Question

Tan+sec-1/tan-sec+1=sec+tan

Answer 1

Answer:

$\: \: \: \: \: \frac{ \tanθ + \secθ - 1}{ \tanθ - \sec θ + 1 } \\ = \frac{ \tan θ + \secθ - ( { \sec }^{2}θ - \ { \tan }^{2} θ) }{ \tanθ - \secθ + 1 } \\ = \frac{ \tanθ + \secθ - ( \secθ + \tan θ)( \secθ - \tanθ) }{ \tanθ - \secθ + 1 } \\ = \frac{ (\secθ + \tanθ)(1 - \secθ + \tanθ) }{(1 - \sec θ + \tanθ)} \\ = \secθ + \tanθ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (proved)$

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