Question

plz answer the question
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Answer 1

2)tan is the answer of the question

Answer 2

$we \: have \: an \: identity \\ \\ \\ \\ {sec}^{2} \theta \: - {tan}^{2} \theta \: = 1 \\ \\ \\ and \\ \\ {x}^{2} - {y}^{2} = (x - y)(x + y) \\ \\ \\ now \\ \\ \\ \sqrt{(sec \theta - 1)(sec \theta + 1)} \\ \\ = \sqrt{ {sec}^{2} \theta - 1} \\ \\ = \sqrt{ { \tan}^{2} \theta } \\ \\ = tan \theta$

hope it helps