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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 | 
 
	