Question

Prove that:
tan A / sec A+ 1= sec A- 1 / Tan A

Refer attachment​

Answer 1

${ \tan }^{2} x = { \sec }^{2} x - 1 \\ = > \tan{}^{2} x = ( \sec{}x + 1)( { \sec }x - 1) \\ = > \frac{ \tan(x) }{ \sec(x + 1) } = \frac{ \sec(x - 1) }{ \tan(x) }$