Question

1 - sinA/ 1 + sinA =( sec A- tan A)^2

Answer 1

➡HERE IS YOUR ANSWER⬇

■] LONG WAY :

L.H.S.

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

= R.H.S. (Proved)

■] EASY WAY :

L.H.S.

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

= R.H.S. (Proved)

⬆HOPE THIS HELPS YOU⬅