Show that (tanx+secx)=1/secx-tanx

(tanx+secx)=1/[secx-tanx]

= 1/[1/cosx-sinx/cosx]

= 1/[(1-sinx)/cosx]

= cosx/[1-sinx]*[1+sinx]/[1+sinx]

=[cosx+cosxsinx]/[1-sin^2x]

= [cosx(1+sinx)]/cos^2x

= [1+sinx]/cosx

= 1/cosx+sinx/cosx

=secx+ tanx

Does it work? Thank You.

Yes, that works. Congratulations!

:D

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Yes, it works. But it can be done way simpler.
(tanx+secx)(-tanx+secx)/(secx - tanx)
= (sec^2x - tan^2x)/(secx - tanx)
= 1/(secx - tanx)
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Attn: sec^2x - tan^2x = 1 since sec^2x = 1 + tan^2x