Question

Show that ( sec∅- tan∅) ^2 = 1 - sin∅ / 1 + sin∅

Answer 1

HELLO DEAR,

$\frac{1 - sin \alpha }{1 + sin \alpha } \\ \\ = > \frac{1 - sin \alpha }{1 + sin \alpha } \times \frac{1 - sin \alpha }{1 - sin \alpha } \\ \\ = > \frac{1 + {sin}^{2} \alpha - 2sin \alpha }{1 - {sin}^{2} \alpha } \\ \\ = > \frac{1 + {sin}^{2} \alpha - 2sin \alpha }{ {cos}^{2} \alpha } \\ \\ = > \frac{1}{ {cos}^{2} \alpha } + \frac{ {sin}^{2} \alpha }{ {cos}^{2} \alpha } - \frac{2sin \alpha }{ {cos}^{2} \alpha } \\ \\ = > {sec}^{2} \alpha + {tan}^{2} \alpha - 2sec \alpha \tan \alpha \\ \\ = > (sec \alpha - tan \alpha ) ^{2}$


I HOPE ITS HELP YOU DEAR,
THANKS