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Elementary Mathematics

11). What is \( \Large sin 25^{\circ} sin 35^{\circ} sec 65^{\circ} sec 55^{\circ}\) equal to?

A). -1

B). 0

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

D). 1

View Answer

Correct Answer: 1

sin 25 \( ^{\circ}\) sin 35 \( ^{\circ}\) sec 65 \( ^{\circ}\) sec 55 \( ^{\circ}\)

=> \( \Large sin \ 25 ^{\circ} \ sin \ 35 ^{\circ} .\frac{1}{cos \ 65 ^{\circ} }.\frac{1}{cos \ 55 ^{\circ} } \)

=> \( \Large sin \ 25 ^{\circ} . \ sin \ 35 ^{\circ} \times \frac{1}{cos(90 ^{\circ} -25 ^{\circ} )} \times \frac{1}{cos(90 ^{\circ} -35 ^{\circ} )} \)

=> \( \Large sin \ 25 ^{\circ} \times sin \ 35 ^{\circ} \times \frac{1}{sin25 ^{\circ} } \times \frac{1}{sin35 ^{\circ} } \)

=> 1

12). If p = a sin x + b cos x and q = a cos x - b sin x, then what is the of \( \Large p ^{2}+q^{2} \) ?

A). a + b

B). ab

C). \( \Large a ^{2}+b^{2} \)

D). \( \Large a ^{2}-b^{2} \)

View Answer

Correct Answer: \( \Large a ^{2}+b^{2} \)

Given, p = asin x + bcos x ...(i)

and q = acos x - bsin x ....(ii)

On squaring both the equations

\( \Large p^{2}=a^{2}sin^{2}x+b^{2}cos^{2}x \)+2ab sinx cosx

and \( \Large q^{2}=a^{2}cos^{2}x+b^{2}sin^{2}x \)-2ab sinx cosx

\( \Large p^{2}+q^{2}=a^{2}sin^{2}x+b^{2}cos^{2}x \)

\( \Large +2ab \ sinx \ cosx

+a^{2} \ cos^{2}x+b^{2}sin^{2}x-2ab \ sinx \ cosx \)

\( \Large p^{2}+q^{2}=a^{2}(sin^{2}x+cos^{2}x)\)

\(+b^{2}(cos^{2}x+sin^{2}x) \)

= \( \Large a^{2}+b^{2} \)

13). What is the value of \( \Large sec(90 ^{\circ} - \theta) \ sin \theta sec45^{\circ}\)?

A). 1

B). \( \Large \frac{\sqrt{3}}{2} \)

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

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

View Answer

14). If \( \Large tan \theta =\frac{4}{3} \) then the value of \( \Large \frac{3 \ sin \theta +2 \ cos \theta }{3 \ sin \theta -2 \ cos \theta } \) is

A). 0.5

B). -0.5

C). 3

D). -3

View Answer

Correct Answer: 3

\( \Large tan \theta =\frac{4}{3} \)

\( \Large \frac{3 \ sin \theta +2 \ cos \theta }{3 \ sin \theta -2 \ cos \theta }=\frac{3 \ tan \theta +2}{3 \ tan \theta -2} \)

\( \Large \frac{3 \times \frac{4}{3}+2}{3 \times \frac{4}{3}-2} =\frac{4+2}{4-2}=3 \)

15). If \( \Large sec \theta -cosec \theta =0 \), then the value of \( \Large tan \theta +cot \theta \) is

A). 0

B). 1

C). -1

D). 2

View Answer

Correct Answer: 2

\( \Large sec \theta -cosec \theta =0 \)

=> \( \Large sec=cosec \theta \)

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

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

=> \( \Large tan \theta =tan \ 45 ^{\circ} \)

=> \( \Large \theta =45 ^{\circ} \)

\( \Large tan \theta +cot \theta =tan \ 45 ^{\circ} +cot \ 45 ^{\circ} \)

= 1+1=2

16). If \( \Large tan \ 7 \theta \ tan \ 2 \theta=1 \),then the value of \( \Large tan \ 3 \theta \) is

A). \( \Large \sqrt{3} \)

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

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

D). \( \Large -\sqrt{3} \)

View Answer

Correct Answer: \( \Large \frac{1}{\sqrt{3}} \)

Given, \( \Large tan \ 7 \theta .tan \ 2 \theta =1 \)

=> \( \Large tan7 \theta =cot2 \theta \)

=> \( \Large tan7 \theta =tan(90 ^{\circ} -2 \theta ) \)

=> \( \Large 7 \theta =90 ^{\circ} -2 \theta \)

=> \( \Large 9 \theta =90 ^{\circ} \) => \( \Large \theta =10 ^{\circ} \)

\( \Large tan3 \theta =tan \ 30 ^{\circ} =\frac{1}{\sqrt{3}} \)

17). The expression \( \Large \frac{tan 57 ^{\circ} +cot 37 ^{\circ} }{tan 33 ^{\circ} + cot 53 ^{\circ} } \) is equal to

A). tan 33 \(^{\circ}\) cot 57\(^{\circ}\)

B). tan 57 \(^{\circ}\)cot 37 \(^{\circ}\)

C). tan 33 \(^{\circ}\)cot 53 \(^{\circ}\)

D). tan 53 \(^{\circ}\) cot 37 \(^{\circ}\)

View Answer

Correct Answer: tan 57 \(^{\circ}\)cot 37 \(^{\circ}\)

\( \Large \frac{tan \ 57 ^{\circ} +cot \ 37 ^{\circ} }{tan \ 33 ^{\circ} +cot \ 53 ^{\circ} }=

\frac{tan \ 57 ^{\circ} +cot \ 37 ^{\circ} }{cot \ 57 ^{\circ} +tan \ 37 ^{\circ} } \)

= \( \Large \frac{tan57 ^{\circ} +\frac{1}{tan37 ^{\circ} }}{\frac{1}{tan57 ^{\circ} }+tan37 ^{\circ} } \)

= \( \Large \frac{1+tan57 ^{\circ} tan37 ^{\circ} }{1+tan37 ^{\circ} tan57 ^{\circ} } \times \frac{tan57 ^{\circ} }{tan37 ^{\circ} } \)

= \( \Large \frac{tan57 ^{\circ} }{tan37 ^{\circ} }=tan \ 57 ^{\circ} .\ cot \ 37 ^{\circ} \)

18). The numerical value of \( \Large \frac{5}{sec^{2} \theta }+\frac{2}{1+cot^{2} \theta }+3 \ sin^{2} \theta \) is

A). 5

B). 2

C). 3

D). 4

View Answer

Correct Answer: 5

\( \Large \frac{5}{sec^{2} \theta }+\frac{2}{1+cot^{2} \theta }+3 \ sin^{2} \theta \)

= \( \Large 5 \ cos^{2} \theta +\frac{2}{cosec^{2} \theta }+3 \ sin^{2} \theta \)

= \( \Large 5 \ cos^{2} \theta +2 \ sin^{2} \theta +3 \ sin^{2} \theta \)

= \( \Large 5 \ cos^{2} \theta +5 \ sin^{2} \theta \)

= \( \Large 5(sin^{2} \theta +cos^{2} \theta ) \)

= \( \Large 5 \times 1=5 \)

19). The value of tan 4\(^{\circ}\).tan 43\(^{\circ}\).tan 47 \(^{\circ}\).tan 86 \(^{\circ}\) is

A). 2

B). 3

C). 1

D). 4

View Answer

20). Consider the following equations

\( \Large cosec^{2}x+sec^{2}x=cosec^{2}x \ sec^{2}x \)

\( \Large sec \ x+tan^{2}x=sec \ x \ tan^{2}x \)

\( \Large cosec \ x+tan^{2}x=cot^{2}x+sec \ x \)

Which of the above equations are correct?

A). (i) and (ii)

B). (ii) and (iii)

C). (i) and (iii)

D). (i), (ii) and (iii)

View Answer

Correct Answer: (i) and (iii)

(i) LHS = \( \Large cosec^{2}x+sec^{2}x \)

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

= \( \Large \frac{1}{sin^{2}x \ cos^{2}x}=\frac{1}{sin^{2}x}.\frac{1}{cos^{2}x} \)

= \( \Large cosec^{2}x \ sec^{2}x=RHS \)

(ii)\( \Large LHS = sec^{2}x + tan^{2}x \)

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

\( \Large \frac{1 + sin^{2}}{cos^{2}x} \ne sec^{2}x tan^{2}x = RHS\)

(iii) LHS = \( \Large cosec^{2}x+tan^{2}x \)

= \( \Large 1+cot^{2}x+sec^{2}x-1 \)

\( \Large cosec^{2}x=1+cot^{2}x \)

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

= \( \Large cot^{2}x+sec^{2}x \)

so, Eqs. (i) and (iii) are correct.

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