Trigonometry Questions and Answers for Effective Learning

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:

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

$(because alpha + beta = 90degree )$

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

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

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