Question

please solve this question Friends

Answer 1

HELLO DEAR,

$\frac{ \tan( \alpha ) + \sec( \alpha ) - 1 }{ \tan( \alpha ) - \sec( \alpha ) + 1} \\ = > \frac{ \tan( \alpha ) + \sec( \alpha ) - ( {sec}^{2} \alpha - {tan}^{2} \alpha )}{ \tan( \alpha ) - \sec( \alpha ) + 1} \\ = > \frac{ ( \tan \alpha + \sec \alpha ) - ( \tan\alpha + sec\alpha )( sec \alpha - \tan \alpha )}{\tan( \alpha ) - \sec( \alpha ) + 1} \\ = > \frac{( \tan \alpha + sec \alpha )(1 + tan \alpha - \sec \alpha ) }{(\tan \alpha - \sec \alpha + 1)} \\ = > \tan \alpha + sec\alpha \\ = > \frac{ \sin \alpha }{ \cos \alpha } + \frac{1}{ \cos \alpha } \\ = > \frac{1 + \sin\alpha }{ \cos \alpha }$
I HOPE ITS HELP YOU DEAR,
THANKS