Question

1/secA-tanA +1/secA+tanA=2secA

Answer 1

Hey user

Here is your answer :-

Let alpha be A .

$\frac{1}{ \sec( \alpha ) - \tan( \alpha ) } + \frac{1}{ \sec( \alpha ) + \tan( \alpha ) } \\ \\ by \: cross \: multiplication \\ \\ \frac{ \sec( \alpha ) + \tan( \alpha ) + \sec( \alpha ) - \tan( \alpha ) }{( \sec( \alpha ) + \tan( \alpha ))( \sec( \alpha ) - \tan( \alpha )) } \\ \\ \frac{2 \sec( \alpha ) }{ { \sec }^{2} (\alpha ) - { \tan}^{2} ( \alpha )} \\ \\ 2 \sec( \alpha ) \: \: \: \: \: \: \: \: \: \: by \: identity$

Thank you.