# maths - trigonometry

posted by on .

I've asked about this same question before, and someone gave me the way to finish, which I understand to some extent. I need help figuring out what they did in the second step though. How they got to the third step from the second.

[sinx + tanx]/[cosx + 1] = tanx

LS=[sinx+tanx]/[cosx+1]
=[sinx+tanx]/[cox+(cosx/cosx)]

makes sense so far, but I don't know how they got from here to the next one

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

or to the next

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

but from there I understand

=[sinx(cos^2x-1)/cosx]*[1/cos^2x-1]
=sinx/cosx
=tanx
=RS

Help?

• maths - trigonometry - ,

It looks like you cut-and-pasted part of my solution I gave you for this question

http://www.jiskha.com/display.cgi?id=1276915220

I multiplied top and bottom by (cosx -1)/cosx -1) , which of course has a value of 1 and does not change the value of the Left Side.

It is a step similar to the one we use in "rationalizing a denominator", I noticed that this would yield cos^2x -1 , which then could be replaced by the single term sin^2x

• maths - trigonometry - ,

Wow that was a tough one... It took me so long to figure out, even with your help! Thanks Reiny! :)

• maths - trigonometry - ,

Prov that 1-(sinxtanx) / 1+sec x