Question

prove that tan A ÷ 1 +sec A - tan A ÷ 1 - sec A = 2 cosec A

Answer 1

♧♧HERE IS YOUR ANSWER♧♧

Now,

L.H.S.

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

= R.H.S.

♧♧HOPE THIS HELPS YOU♧♧

Answer 2

Answer:

Hope it helps you. See the attachment please.