If secθ+tanθ=p, (p≠0) then secθ is equal to

  1. 2[p-1/p], p≠0
  2. 1/2[p+1/p], p≠0
  3. p+1/p, p≠0
  4. p-1/p, p≠0
Monis Rasool Professor Asked on 31st August 2015 in Maths.
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  • 1 Answer(s)

     

    Answer: (2) 1/2[p+1/p], p≠0

    Explanation:-
    secθ + tanθ = p   ……………………..(1)
    As we know that, sec2θ – tan2θ = 1
    (secθ – tanθ) (secθ + tanθ) = 1
    (secθ – tanθ) p = 1           (From equation 1)
    secθ – tanθ = /1p                  ……………… (2)
    Add equation (1) & (2),
    2 secθ = p + 1/p
    secθ = (p + 1/p)/2

    Then, the secθ is equal to (p + 1/p)/2 and p does not equal to zero.

    Hence, the answer is (2) 1/2[p+1/p], p≠0.

    Anurag Mishra Professor Answered on 2nd September 2015.
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