Question

secQ/tanQ+cotQ=SENQ
ALGUIEN ME AYUDA

Answer 1

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