A question on AP.

BITSAT 2015 Preparation › Algebra questions › A question on AP.

If sec(x-y), secx , sec(x+y) are in AP, then (cosx)(secy2)=
a)Â±âˆš3
b)Â±âˆš5
c)Â±âˆš7
d)Â±âˆš2

Plz show the steps for this prob also..

#Mathematics #Algebra

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4 Answers

1

Abhisek ·2010-10-01 11:16:09

sec(x-y) secx sec(x+y) â†’ AP
1cos(x-y) 1cos x 1cos(x+y) â†’AP

Therefore,

2cos x = 1cos(x-y) + 1cos(x+y)

2cos x = 2.cos x.cos ycos(x-y).cos(x+y)

1cos x = 2.cos x.cos ycos²x-sin²y

cos²x-sin²y = cos²x.cosy

cos²x (1 - cos y) = sin²y

cos²x (1 - cos y) = (1 - cos²y)

cos²x = 1 + cos y

cos²x - cos y = 1

Since maximum value of cos²x is (+1)
or cos y = Â±1................................(1)
Therefore, cos y = 0 in order to fulfill the above equation.

cos y = 0

y= Ï€2

y2 = Ï€4

cos y2 = cosÏ€4 = 1√2

secy2 = √2.....................................(2)

Multiplying (1) and (2), we get
cos y . sec y2 = Â±1 x √2 = Â±√2 [ Option (d) ]

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Avinav Prakash ·2010-10-01 11:23:12

TO CUT IT SHORT
cos²x = 1 + cos y=2cos²y/2
from here directly get the answeer

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Abhisek ·2010-10-01 11:27:24

Yup. !

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chessenthus ·2010-10-01 12:17:53

Thank you very much for the quick reply!

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