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January 27, 2015

January 27, 2015

Posted by **anonymous** on Saturday, June 19, 2010 at 11:56am.

[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 -
**Reiny**, Saturday, June 19, 2010 at 12:31pmIt 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 -
**anonymous**, Saturday, June 19, 2010 at 1:47pmWow that was a tough one... It took me so long to figure out, even with your help! Thanks Reiny! :)

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