If $\sec \theta + \tan \theta = p,$ then $\tan \theta $ is equal to
If $\sec \theta + \tan \theta = p,$ then $\tan \theta $ is equal to
- A
$\frac{{2p}}{{{p^2} - 1}}$
- B
$\frac{{{p^2} - 1}}{{2p}}$
- C
$\frac{{{p^2} + 1}}{{2p}}$
- D
$\frac{{2p}}{{{p^2} + 1}}$
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