Trigonometric ratio Questions and answers

Topics

General Knowledge
General Science
General English
Aptitude
General Computer Science
General Intellingence and Reasoning
Current Affairs
Exams
Elementary Mathematics
English Literature

Elementary Mathematics

31). The value of \( \Large \sin 36 ^{\circ} \sin 72 ^{\circ} \sin 108 ^{\circ} \sin 144 ^{\circ} \) is equal to:

A). \( \Large \frac{1}{4} \)

B). \( \Large \frac{1}{16} \)

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

D). \( \Large \frac{5}{16} \)

View Answer

Correct Answer: \( \Large \frac{5}{16} \)

\( \Large \sin 36 ^{\circ} \sin 72 ^{\circ} \sin 108 ^{\circ} \sin 144 ^{\circ} \)

= \( \Large \sin^{2}36 ^{\circ} \sin^{2}72 ^{\circ} = \frac{1}{4} \left(2 \sin^{2}36 ^{\circ} \right) \left(2 \sin^{2}72 ^{\circ} \right) \)

= \( \Large \frac{1}{4} \left(1-\cos 72 ^{\circ} \right) \left(1-\cos 144 ^{\circ} \right) \)

= \( \Large \frac{1}{4} \left(1-\sin 18 ^{\circ} \right) \left(1+\cos 36 ^{\circ} \right) \)

= \( \Large \frac{1}{4}\left[ \left(1-\frac{\sqrt{5}-1}{4}\right) \left(1+\frac{\sqrt{5}+1}{4}\right) \right] \)

= \( \Large \frac{1}{4}\left[ 1 + \left(\frac{\sqrt{5}+1}{4}\right)- \left(\frac{\sqrt{5}-1}{4}\right)- \left(\frac{4}{16}\right) \right] \)

= \( \Large \frac{1}{4} \left[ 1+\frac{1}{2}-\frac{1}{4} \right] = \frac{5}{16} \)

32). If \( \Large \cos \left( \theta - \alpha \right),\ \cos \theta\ and\ \cos \left( \theta + \alpha \right) \) are in HP, the: \( \Large \cos \theta \sec \frac{ \alpha }{2} \) is equal to:

A). \( \Large \pm \sqrt{2} \)

B). \( \Large \pm \sqrt{3} \)

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

D). none of these

View Answer

Correct Answer: \( \Large \pm \sqrt{2} \)

Given \( \Large \cos \left( \theta - \alpha \right) \cos \theta\ and\ \cos \left( \theta + \alpha \right)\ \) are in HP

=> \( \Large \frac{1}{\cos \left( \theta - \alpha \right) },\ \frac{1}{\cos \theta },\ \frac{1}{\cos \left( \theta + \alpha \right) } \) will be in AP.

=> \( \Large \frac{2}{\cos \theta }=\frac{1}{\cos \left( \theta - \alpha \right) }+\frac{1}{\cos \left( \theta + \alpha \right) } \)

= \( \Large \frac{\cos \left( \alpha + \theta \right)+\cos \left( \theta - \alpha \right) }{\cos^{2} \theta - \sin^{2} \alpha } \)

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

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

=> \( \Large \cos^{2} \theta \left(1-\cos \alpha \right) = \sin^{2} \alpha \)

=> \( \Large \cos^{2} \theta \left(2 \sin^{2}\frac{ \alpha }{2}\right) = 4 \sin^{2}\frac{ \alpha }{2} \cos^{2}\frac{ \alpha }{2} \)

=> \( \Large \cos^{2} \theta \sec^{2}\frac{ \alpha }{2} = 2 \)

=> \( \Large \cos \theta \sec^{2}\frac{ \alpha }{2} = \pm \sqrt{2} \)

33). If \( \Large \cos x + \cos y + \cos \alpha = 0 \) and \( \Large \sin x + \sin y + \sin \alpha = 0,\) then \( \Large \cot \left(\frac{x+y}{2}\right) \) is equal to:

A). \( \Large \sin \alpha \)

B). \( \Large \cos \alpha \)

C). \( \Large \cot \alpha \)

D). \( \Large \sin \left(\frac{x+y}{2}\right) \)

View Answer

Correct Answer: \( \Large \cot \alpha \)

Given equation are \( \Large \cos x + \cos y + \cos \alpha = 0 \) and \( \Large \sin x + \sin y + \sin \alpha = 0 \). Then given equations may be written as

\( \Large \cos x + \cos y = - \cos \alpha \)

and \( \Large \sin x + \sin y = - \sin \alpha \)

=> \( \Large 2 \cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right) = - \cos \alpha \) ...(i)

and \( \Large 2 \sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right) = - \sin \alpha \) ...(ii)

From equations (i) and (ii) we get

\( \Large \frac{2 \cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right) }{2 \sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right) } = \frac{\cos \alpha }{\sin \alpha } \)

=> \( \Large \cot \left(\frac{x+y}{2}\right) = \cot \alpha \)

34). If \( \Large \tan \alpha = \left(1+2^{-x}\right)^{-1},\ \tan \beta = \left(1-2^{x+1}\right)^{-1},\ then\ \alpha - \beta \) equal

A). \( \Large \frac{ \pi }{6} \)

B). \( \Large \frac{ \pi }{4} \)

C). \( \Large \frac{ \pi }{3} \)

D). \( \Large \frac{ \pi }{2} \)

View Answer

Correct Answer: \( \Large \frac{ \pi }{4} \)

We have,

\( \Large \tan \left( \alpha + \beta \right) = \frac{\tan \alpha + \tan \beta }{1-\tan \alpha \tan \beta } \)

\( \Large \therefore \tan \alpha = \frac{1}{1+2^{-x}} \)

and \( \Large \tan \beta = \frac{1}{1+2^{x+1}} \)

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

\( \Large \therefore \tan \left( \alpha + \beta \right) = \frac{2}{1+\frac{1}{2x}.\frac{1}{1+2^{x+1}}} \)

=> \( \Large \tan \left( \alpha + \beta \right) = \frac{2^{x}+2.2^{x+x}+2^{x}+1}{1+2^{x}+2.2^{x}+2.2^{x+x}-2^{x}} \)

=> \( \Large \tan \left( \alpha + \beta \right) = 1 \)

=> \( \Large \alpha + \beta = \frac{ \pi }{4} \)

35). If \( \Large \tan A + \sin A = m\ and\ \tan A - \sin A = n,\ then\ \frac{ \left(m^{2}-n^{2}\right)^{2} }{mn} \) is equal to:

A). 4

B). 3

C). 16

D). 9

View Answer

Correct Answer: 16

Since, \( \Large \tan A + \sin A = m \)

and \( \Large \tan A - \sin A = n \)

\( \Large \therefore m+n = 2 \tan A \)

and \( \Large m-n = 2 \sin A \)

Also \( \Large mn = \left(\tan A + \sin A\right) \left(\tan A - \sin A\right) \)

= \( \Large \tan^{2}A - \sin^{2}A \)

Now \( \Large \frac{m^{2}-n^{2}}{mn} = \frac{ \left(m+n\right)^{2} \left(m-n\right)^{2} }{mn} \)

=\( \Large \frac{ \left(2 \tan A\right)^{2} \left(2 \sin A\right)^{2} }{\tan^{2}A-\sin^{2}A} \)

= \( \Large \frac{16 \tan^{2}A \sin^{2}A}{\sin^{2}A \tan^{2}A} = 16 \)

36). \( \Large 3 \left(\sin x - \cos x\right)^{4} + 6 \left(\sin x + \cos x\right)^{2} + 4 \left(\sin^{6}x + \cos^{6}x\right) \) is equal to:

A). 11

B). 12

C). 13

D). 14

View Answer

Correct Answer: 13

\( \Large 3 \left(\sin x - \cos x\right)^{4}+6 \left(\sin x + \cos x\right)^{2} + 4 \left(\sin^{6}x+\cos^{6}x\right) \)

= \( \Large 3 \left(1- \sin 2x\right)^{2} + 6 \left(1+ \sin 2x\right) \) \( \Large + 4 \{ \left(\sin^{2}x+\cos^{2}x\right)^{3} \) \( \Large -3 \sin^{2}x \cos^{2}x \left(\sin^{2}x+\cos^{2}x\right) \}\)

\(= \Large 3 \left(1-\sin 2x + \sin^{2}2x\right)+6+6 \sin 2x + 4\{ 1-3 \sin^{2}x \cos^{2}x \} \)

= \( \Large 3 \left(1-2 \sin 2x + \sin^{2}2x+2+2 \sin 2x\right)+4\{ 1 - \frac{3}{4}\sin^{2} 2x \} \)

\( \Large 13 + 3 \sin^{2} 2x - 3 \sin^{2} 2x = 13 \)

37). If \( \Large \cos \theta = \cos \alpha \cos \beta .\ then\ \tan \frac{ \theta + \alpha }{2} \tan \frac{ \theta - \alpha }{2} \) is equal to:

A). \( \Large \tan^{2}\frac{ \alpha }{2} \)

B). \( \Large \tan^{2}\frac{ \beta }{2} \)

C). \( \Large \tan^{2}\frac{ \theta }{2} \)

D). \( \Large \cot^{2}\frac{ \beta }{2} \)

View Answer

Correct Answer: \( \Large \tan^{2}\frac{ \beta }{2} \)

Since, \( \Large \tan \frac{ \theta + \alpha }{2}\tan \frac{ \theta \alpha }{2} = \frac{\tan^{2}\frac{ \theta }{2}-\tan^{2}\frac{ \alpha }{2}}{1-\tan^{2}\frac{ \theta }{2}\tan^{2}\frac{ \theta }{2}} \)

= \( \Large \frac{\frac{1-\cos \theta }{1+\cos \theta } - \frac{1-\cos \alpha }{1+\cos \alpha }}{1 - \frac{1-\cos \theta }{1+\cos \theta }.\frac{1-\cos \alpha }{1+\cos \alpha }} = \frac{2 \left(\cos \alpha - \cos \theta \right) }{2 \left(\cos \alpha + \cos \theta \right) } \)

= \( \Large \frac{\cos \alpha \left(1-\cos \beta \right) }{\cos \alpha \left(1+\cos \beta \right) } \)

\( \Large \left(\because \cos \theta = \cos \alpha \cos \beta \right) \)

= \( \Large \tan^{2}\frac{ \beta }{2} \)

38). The greatest value of \( \Large \cos \theta \) for which \( \Large \cos 5 \theta = 0 \), is:

A). 0

B). \( \Large \frac{1+\sqrt{5}}{4} \)

C). \( \Large \sqrt{\frac{5+\sqrt{5}}{8}} \)

D). \( \Large \sqrt{\frac{5+\sqrt{1}}{4}} \)

View Answer

39). A solution \( \Large \left(x, y\right)\ of\ x^{2}+2x \sin xy + 1 = 0 \) is:

A). (1,0)

B). \( \Large \left(1,\ \frac{7 \pi }{2}\right) \)

C). \( \Large \left(-1, \frac{7 \pi }{2}\right) \)

D). (-1, 0)

View Answer

40). If \( \Large \tan \theta ,\ 2 \tan \theta + 2,\ 3 \tan \theta + 3 \) are in GP, then the value of \( \Large \frac{7-5 \cot \theta }{9-4\sqrt{sec ^{2} \theta - 1}} \) is:

A). \( \Large \frac{12}{5} \)

B). \( \Large -\frac{33}{28} \)

C). \( \Large \frac{33}{100} \)

D). \( \Large \frac{12}{13} \)

View Answer

1	2	3	4

Go to :

Total Pages : 4