urgent

Asked by 2806rahul | 15th May, 2010, 07:00: PM

Expert Answer:

secθ + tanθ = x,

secθ + (sec2θ - 1)1/2 = x,            ...........1 + tan2θ = sec2θ

  (sec2θ - 1)1/2 = x - secθ

square on both sides,

 (sec2θ - 1) = x2 - 2x secθ + sec2θ

x2 - 2x secθ + 1 = 0

secθ = (x2 + 1)/2x

Regards,

Team,

TopperLearning.

Answered by  | 17th May, 2010, 12:05: PM

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