Question

1+tan square a/1+seca=seca​

Answer 1

Answer:

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Answer 2

$= > 1 + \frac{ {tan}^{2}a }{sec \: a \: + 1}$

$= > 1 + \frac{ {sec}^{2} a \: - 1}{sec \: a \: + 1}$

$= > 1 + (sec \: a \: - 1)$

$= > sec \: a$