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Elementary Mathematics

51). If \( \Large \theta \) be acute and \( \Large tan \theta+cot \theta =2 \), then the value of \( \Large tan^{5} \theta +cot^{10} \theta \) is

A). 1

B). 2

C). 3

D). 4

View Answer

Correct Answer: 2

\( \Large tan \theta +cot \theta =2 \)

\( \Large => \ tan \theta +\frac{1}{tan \theta }=2 \)

\( \Large => \ tan^{2} \theta +1=2 \ tan \theta \)

\( \Large => \ tan^{2} \theta -2tan \theta +1=0 \)

\( \Large => \ (tan \theta -1)^{2}=0 \)

\( \Large => \ tan \theta =1 \)

and \( \Large cot \theta =1 \)

\( \Large tan^{5} \theta +cot^{10} \theta =(1)^{5}+(1)^{10}=1+1=2 \)

52). If \( \Large 0 ^{\circ} < \theta <90 ^{\circ} \), then all the trigonometric ratios can be obtained when

A). only \( \Large sin \theta \) is given

B). only \( \Large cos \theta \) is given

C). only \( \Large tan \theta \) is given

D). any one of the six ratios is given

View Answer

Correct Answer: any one of the six ratios is given

If \( \Large 0 ^{\circ} < \theta <90 ^{\circ} \), then all the trigonomat ratios can be obtained

when any one of the six ratios is given.

We use any of the following identity to get any trigonometric ratios

\( \Large sin^{2} \theta +cos^{2} \theta =1, \ 1+tan^{2} \theta =sec^{2} \theta \)

and \( \Large 1+cot^{2} \theta +cosec \ \theta \)

53). What is the value of \( \Large \frac{sin \theta }{1+cos \theta } +\frac{1+cos \theta }{sin \theta } \)?

A). \( \Large 2 \ cosec \theta \)

B). \( \Large 2 \ sec \theta \)

C). \( \Large sec \theta \)

D). \( \Large cosec \theta \)

View Answer

Correct Answer: \( \Large 2 \ cosec \theta \)

\( \Large Let \ f( \theta )=\frac{sin \theta }{1+cos \theta }+\frac{1+cos \theta }{sin \theta } \)

=\( \Large \frac{sin^{2} \theta +(1+cos \theta )^{2}}{sin \theta (1+cos \theta )} \)

=\( \Large \frac{sin^{2} \theta +1+cos^{2} \theta +2cos \theta }{sin \theta (1+cos \theta )} \)

=\( \Large \frac{2+2cos \theta }{sin \theta (1+cos \theta )} \)

=\( \Large \frac{2(1+cos \theta )}{sin \theta (1+cos \theta )}=2 \ cosec \theta \)

54). If \( \Large sin \theta cos \theta =\frac{\sqrt{3}}{4} \), then the value of \( \Large sin^{4} \theta + cos^{4} \theta \) is

A). \( \Large 2 cosec \theta \)

B). \( \Large 2 sec \theta \)

C). \( \Large sec \theta \)

D). \( \Large cosec \theta \)

View Answer

Correct Answer: \( \Large 2 sec \theta \)

Given that, \( \Large sin \theta .cos \theta =\frac{\sqrt{3}}{4} \) .... (i)

Now, we have \( \Large sin^{4} \theta +cos^{4} \theta \)=

\((sin^{2} \theta +cos^{2} \theta )^{2}-2sin^{2} \theta .cos^{2} \theta \)

= \( \Large (1)^{2}-2(sin \theta .cos \theta )^{2} \)

= \( \Large 1-2 \left(\frac{\sqrt{3}}{4}\right)^{2}=1-2.\frac{3}{16}=1-\frac{3}{8}=\frac{5}{8} \)

55). If \( \Large 2 \ cot \theta =3 \), then what is \( \Large \frac{2 \ cos \theta -sin \theta }{2 \ cos \theta +sin \theta } \) equal to?

A). \( \Large \frac{2}{3} \)

B). \( \Large \frac{1}{3} \)

C). \( \Large \frac{1}{2} \)

D). \( \Large \frac{3}{4} \)

View Answer

Correct Answer: \( \Large \frac{1}{2} \)

In \( \Large \triangle ABC, \ cot=\frac{3}{2} \)

AB=3 and AC=2

By Pythagoras theorem, \( \Large BC^{2}=(2)^{2}+(3)^{2} \)

\( \Large BC=\sqrt{13} \)

Now, \( \Large cos \theta =\frac{3}{\sqrt{13}} \) and \( \Large sin \theta =\frac{2}{\sqrt{13}} \)

\( \Large \frac{2cos \theta -sin \theta }{2cos \theta +sin \theta }=\frac{\frac{6}{\sqrt{13}}-\frac{2}{\sqrt{13}}}{\frac{6}{\sqrt{13}}+\frac{2}{\sqrt{13}}}=\frac{4}{8}=\frac{1}{2} \)

56). If \( \Large cos \theta +sin \theta =\sqrt{2}cos \theta \), then \( \Large cos \theta -sin \theta \) is

A). \( \Large -\sqrt{2} cos \theta \)

B). \( \Large -\sqrt{2} sin \theta \)

C). \( \Large \sqrt{2} sin \theta \)

D). \( \Large \sqrt{2} tan \theta \)

View Answer

Correct Answer: \( \Large \sqrt{2} sin \theta \)

Given, \( \Large cos \theta +sin \theta =\sqrt{2}cos \theta \) .....(i)

On squaring both sides, we get

\( \Large (cos \theta+sin \theta )^{2}=(\sqrt{2}cos \theta )^{2} \)

\( \Large cos^{2} \theta +sin^{2} \theta +2sin \theta cos \theta =2cos^{2} \theta \)

\( \Large 2sin \theta cos \theta =cos^{2} \theta -sin^{2} \theta \)

\( \Large 2sin \theta cos \theta =(cos \theta -sin \theta )(cos \theta +sin \theta ) \)

\( \Large cos \theta -sin \theta =\frac{2sin \theta cos \theta }{(cos \theta +sin \theta )}=\frac{2sin \theta cos \theta }{\sqrt{2}cos \theta } \)

=\( \Large \sqrt{2}sin \theta \)

57). If 3sinx+5cosx=5, then what is the value of (3cosx - 5sinx)?

A). 0

B). 2

C). 3

D). 5

View Answer

Correct Answer: 3

Given that, 3sinx + 5cosx = 5

On sequaring both side, we get

\( \Large 9sin^{2}x+25cos^{2}x+30sinxcosx=25 \)

\( \Large => \ 9(1-cos^{2}x)+25(1-sin^{2}x)+30sinxcosx=25 \)

\( \Large => \ 9-9cos^{2}x+25-25sin^{2}x+30sinx.cosx=25 \)

\( \Large => \ 9+25-\{ 9cos^{2}x+25sin^{2}x-30sinxcosx \}=25 \)

\( \Large => \ 9=(3cosx-5sinx)^{2} \)

\( \Large => \ 3cosx-5sinx=3 \)

58). If \( \Large 0 < x < \frac{\pi}{2} \) and sec x = cosec y, then the value of sin(x + y) is

A). 0

B). 1

C). \( \Large \frac{1}{2} \)

D). \( \Large \frac{1}{\sqrt{3}} \)

View Answer

Correct Answer: 1

\( \Large sec \ x=cosec \ y \ => \ \frac{1}{cos \ x}=\frac{1}{sin \ y} \)

\( \Large cosx=siny \)

\( \Large sin \left(\frac{ \pi }{2}-x\right) =sin \ y \)

\( \Large \frac{ \pi }{2}-x=y \)

\( \Large (x+y)=\frac{ \pi }{2} \)

Now, value of \( \Large sin(x+y)=sin\frac{ \pi }{2}=1 \)

59). If sin 17 \( \Large ^{\circ} \)=\( \Large \frac{x}{y} \), then the value of sec 17 \( \Large ^{\circ} \)- sin 73 \( \Large ^{\circ} \)is

A). \( \Large \frac{y^{2}-x^{2}}{xy} \)

B). \( \Large \frac{x^{2}}{\sqrt{y^{2}-x^{2}}} \)

C). \( \Large \frac{x^{2}}{y\sqrt{y^{2}+x^{2}}} \)

D). \( \Large \frac{x^{2}}{y\sqrt{y^{2}-x^{2}}} \)

View Answer

Correct Answer: \( \Large \frac{x^{2}}{y\sqrt{y^{2}-x^{2}}} \)

Given, sin17 \( \Large ^{\circ} \)=x/y

Now, sec 17 \( \Large ^{\circ} \)- sin 73 \( \Large ^{\circ} \)

= sec 17 \( \Large ^{\circ} \)- sin (90 \( \Large ^{\circ} \)-17 \( \Large ^{\circ} \))

= sec 17 \( \Large ^{\circ} \)- cos 17 \( \Large ^{\circ} \)

= \( \Large \frac{1}{cos \ 17 ^{\circ} }-cos \ 17 ^{\circ} \)

=\( \Large \frac{1-cos^{2}17 ^{\circ} }{cos \ 17 ^{\circ} }=\frac{sin^{2}17}{cos \ 17 ^{\circ} } \)

=\( \Large \frac{\frac{x^{2}}{y^{2}}}{\sqrt{1-\frac{x^{2}}{y^{2}}}} \)

=\( \Large \frac{\frac{x^{2}}{y^{2}}}{\sqrt{1-\frac{x^{2}}{y^{2}}}}=\frac{\frac{x^{2}}{y^{2}}}{\sqrt{\frac{y^{2}-x^{2}}{y}}} =\frac{x^{2}}{y\sqrt{y^{2}-x^{2}}}\)

60). If \( \Large 2 \ sin \left(\frac{ \pi x}{2}\right)=x^{2}+\frac{1}{x^{2}} \), then the value of \( \Large \left(x-\frac{1}{x}\right) \) is

A). -1

B). 2

C). 1

D). 0

View Answer

Correct Answer: 0

\( \Large x^{2}+\frac{1}{x^{2}}=2sin \left(\frac{ \pi x }{2}\right) \)

\( \Large => \ \left(x-\frac{1}{x}\right)^{2}+2=2sin \left(\frac{ \pi x}{2}\right) \)

\( \Large x-\frac{1}{x}=0 \)

\( \Large \{ sin\frac{ \pi x}{2}=1 for \ all \ integer \ values \ of \ x \} \)

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