A = 2p/3
So, the greatest angle is
= 2p/3


Question 10.
The value of tan 20 Ũ tan 40 Ũ tan 80 is
(a) tan 30
(b) tan 60
(c) 2 tan 30
(d) 2 tan 60

Answer

Answer: (b) tan 60
Hint:
Given, tan 20 Ũ tan 40 Ũ tan 80
= tan 40 Ũ tan 80 Ũ tan 20
= [{sin 40 Ũ sin 80}/{cos 40 Ũ cos 80}] Ũ (sin 20/cos 20)
= [{2 * sin 40 Ũ sin 80}/{2 Ũ cos 40 Ũ cos 80}] Ũ (sin 20/cos 20)
= [{cos 40 – cos 120}/{cos 120 + cos 40}] Ũ (sin 20/cos 20)
= [{cos 40 – cos (90 + 30)}/{cos (90 + 30) + cos 40}] Ũ (sin 20/cos 20)
= [{cos 40 + sin30}/{-sin30 + cos 40}] Ũ (sin 20/cos 20)
= [{(2 Ũ cos 40 + 1)/2}/{(-1 + cos 40)/2}] Ũ (sin 20/cos 20)
= [{2 Ũ cos 40 + 1}/{-1 + cos 40}] Ũ (sin 20/cos 20)
= [{2 Ũ cos 40 Ũ sin 20 + sin 20}/{-cos 20 + cos 40 Ũ cos 20}]
= (sin 60 – sin 20 + sin 20)/(-cos 20 + cos 60 + cos 20)
= sin 60/cos 60
= tan 60
So, tan 20 Ũ tan 40 Ũ tan 80 = tan 60


Question 11.
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 : (v5 – 1)
(b) 5 : 4
(c) (v5 – 1) : 4
(d) none of these

Answer

Answer: (a) 4 : (v5 – 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 = (v5 – 1)/4 : {v(10 + 2v5)}/4 : 1
? a : b : c = (v5 – 1) : {v(10 + 2v5)} : 4
Now, c /a = 4/(v5 – 1)
? c : a = 4 : (v5 – 1)


Question 12.
The general solution of v3 cos x – sin x = 1 is
(a) x = n Ũ p + (-1)n Ũ (p/6)
(b) x = p/3 – n Ũ p + (-1)n Ũ (p/6)
(c) x = p/3 + n Ũ p + (-1)n Ũ (p/6)
(d) x = p/3 – n Ũ p + (p/6)

Answer

Answer: (c) x = p/3 + n Ũ p + (-1)n Ũ (p/6)
Hint:
v3 cos x-sin x=1
? (v3/2)cos x – (1/2)sin x = 1/2
? sin 60 Ũ cos x – cos 60 Ũ sin x = 1/2
? sin (x – 60) = 1/2
? sin (x – p/3) = sin 30
? sin (x – p/3) = sinp/6
? x – p/3 = n Ũ p + (-1)n Ũ (p/6) {where n ? Z}
? x = p/3 + n Ũ p + (-1)n Ũ (p/6)



Question 13.
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

Answer

Answer: (d) 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 Ũ cot B) = x
? y/(cot A Ũ cot B) = x {from equation 2}
? y = x Ũ (cot A Ũ cot B)
? cot A Ũ cot B = y/x
Now, cot (A – B) = (cot A Ũ cot B + 1)/(cot B – cot A)
? cot (A – B) = (y/x + 1)/y
? cot (A – B) = (y/x) Ũ (1/y) + 1/y
? cot (A – B) = 1/x + 1/y


Question 14.
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

Answer

Answer: (b) 2 tan 6x
Hint:
Given, (sin 7x + sin 5x) /(cos 7x + cos 5x) + (sin 9x + sin 3x) / (cos 9x + cos 3x)
? [{2 Ũ sin(7x+5x)/2 Ũ cos(7x-5x)/2}/{2 Ũ cos(7x+5x)/2 Ũ cos(7x-5x)/2}] + [{2 Ũ sin(9x+3x)/2 Ũ cos(9x-3x)/2}/{2 Ũ cos(9x+3x)/2 Ũ cos(9x-3x)/2}]
? [{2 Ũ sin 6x Ũ cosx}/{2 Ũ cos 6x Ũ cosx}] + [{2 Ũ sin 6x Ũ cosx}/{2 Ũ cos 6x Ũ cosx}]
? (sin 6x/cos 6x) + (sin 6x/cos 6x)
? tan 6x + tan 6x
? 2 tan 6x


Question 15.
The value of 4 Ũ sin x Ũ sin(x + p/3) Ũ sin(x + 2p/3) is
(a) sin x
(b) sin 2x
(c) sin 3x
(d) sin 4x

Answer

Answer: (c) sin 3x
Hint:
Given, 4 Ũ sin x Ũ sin(x + p/3) Ũ sin(x + 2p/3)
= 4 Ũ sin x Ũ {sin x Ũ cos p/3 + cos x Ũ sin p/3} Ũ {sin x Ũ cos 2p/3 + cos x Ũ sin 2p/3}
= 4 Ũ sin x Ũ {(sin x)/2 + (v3 Ũ cos x)/2} Ũ {-(sin x)/2 + (v3 Ũ cos x)/2}
= 4 Ũ sin x Ũ {-(sin 2x)/4 + (3 Ũ cos 2x)/4}
= sin x Ũ {-sin 2x + 3 Ũ cos 2x}
= sin x Ũ {-sin 2x + 3 Ũ (1 – sin 2x)}
= sin x Ũ {-sin 2x + 3 – 3 Ũ sin 2x}
= sin x Ũ {3 – 4 Ũ sin 2x}
= 3 Ũ sin x – 4 sin 3x
= sin 3x
So, 4 Ũ sin x Ũ sin(x + p/3) Ũ sin(x + 2p/3) = sin 3x


Question 16.
The value of cos 20 + 2sinē 55 – v2 sin65 is
(a) 0
(b) 1
(c) -1
(d) None of these

Answer

Answer: (b) 1
Hint:
Given, cos 20 + 2sinē 55 – v2 sin65
= cos 20 + 1 – cos 110 – v2 sin65 {since cos 2x = 1 – 2sinē x}
= 1 + cos 20 – cos 110 – v2 sin65
= 1 – 2 Ũ sin {(20 + 110)/2 Ũ sin{(20 – 110)/2} – v2 sin65 {Apply cos C – cos D formula}
= 1 – 2 Ũ sin 65 Ũ sin (-45) – v2 sin65
= 1 + 2 Ũ sin 65 Ũ sin 45 – v2 sin65
= 1 + (2 Ũ sin 65)/v2 – v2 sin65
= 1 + v2 ( sin 65 – v2 sin 65
= 1
So, cos 20 + 2sinē 55 – v2 sin65 = 1


Question 17.
If the radius of the circumcircle of an isosceles triangle PQR is equal to PQ ( = PR), then the angle P is
(a) 2p/3
(b) p/3
(c) p/2
(d) p/6

Answer

Answer: (a) 2p/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°


Question 18.
If cos a + 2cos b + cos c = 2 then a, b, c are in
(a) 2b = a + c
(b) bē = a Ũ c
(c) a = b = c
(d) None of these

Answer

Answer: (a) 2b = a + c
Hint:
Given, cos A + 2 cos B + cos C = 2
? cos A + cos C = 2(1 – cos B)
? 2 cos((A + C)/2) Ũ cos((A-C)/2 = 4 sinē (B/2)
? 2 sin(B/2)cos((A-C)/2) = 4sinē (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{v(s-b)(s-c)vbc} Ũ {v(s-a)(s-b)vab} = v(s-a)(s-c)vac
? 2(s – b) = b
? a + b + c – 2b = b
? a + c – b = b
? a + c = 2b


Question 19.
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 : (v5 – 1)
(b) 5 : 4
(c) (v5 – 1) : 4
(d) none of these

Answer

Answer: (a) 4 : (v5 – 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 = (v5 – 1)/4 : {v(10 + 2v5)}/4 : 1
? a : b : c = (v5 – 1) : {v(10 + 2v5)} : 4
Now, c /a = 4/(v5 – 1)
? c : a = 4 : (v5 – 1)


Question 20.
The value of cos 180° is
(a) 0
(b) 1
(c) -1
(d) infinite

Answer

Answer: (c) -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


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