Question

$\frac{secx + 1}{secx} = \frac{senx^{2}}{1-cosx}$

Answer 1

${\frac{secx+1}{secx}\times \frac{secx-1}{secx-1}=}\\\\{\frac{sec^2x-1}{sec^2x-secx}=}\\\\{\frac{\frac{1}{cos^2x} -1}{\frac{1}{cos^2x}-\frac{1}{cosx}}=}\\\\{\frac{\frac{1-cos^2x}{cos^2x}}{\frac{1-cosx}{cos^2x}}=}\\\\{\frac{\frac{sen^2x}{cos^2}}{\frac{1-cosx}{cos^2x}}=}\\\\{\frac{sen^2x}{cos^2x}\times \frac{cos^2x}{1-cos^2}=}\\\\{\boxed{\frac{sen^2x}{1-cosx}}}\\\\{\mathbf{salu2.!!\ :)}}\\{\mathbf{Wellington}}$