Posted by anonymous on Saturday, June 19, 2010 at 11:56am.
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
=[sinxcosxsinx+sinxsinx/cosx]/[cosx1]
or to the next
=[(sinxcos^2xsinx)/cosx]/[cos^2x1]
but from there I understand
=[sinx(cos^2x1)/cosx]*[1/cos^2x1]
=sinx/cosx
=tanx
=RS
Help?

maths  trigonometry  Reiny, Saturday, June 19, 2010 at 12:31pm
It looks like you cutandpasted 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  anonymous, Saturday, June 19, 2010 at 1:47pm
Wow that was a tough one... It took me so long to figure out, even with your help! Thanks Reiny! :)

maths  trigonometry  Loved Chilapu, Tuesday, March 29, 2016 at 2:58am
Prov that 1(sinxtanx) / 1+sec x
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