Trigonometry Questions and Answers

Q1 :

The value of $theta left(0 leq theta leq frac{pi}{2} right)$ satisfying the equation

$text{sin}^2 theta - 2 text{ cos} theta + frac{1}{4} = 0 text{ , is}$

A

$frac{pi}{2}$

B

$frac{pi}{3}$

C

$frac{pi}{4}$

D

$frac{pi}{6}$

View Answer

Correct Answer: B

$frac{pi}{3}$

Description:

The equation can be written as

$1 - text{cos}^2 theta - 2 text{ cos } theta + frac{1}{4} = 0 text{ is}$

$Rightarrow text{ cos }^2 theta - 2 text{ cos } theta - frac{5}{4} = 0$

$Rightarrow text{ cos } theta = frac{-2 pm sqrt{4 + 5}}{2}$

$left[therefore x - frac{-b pm sqrt {D}}{2a} right]$

$= frac{-2 pm 3}{2} = frac{1}{2} Rightarrow theta = frac{pi}{3}$

Q2 :

The expression for sin $theta$ in terms of sec $θ$

A

$frac{1}{sqrt{text{ sec}^2 theta - 1 }}$

B

$frac{text{ sec}^2 theta}{sqrt{text{ sec}^2 theta - 1 }}$

C

$frac{sqrt{text{ sec}^2 theta - 1 }}{text{ sec} theta}$

D

$sqrt{text{ sec}^2 theta - 1 }$

View Answer

Correct Answer: C

$frac{sqrt{text{ sec}^2 theta - 1 }}{text{ sec} theta}$

Description:

$text{sin} theta = sqrt{ 1 - text{cos}^2 theta }$

$= sqrt{ 1 - frac{1}{text{sec}^2 theta} }$

$= sqrt{frac{ text{sec}^2 theta - 1}{text{sec}theta}}$

Q3 :

The value of $frac{text{sin } 45^circ + text{ sin } 105^circ + text{ cos } 105^circ text{ cos } 45^circ} { text{sin } 75^circ + text{ cos } 75^circ }$ is equal to

A

$frac{1}{sqrt6}$

B

$frac{1}{sqrt3}$

C

$frac{1}{2sqrt3}$

D

$frac{1}{3sqrt2}$

View Answer

Correct Answer: A

$frac{1}{sqrt6}$

Description:

$=frac{text{cos } (105 - 45)}{sqrt2 text{ cos } 30^circ } = frac{text {cos } 60}{sqrt2 text{ cos } 30^circ}$

$=dfrac{frac{1}{2}}{sqrt2 times frac{sqrt3}{2}} = dfrac{1}{sqrt6}$

Q4 :

If $text{ tan}x + text{cot}x = 3, text{then sec}^2x + text{cosec}^2x$ is equal to

A

3

B

9

C

12

D

15

View Answer

Correct Answer: B

9

Description:

$= (text{tan }x + text{cot }x)^2 = (3)^2 = 9$

$= text{tan}^2x + text{cot}^2x + 2 text{ tan}x . text{cot}x = 9$

$= text{sec}^2x + text{cosec}^2x = 9$

Q5 :

Two boys are on opposite sides of a tower of 100 meter height. If they measure the elevation of the top of the tower as $30^circ$ and $45^circ$ respectively, the distance between the tower is 200 metres, then the distance between the boys is

A

$100sqrt3 text{ metres}$

B

$100(sqrt3 +1) text{ metres}$

C

$100(sqrt3 -1) text{ metres}$

D

$100sqrt3 -1 text{ metres}$

View Answer

Correct Answer: C

$100(sqrt3 -1) text{ metres}$

Description:

$AQ = DA = 100 text{ m }$

$PB = 100sqrt3 text{ m }$

$text{ Distance between } PQ = AQ + PB -AB$

$= 100 + 100sqrt3 - 200$

$= 100sqrt3 - 100$

$= 100(sqrt3 - 1) text{ metre }$

Q6 :

For $lambda ne 0$ , the angle between the lines given by the equation $lambda y^2 + (1 - lambda^2) ; xy - lambda x^2 = 0$ is

A

$30degree$

B

$45degree$

C

$60degree$

D

$90degree$

View Answer

Correct Answer: D

$90degree$

Description:

The equation of the pair of straight lines given by

$ax^2 + 2hxy + by^2 = 0$

the angle $theta$ between them is given by

$text{tan } theta = dfrac{2sqrt{h^2 - ab}}{a + b}$

According to question, $a = -lambda, b = lambda$

$Rightarrow a + b = 0$

$therefore theta = dfrac{pi}{2}$

Q7 :

If $text{cos } alpha + text{sec } alpha = 2$ , then the value of $text{cos}^8 alpha + text{sec}^8 alpha$ is equal to

A

$2$

B

$2^2$

C

$2^4$

D

$2^8$

View Answer

Correct Answer: A

$2$

Description:

Expression, $text{cos } x + text{sec } x = 2$

$Rightarrow (text{cos } alpha + text{sec } alpha)^2 = 4$

$Rightarrow text{cos}^2 alpha + text{sec}^2 alpha + 2 = 4$

$Rightarrow text{cos}^2 alpha + text{sec}^2 alpha = 2$

Squaring again, we get

$text{cos}^4 alpha + text{sec}^4 alpha = 2$

Squaring again, we get

$text{cos}^8 alpha + text{sec}^8 alpha = 2$

Q8 :

The numerical value of the expression

$dfrac{text{sin}9degree}{text{sin}48degree} - dfrac{text{cos}81degree}{text{cos}42degree}$ $frac{text{sin}9degree}{text{sin}48degree} - frac{text{cos}81degree}{text{cos}42degree}$ is

A

1

B

1/2

C

0

D

-1

View Answer

Correct Answer: C

0

Description:

$sin(90 - theta) = costheta$

$therefore sin9degree = sin (90degree - 81degree) = cos81degree$

$sin48degree = sin (90degree - 42degree) = cos42degree$

$therefore$ Given expression

$= frac{cos81degree}{cos42degree} - frac{cos81degree}{cos42degree} = 0$

Q9 :

An angle is divided into two parts $alpha$ and $beta$ in such $alpha$ way that $tan alpha = frac{1}{2}$ and $tan beta = 2$ . The measure of the angle is

A

$frac{2pi}{3}$

B

$frac{pi}{2}$

C

$pi$

D

$frac{3pi}{4}$

View Answer

Correct Answer: B

$frac{pi}{2}$

Description:

$tan (alpha + beta) = frac{tan alpha + tan beta}{1 - tan alpha tan beta}$

$= dfrac{frac{1}{2} + 2}{1 - frac{1}{2} times 2} = infty$

Hence $alpha + beta = frac{pi}{2}$

Q10 :

If $alpha + beta = 90degree$ , then $text{cosec}^2alpha + text{cosec}^2beta$ is equal to

A

$text{cosec}^2alpha + text{cosec}^2beta$

B

$text{sin}^2alpha + text{sin}^2beta$

C

$text{tan}^2alpha + text{tan}^2beta$

D

$text{sec}^2alpha + text{sec}^2beta$

View Answer

Correct Answer: A

$text{cosec}^2alpha + text{cosec}^2beta$

Description:

$text{cosec}^2 alpha + cos^2 beta = dfrac{1}{sin^2 alpha} + dfrac{1}{sin^2 beta}$

$(because alpha + beta = 90degree )$

$=dfrac{1}{sin^2 alpha} + dfrac{1}{cos^2 alpha} = dfrac{1}{sin^2 alpha cos^2 alpha}$

$=dfrac{1}{sin^2 alpha sin^2beta} quadquad [because cos^2 alpha = sin^2 beta]$

$=text{cosec}^2 alpha text{ cosec}^2 beta$

Q11 :

If $sin 2theta =cos 3theta$ and $theta$ is an acute angle, then $theta$ is equal to

A

$18degree$

B

$27degree$

C

$36degree$

D

$45degree$

View Answer

Correct Answer: A

$18degree$

Description:

We know that $sintheta = cos (90degree - theta)$

$therefore sin 2theta = cos 3theta$

$Rightarrow cos (90degree - 2theta) = cos 3theta$

$Rightarrow 2theta = 90degree - 3theta$

$Rightarrow theta = 18degree$

Q12 :

If sec $11theta = text{ cosec } 7theta (0degree lt theta lt 20degree)$ , then the value of $theta$ is

A

$5degree$

B

$10degree$

C

$15degree$

D

$18degree$

View Answer

Correct Answer: A

$5degree$

Description:

Given $dfrac{1}{cos 11theta} = dfrac{1}{sin 7theta} = dfrac{1}{cos (90- 7theta) }$

$Rightarrow 11theta = 90 - 7theta$

$Rightarrow theta = 5degree$

Q13 :

The maximum value of $sin theta cdot cos theta$ is

A

1

B

$frac{1}{2}$

C

$frac{1}{sqrt{2}}$

D

$frac{sqrt{3}}{2}$

View Answer

Correct Answer: B

$frac{1}{2}$

Description:

By definition, for Maxima or Minima,

$frac{d}{dtheta} (sin theta cos theta ) = 0$

$Rightarrow cos^2 theta - sin^2 theta = 0$

$Rightarrow cos theta = sin theta Rightarrow theta = 45$

$therefore text{max } (sin theta cos theta )$

$=sin 45^circ cos 45^circ = frac{1}{2}$

Q14 :

The value of $sin^3 (15degree) - cos^3 (15degree)$ is

A

$frac{3}{4}(sin 15degree - cos 15degree )$

B

$frac{5}{8sqrt{2}}$

C

$-frac{5}{8sqrt{2}}$

D

$-frac{5}{4sqrt{2}}$

View Answer

Correct Answer: D

$-frac{5}{4sqrt{2}}$

Description:

$sin^3 15 - cos^3 15 = (sin 15 - cos 15)(1 + sin 15 cos 15)$

$= (sin 15 - cos 15) Big(1 + frac{1}{2}sin 30Big)$

$=frac{5}{4}(sin 15 - cos 15)$

$=frac{5}{4}Big[pmsqrt{1 - sin 30}Big]$

$=pmfrac{5}{4sqrt{2}} = - frac{5}{4sqrt{2}} ; text{(Neglecting +ve sign)}$

Q15 :

The value of $frac{3 tan 20degree - tan^3 20 degree}{1 - 3tan^2 20}$ is equal to

A

$frac{1}{sqrt{3}}$

B

1

C

$sqrt{3}$

D

$frac{2}{sqrt{3}}$

View Answer

Correct Answer: C

$sqrt{3}$

Description:

We know that $tan 3theta =frac{3 tan theta - tan^3 theta}{1 - 3 tan^2theta}$

Here, $theta = 20degree$

$therefore frac{3 tan 20degree - tan^3 20degree }{1 - 3 tan^2 20degree}$

$=tan 3theta = tan 60degree = sqrt{3}$

Q16 :

If $2 cos^2 theta + 11 sin theta - 7 = 0$ , then the value of $sin theta$ is equal to

A

$frac{-1}{2}$

B

$frac{1}{2}$

C

5

D

$frac{1}{sqrt{2}}$

View Answer

Correct Answer: B

$frac{1}{2}$

Description:

The given equation can be written as

$2 - 2 sin^2 theta + 11 sin theta - 7 = 0$

$Rightarrow 2 sin^2 theta - 11 sin theta + 5 = 0$

$Rightarrow sin theta = frac{11 pm sqrt{121-40}}{4}$

$=frac{11 pm 9}{4} = 5 text{ or } frac{1}{2}$

Since $sin theta$ can not be greater than 1,

$therefore sin theta = frac{1}{2}$

Q17 :

The angle between the hour and minute hands of a clock at 02 : 15 hour is

A

$15degree$

B

$7frac{1}{2}degree$

C

$22frac{1}{2}degree$

D

$30degree$

View Answer

Correct Answer: C

$22frac{1}{2}degree$

Description:

The angle between two consecutive hour marks $=frac{360}{12} = 30degree$
By question, minute hand is at 3 and hour hand slightly ahead of 2.

$therefore$ In 15 minutes hour hand moves by an angle of $frac{30}{60} times 15 = 7frac{1}{2}degree$

$therefore$ Required angle $= 30degree - 7frac{1}{2}degree =22frac{1}{2}degree$

Q18 :

An aeroplane at a height of 3000 m, passes vertically above another aeroplane at an instant. If the angles of elevation of the two aeroplanes from some point on the ground are $60degree$ and $45degree$ , respectively, then the vertical distance between the two planes is:

A

$1000 Big(sqrt{3 - 1}Big) text{ m }$

B

$1000 sqrt{3} text{ m }$

C

$1000 Big(3 - sqrt{3}Big) text{ m }$

D

$1000 sqrt{3} text{ m }$

View Answer

Correct Answer: C

$1000 Big(3 - sqrt{3}Big) text{ m }$

Description:

$begin{aligned} text{Let} quad text{AD} &= h therefore quad text{BD} &= (3000 - h) text{Now, } tan 45degree &= frac{text{BD}}{text{CB}} therefore quad text{ BD} &= text{CB} = (3000 - h) text{m} end{aligned}$

Again

$quad text{tan } 60degree = frac{text{AB}}{text{BC}}$

$Rightarrow quad text{V}_3 = frac{3000}{3000 - h}$

$Rightarrow quad h =frac{3000}{1} - frac{3000}{sqrt{3}} =frac{3000sqrt{3} - 3000}{sqrt{3}}$

$= 1000 (3 - sqrt3)$

Q19 :

If $tan theta + sec theta = 2, 0 le theta le frac{pi}{2}$ ; then the value of $tantheta$ is equal to :

A

$frac{3}{4}$

B

$frac{5}{4}$

C

$frac{3}{2}$

D

$frac{5}{2}$

View Answer

Correct Answer: A

$frac{3}{4}$

Description:

We know that

$sec^2theta - tan^2 theta = 1$
and $sec theta + tan theta = 2$ (given) ...(i)

then, $frac{sec^2 theta - tan^2 theta}{sec theta + tan theta} = frac{1}{2}$

$therefore sec theta - tan theta = frac{1}{2}$ ...(ii)

Subtracting (ii) from (i)

$2 tan theta = frac{3}{2} Rightarrow tan theta = frac{3}{4}$

Q20 :

The value of $(cos^2 25degree + cos^2 65degree)$ is:

A

0

B

$sin^2 40degree$

C

$cos^2 40degree$

D

1

View Answer

Correct Answer: D

1

Description:

$cos^2 25degree = cos^2 (90degree - 25degree)$

$= sin^2 65degree$

$therefore cos^2 25degree + cos^2 65degree = sin^2 65degree + cos^2 65degree$

$= 1 (because sin^2 theta + cos^2 theta = 1)$

Q21 :

If $frac{cos^2theta - 3costheta + 2}{sin^2theta} = 1, theta, 0$ then $theta$ is :

A

$30degree$

B

$60degree$

C

$75degree$

D

$90degree$

View Answer

Correct Answer: B

$60degree$

Description:

Given, $dfrac{cos^2theta - 3costheta + 2}{sin^2theta} = 1$

$Rightarrow cos^2theta - 3costheta + 2 = sin^2theta = 1 - cos^2 theta$

$Rightarrow 2 cos^2 theta - 3 cos theta + 1 = 0$

$Rightarrow 2 cos^2 theta - 2 cos theta - cos theta + 1 = 0$

$Rightarrow (2 cos theta - 1)(cos theta - 1) = 0$ either $cos theta = 1 quad text{ or } (2 cos theta - 1) = 0$ but $cos theta ne 1$

$therefore cos theta = 1$

$therefore cos theta = frac{1}{2} = cos 60degree$

$therefore theta = 60^degree$

Q22 :

If a and b are positive, then the relation $sin theta = (text{a} + 2text{b})/2text{b }$ is

A

not possible

B

possible only if a = b

C

possible if a < b

D

possible if b > a

View Answer

Correct Answer: A

not possible

Description:

$sin theta = frac{text{a} + 2text{b}}{2text{b}} = frac{text{a}}{2text{b}} + 1 gt 1$

$[because text{ a } text{ and } text{ b } text{ are the } therefore frac{text{a}}{2text{b}} text{ in the} ]$

But $sin theta$ lies between -1 and 1

$therefore$ Not possible

Q23 :

If the angle of elevation of the top of a tower at a distance 100 metres from its foot is $45degree$ , then the height, in metres, of the tower is equal to

A

100

B

75

C

4500

D

50

View Answer

Correct Answer: A

100

Description:

$frac{h}{100} = tan 45degree$

$Rightarrow h = 100 text{ m }$

Q24 :

The expression $frac{cos^3 theta - sin^3 theta}{cos theta - sin theta}$ simplifies to

A

$1 + frac{sin theta}{2}$

B

$1 + frac{sin 2 theta}{2}$

C

$frac{1 + sin theta}{2}$

D

$frac{1 + sin 2 theta}{2}$

View Answer

Correct Answer: B

$1 + frac{sin 2 theta}{2}$

Description:

$frac{cos^3 theta - sin^3 theta}{cos theta - sin theta}$

$= frac{(cos theta - sin theta) (cos^2 theta + sin^2 theta sin theta cos theta) }{cos theta - sin theta}$

$= 1 + frac{2 sin theta cos theta}{2} (because sin^2 theta + cos^2 theta = 1)$

$= 1 + frac{sin 2 theta}{2}$

Q25 :

If tan A + cot A = 2, then the value of sec A, is equal to

A

$1$

B

$sqrt2$

C

$frac{1}{2}$

D

$frac{sqrt3}{2}$

View Answer

Correct Answer: B

$sqrt2$

Description:

tan A + cot A = 2

$therefore frac{sin A}{cos A} + frac{cos A}{sin A} =2 Rightarrow sin^2 A + cos^2 A$

$=2 sin A cos A Rightarrow 1 = sin 2A Rightarrow sin 2A$

$=sin 90degree Rightarrow 2A = 90 Rightarrow A = cos 45degree$

$= sec A = sec 45 = sqrt2$