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51.

sin 75° – sin 15° =_________________(a) \(\frac{\sqrt{3}}{\sqrt{2}}\)(b) \(\frac{\sqrt{2}}{\sqrt{3}}\)(c) \(\frac{\sqrt{3}}{2}\)(d) \(\frac{1}{\sqrt{2}}\)I have been asked this question in final exam.This is a very interesting question from Trigonometric Functions of Sum and Difference of Two Angles-2 topic in chapter Trigonometric Functions of Mathematics – Class 11

Answer» RIGHT CHOICE is (d) \(\frac{1}{\sqrt{2}}\)

To EXPLAIN: We KNOW, sin x – sin y = 2 cos (x+y)/2 sin (x-y)/2

sin 75° – sin 15° = 2 cos (75°+15°)/2SIN(75°-15°)/2

= 2 cos 45° sin 30°

= 2*1/\(\sqrt{2}\)*1/2 = 1/\(\sqrt{2}\).
Discussion
52.

Find cos 2x if tan x=1/\(\sqrt{3}\).(a) 1/2(b) 0(c) \(\sqrt{3}\)/2(d) 1This question was posed to me in an internship interview.The question is from Trigonometric Functions of Sum and Difference of Two Angles-1 in chapter Trigonometric Functions of Mathematics – Class 11

Answer»

Right answer is (a) 1/2

Easy explanation: We know, COS 2X = cos^2x – sin^2x = (cos^2x – sin^2x)/(cos^2x+sin^2x) {1 = sin^2x + cos^2x}

= (1-tan^2x)/(1+tan^2x)

= (1-(1/\(\sqrt{3}\))^2)/(1+(1/\(\sqrt{3}\))^2)

= (1-1/3)/(1+1/3) = (2/3)/(4/3) = 1/2.
Discussion
53.

cot 15° =______________(a) 2+\(\sqrt{3}\)(b) 2-\(\sqrt{3}\)(c) 1+\(\sqrt{3}\)(d) \(\sqrt{3}\)-1I have been asked this question during an interview for a job.My enquiry is from Trigonometric Functions of Sum and Difference of Two Angles-1 topic in portion Trigonometric Functions of Mathematics – Class 11

Answer»

Correct option is (a) 2+\(\SQRT{3}\)

The explanation: We KNOW, cot (X – y) = (cot x cot y +1)/cot y – cot x)

cot(45°-30°) = (cot 45° cot 30°+1)/(cot 45° – cot 30°)

cot 15° = (\(\sqrt{3}\) + 1)/(\(\sqrt{3}\) – 1) = 2+\(\sqrt{3}\).
Discussion
54.

1-cos^2x=_________(a) sin x(b) cos x(c) sin 2x(d) sin^2xI have been asked this question in a job interview.I'd like to ask this question from Trigonometric Functions in portion Trigonometric Functions of Mathematics – Class 11

Answer» RIGHT CHOICE is (d) sin^2x

Explanation: We KNOW, sin^2x+ cos^2x=1

So, 1-cos^2x=sin^2x.
Discussion
55.

If minute hand covers 24 cm length in 30 minutes, then how much length minute hand have?(a) 19.1 cm(b) 38.2 cm(c) 57.3 cm(d) 45 cmI got this question during a job interview.The doubt is from Trigonometric Functions topic in chapter Trigonometric Functions of Mathematics – Class 11

Answer»

Right option is (b) 38.2 cm

The best I can EXPLAIN: In 60 minutes, ANGLE covered by MINUTE hand is 360°.

In 30 minutes, angle covered by minute hand is 180°.

L=24 cm, θ=180°=π

θ=L/r => π=24/r => r=24/π = 38.2 cm.
Discussion
56.

sin (75°) =__________________(a) (1 – \(\sqrt{3}\))/2\(\sqrt{2}\)(b) (\(\sqrt{3}\) + 1)/2\(\sqrt{2}\)(c) (\(\sqrt{3}\) – 1)/2\(\sqrt{2}\)(d) (- \(\sqrt{3}\) – 1)/2\(\sqrt{2}\)I got this question during an online interview.My question comes from Trigonometric Functions of Sum and Difference of Two Angles-1 topic in chapter Trigonometric Functions of Mathematics – Class 11

Answer»

The CORRECT OPTION is (B) (\(\sqrt{3}\) + 1)/2\(\sqrt{2}\)

The explanation: sin (75°) = sin (45°+30°) = sin45° cos30° + cos45° sin30°

= (1/\(\sqrt{2}\) * \(\sqrt{3}\)/2) + (1/\(\sqrt{2}\) * 1/2) {sin(X + y)=sin x cos y + cos x sin y}

= (\(\sqrt{3}\) + 1)/2\(\sqrt{2}\).
Discussion
57.

Find the principal solutions of the equation sec x = -2.(a) 2π/3, 4π/3(b) 2π/3, 5π/3(c) 4π/3, 5π/3(d) π/3, 5π/3This question was addressed to me in semester exam.The doubt is from Trigonometric Equations in section Trigonometric Functions of Mathematics – Class 11

Answer»

Right OPTION is (a) 2π/3, 4π/3

Best explanation: SINCE SEC X is negative in 2^nd and 3^rd QUADRANT so, there are two principal solutions to above equation i.e. x = π – π/3, π + π/3. So, x=2π/3, 4π/3.
Discussion
58.

Find general solution to equation cos x = – 1/√2.(a) x = 2nπ±7π/4(b) x = 2nπ±3π/4(c) x = 2nπ±π/4(d) x = 2nπ±π/3I have been asked this question in a job interview.My query is from Trigonometric Equations topic in chapter Trigonometric Functions of Mathematics – Class 11

Answer»

Correct option is (c) X = 2nπ±π/4

The BEST EXPLANATION: cos x = – 1/√2

cos x = – cos π/4 = cos (π- π/4) = cos 3π/4

So, x = 2nπ±π/4.
Discussion
59.

Find the principal solutions of the equation tan x = – √3.(a) π/6, 5π/6(b) π/3, 5π/3(c) π/3, 2π/3(d) 2π/3, 5π/3This question was posed to me in a job interview.My query is from Trigonometric Equations in section Trigonometric Functions of Mathematics – Class 11

Answer»

The CORRECT CHOICE is (d) 2π/3, 5π/3

Explanation: Since tan X is negative in 2^nd and 4^th quadrant so, there are two PRINCIPAL solutions to above equation i.e. x = π – π/3, 2π – π/3. So, x=2π/3, 5π/3.
Discussion
60.

cos 75° – cos 15° =___________________(a) \(\frac{\sqrt{3}}{\sqrt{2}}\)(b) \(\frac{\sqrt{2}}{\sqrt{3}}\)(c) \(\frac{1}{\sqrt{2}}\)(d) –\(\frac{1}{\sqrt{2}}\)The question was asked in unit test.Enquiry is from Trigonometric Functions of Sum and Difference of Two Angles-2 in division Trigonometric Functions of Mathematics – Class 11

Answer» RIGHT choice is (d) –\(\frac{1}{\SQRT{2}}\)

The EXPLANATION: cos X – cos y = – 2 sin (x+y)/2 sin (x-y)/2

cos 75° – cos 15° = – 2 sin (75°+15°)/2sin(75°-15°)/2

= – 2 sin 45° sin 30°

= – 2*1/\(\sqrt{2}\)*1/2= – 1/\(\sqrt{2}\).
Discussion
61.

Find the principal solutions of the equation cos x = 1/2.(a) π/6, 5π/6(b) π/3, 5π/3(c) π/3, 2π/3(d) 2π/3, 5π/3The question was asked during an online interview.Question is taken from Trigonometric Equations in section Trigonometric Functions of Mathematics – Class 11

Answer»

Correct answer is (b) π/3, 5π/3

Explanation: Since COS x is POSITIVE in 1^st and 4^th quadrant, so there are two PRINCIPAL solutions to above EQUATION i.e. x = π/3, 2π-π/3. So, x = π/3, 5π/3.
Discussion
62.

Find cos 2x if sin x=1/2.(a) 1/2(b) 1/\(\sqrt{2}\)(c) \(\sqrt{3}\)/2(d) 1The question was asked during an internship interview.The query is from Trigonometric Functions of Sum and Difference of Two Angles-1 topic in division Trigonometric Functions of Mathematics – Class 11

Answer» CORRECT CHOICE is (c) \(\SQRT{3}\)/2

The best EXPLANATION: We know, cos 2X = cos^2x – sin^2x = 1-2sin^2x {cos^2x = 1-sin^2x}

= 1-2(1/2)^2 = 1-2(1/4) = 1-1/2 = 1/2.
Discussion
63.

If sin x= 1/2, then sin 3x =_______________(a) \(\sqrt{3}\)/2(b) 1/2(c) 1/\(\sqrt{2}\)(d) 1This question was addressed to me by my college professor while I was bunking the class.My question is from Trigonometric Functions of Sum and Difference of Two Angles-2 in portion Trigonometric Functions of Mathematics – Class 11

Answer»

The correct OPTION is (d) 1

The BEST explanation: We know, SIN 3x = 3SIN X – 4sin^3x

= 3(1/2) – 4(1/2)^3

= 3/2 – 4/8 = 3/2 – 1/2 = 1.
Discussion
64.

If cosec x = -5/12 and x lies in 2^nd quadrant, then find cos x.(a) 12/13(b) 5/13(c) -13/5(d) 12/13I have been asked this question in final exam.My enquiry is from Trigonometric Functions in portion Trigonometric Functions of Mathematics – Class 11

Answer» CORRECT answer is (c) -13/5

For EXPLANATION: COSEC x = 5/12

sin x = 1/cosec x = 12/5

We know, sin^2x+COS^2x=1

cos^2 x = 1-(-12/5)^2 = 1+144/25 = 169/25

cos x = ±13/5

cos x is negative in 2^nd quadrant so, cos x=-13/5.
Discussion
65.

tan(75°) =___________________(a) 2+\(\sqrt{3}\)(b) 2-\(\sqrt{3}\)(c) 1+\(\sqrt{3}\)(d) \(\sqrt{3}\)-1The question was posed to me in homework.My enquiry is from Trigonometric Functions of Sum and Difference of Two Angles-1 in chapter Trigonometric Functions of Mathematics – Class 11

Answer» RIGHT choice is (a) 2+\(\sqrt{3}\)

BEST EXPLANATION: tan (X +y) = (tan x + tan y)/(1- tan x tan y)

tan (45°+30°) = (tan 45° + tan 30°)/(1- tan 45° tan 30°)

tan 75° = (1+ 1/\(\sqrt{3}\))/(1-1/\(\sqrt{3}\)) = (\(\sqrt{3}\) + 1)/ (\(\sqrt{3}\) – 1) = 2+\(\sqrt{3}\).
Discussion
66.

sin (-45°) = ______________(a) 1(b) -1(c) 1/√2(d) -1/√2This question was addressed to me during an interview for a job.My query is from Trigonometric Functions topic in division Trigonometric Functions of Mathematics – Class 11

Answer» CORRECT ANSWER is (d) -1/√2

For EXPLANATION: We know, sin(-x) = sin x

So, sin (-45°) = -sin 45° = -1/√2.
Discussion
67.

sin(15°) =_________________(a) (1 – \(\sqrt{3}\))/2\(\sqrt{2}\)(b) (\(\sqrt{3}\) + 1)/2\(\sqrt{2}\)(c) (\(\sqrt{3}\) – 1)/2\(\sqrt{2}\)(d) (- \(\sqrt{3}\) – 1)/2\(\sqrt{2}\)The question was posed to me by my school principal while I was bunking the class.My question is based upon Trigonometric Functions of Sum and Difference of Two Angles-1 in division Trigonometric Functions of Mathematics – Class 11

Answer»

Correct OPTION is (c) (\(\SQRT{3}\) – 1)/2\(\sqrt{2}\)

The best I can explain: SIN (15°) = sin (45°-30°) = sin45° cos30° – cos45° sin30°

= (1/\(\sqrt{2}\) * \(\sqrt{3}\)/2) – (1/\(\sqrt{2}\) * 1/2) {sin(x – y)=sin x cos y – cos x sin y}

= (\(\sqrt{3}\) -1)/2\(\sqrt{2}\).
Discussion
68.

cot^2x – cosec^2x = __________(a) 1(b) -1(c) sin^2x(d) cos^2xI have been asked this question in an online quiz.I'm obligated to ask this question of Trigonometric Functions in section Trigonometric Functions of Mathematics – Class 11

Answer»

The correct choice is (B) -1

Easiest EXPLANATION: We know, cosec^2x – cot^2x = 1

So, cot^2x – cosec^2x = -1.
Discussion
69.

If sin x=0 then x = ________(a) nπ(b) (2n+1) π/2(c) (n+1) π(d) nπ/2The question was asked in final exam.I would like to ask this question from Trigonometric Functions in section Trigonometric Functions of Mathematics – Class 11

Answer» RIGHT ANSWER is (a) nπ

For EXPLANATION: When KNOW, sin x =0 whenever x is 0, π, 2π, 3π,….. i.e. all integral multiples of π so, x=nπ when sin x=0.
Discussion
70.

2 sin 75° sin15° =_____________________________(a) \(\frac{\sqrt{3}}{2}\)(b) \(\frac{\sqrt{2}}{\sqrt{3}}\)(c) \(\frac{1}{2}\)(d) \(\frac{1}{\sqrt{2}}\)This question was addressed to me in my homework.I need to ask this question from Trigonometric Functions of Sum and Difference of Two Angles-2 topic in section Trigonometric Functions of Mathematics – Class 11

Answer»

The correct CHOICE is (c) \(\frac{1}{2}\)

The explanation: We KNOW, 2 sin X sin y = cos (x – y) – cos (x + y)

2 sin 75° sin15° = cos (75°-15°) – cos (75°+15°)

= cos 60° – cos 90° = 1/2 – 0 = 1/2.
Discussion
71.

Which is correct?(a) cos 3x = 3cosx – 4cos^3x(b) cos 3x = 4cosx – 3cos^3x(c) cos 3x = 3cos^3x – 4cosx(d) cos 3x = 4cos^3x – 3cosxI got this question by my school teacher while I was bunking the class.This is a very interesting question from Trigonometric Functions of Sum and Difference of Two Angles-2 topic in division Trigonometric Functions of Mathematics – Class 11

Answer»

The correct option is (d) cos 3X = 4cos^3x – 3cosx

Easy explanation: cos 3x = cos (2x +x)

= cos 2x cos x – sin 2x sin x

= (2COS^2 x – 1) cos x – 2SIN x cos x sin x

= (2cos^2 x – 1) cos x – 2cos x (1 – cos^2 x)

= 2cos^3 x – cos x – 2cos x + 2 cos^3 x

= 4cos^3 x – 3cos x.
Discussion
72.

cosec^2x – 1 = ______________(a) cot^2x(b) -cot^2x(c) tan^2x(d) -tan^2xI had been asked this question in an online quiz.My doubt is from Trigonometric Functions in division Trigonometric Functions of Mathematics – Class 11

Answer» CORRECT ANSWER is (a) cot^2x

To EXPLAIN: We KNOW, cosec^2x – cot^2x = 1

So, cosec^2x – 1 = cot^2x.
Discussion
73.

Is cos (90° – x) = sin x.(a) True(b) FalseThe question was posed to me in quiz.The question is from Trigonometric Functions of Sum and Difference of Two Angles-1 in section Trigonometric Functions of Mathematics – Class 11

Answer» CORRECT option is (a) True

For EXPLANATION: cos (90° – X) = cos 90° cos x + sin 90° sin x {cos(x – y)=cos x cos y + sin x sin y}

= 0*cos x + 1*sin x

= sin x.
Discussion
74.

Which one is correct for Napier’s Analogy?(a) tan (A/2) = (b – c)/(b + c) (tan(B + C)/2)(b) tan (A/2) = (b – c)/(b + c) (cot(B – C)/2)(c) tan (A/2) = (b + c)/(b – c) (cot(B – C)/2)(d) tan (A) = (b – c)/(b + c) (tan(B – C)/2)I have been asked this question in a national level competition.This interesting question is from Trigonometric Equations topic in division Trigonometric Functions of Mathematics – Class 11

Answer»

Correct answer is (b) tan (A/2) = (b – c)/(b + c) (COT(B – C)/2)

For explanation I WOULD say: ACCORDING to the law of sines, in any TRIANGLE ABC,

a/sinA = b/sinB = c/sinC

So, a/b = sinA/sinB

(a + b)/(a – b) = (sinA + sinB)/( sinA – sinB)

=> (a + b)/(a – b) = (2 sin((a + b)/2) cos((a – b)/2))/ (2 sin((a + b)/2) sin((a – b)/2))

=> (a + b)/(a – b) = (tan(A + B)/2)/(tan(A – B)/2)

=> (tan(A + B)/2) = (a + b)/(a – b) (tan(A – B)/2)

=> (tan(π/2 + C/2)) = (a + b)/(a – b) (tan(A – B)/2)

=> cot (C/2) = (a + b)/(a – b) (tan(A – B)/2)

=> tan (C/2) = (a – b)/(a + b) (tan(A – B)/2)

SIMILARLY, tan (A/2) = (b – c)/(b + c) (cot(B – C)/2).
Discussion
75.

What is the value of (1 + cotA)(1 + cotB) for isosceles right triangle ABC right angled at A?(a) 0(b) 1(c) 2(d) Data inadequateThe question was asked in an interview.My question comes from Trigonometric Equations topic in section Trigonometric Functions of Mathematics – Class 11

Answer»

The correct answer is (c) 2

The best explanation: (1 + COTA)(1 + cotB)

SIMPLIFYING the above EQUATION,

= 1 + cotA + cotB + cotAcotB

Putting the values of A and B, we get,

= 1 + cot90 + cot45 + cot90cot45

= 1 + 0 + 1 + 0

= 2.
Discussion
76.

Which is correct?(a) sin 3x = 3sinx – 4sin^3x(b) sin 3x = 4sinx – 3sin^3x(c) sin 3x = 3sin^3x – 4sinx(d) sin 3x = 4sin^3x – 3sinxThis question was addressed to me in an online quiz.The query is from Trigonometric Functions of Sum and Difference of Two Angles-2 topic in chapter Trigonometric Functions of Mathematics – Class 11

Answer»

The CORRECT option is (a) SIN 3x = 3sinx – 4sin^3x

Explanation: sin 3x = sin (2x + x)

= sin 2x cos x + cos 2x sin x

= (2 sin x cos x) cos x + (1 – 2sin^2 x) sin x

= 2 sin x (1 – sin^2 x) + sin x – 2 sin^3 x

= 2 sin x – 2 sin^3 x + sin x – 2 sin^3 x

= 3 sin x – 4 sin^3 x.
Discussion
77.

tan 1560°=_________________(a) -√3(b) √3(c) 1/√3(d) -1/√3This question was posed to me in semester exam.This interesting question is from Trigonometric Functions in portion Trigonometric Functions of Mathematics – Class 11

Answer» RIGHT CHOICE is (a) -√3

To explain: TAN 1560° = tan (360°*4 + 120°) = tan 120° {tan (2nπ+X) = tan x}

= tan (180°-60°) = -tan 60° = – √3. {tan π-x = – tan x}
Discussion
78.

If tan x = -5/12 and x lies in 2^nd quadrant, then find cosec x.(a) 12/5(b) 13/5(c) -13/5(d) -12/5I have been asked this question in exam.This intriguing question originated from Trigonometric Functions topic in portion Trigonometric Functions of Mathematics – Class 11

Answer»

Right CHOICE is (b) 13/5

The explanation: tan X = -5/12

cot x = 1/tan x = 1/ (-5/12) = -12/5

We know, COSEC^2x – cot^2x = 1

cosec^2x =1+(-12/5)^2 = 1+144/25 = 169/25

cosec x = ± 13/5

cosec x is positive in 2^nd QUADRANT so, cosec x = 13/5.
Discussion
79.

If in two circles, arcs of the same length subtend angles 45° and 60° at Centre, find the ratio of their radii.(a) 2:3(b) 2:5(c) 3:4(d) 4:3This question was posed to me in class test.The doubt is from Trigonometric Functions topic in division Trigonometric Functions of Mathematics – Class 11

Answer»

The CORRECT answer is (d) 4:3

The EXPLANATION: Given θ1=45° and θ2=60°, L1=L2

We know, θ = l/r

r1 θ1 = r2 θ2

r1/r2=60/45=4/3.

So, RADII are in RATIO 4:3.
Discussion
80.

1 radian is _______________(a) 54°48’(b) 57°16’(c) 180°(d) 17°46’I have been asked this question in an internship interview.The query is from Trigonometric Functions in division Trigonometric Functions of Mathematics – Class 11

Answer»

Correct OPTION is (b) 57°16’

EXPLANATION: We KNOW, π radian = 180 DEGREE

1 radian = 180/π degree = 57.27° = 57° (0.27*60)’ = 57°16’.
Discussion