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August 30, 2014

Posted by **Jay** on Sunday, December 4, 2011 at 11:42am.

1.1+sinx/1-sinx=cscx+1/cscx-1

2.tanx+sinx/1+cosx=tanx

3.sec^2x/sin^2x=1/sin^2x+1/cos^2x

4.tan^2x/1+tan^2x=1-cos^2x

5.sinx-sin^3x/cos^3x=tanx

- trig -
**Reiny**, Sunday, December 4, 2011 at 11:50amMy method

1. start with the more complicate looking side

2. change everything to sines and cosines, unless I recognize an obvious identity

I will do #2 for you

LS = (tanx + sinx)/(1 + cosx)

= (sinx/cosx + sinx)/(1+cosx)

= [( sinx + sinxcosx)/cosx ]/(1+cosx)

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

= sinx/cosx

= tanx

= RS

by "obvious identity" I notice in #4 that we have the expression

1 + tan^2x

from the Pythagorean identities

1 + tan^2x = sec^2x

so #4

LS = tan^2x/(1+tan^2)

= tan^2x/sec^2x

= tan^2x cos^2x

= sin^2x/cos^2x (cos^2x) = sin^2x

= 1 - cos^2x

= RS

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