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If secθ + tanθ = p, (p ≠ 0) then secθ is equal to
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p - 1 
, p ≠ 0 p -
2 
p - 1 
, p ≠ 0 p -

p + 1 
, p ≠ 0 p -
(1 / 2) 
p + 1 
, p ≠ 0 p
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Correct Option: D
secθ + tanθ = p .....(i)
∵ sec²θ – tan²θ = 1
⇒ (secθ + tanθ) (secθ – tanθ) = 1
| ⇒ secθ – tanθ = | ....(ii) | p |
On adding both the equations
| 2 secθ = p + | p |
| ⇒ secθ = | | ![]() | p + | 1 | |
| 2 | p |