Question

1+cos x/ 1-cosx = tan²x/ (sec x-1)². prove the following identities​

Answer 1

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