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tan^2x-1/1-tanx I need to get it down to one function with no fractions. I got to (sin^2x/cosx)-cosx but I don't know where to go from there or if that is right.
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Use difference of squares.
\[ a^2-b^2 = (a-b)(a+b) \]
could you show me because I don't see how that would work in this problem
\[ a=\tan(x),b=1 \]
\[ \tan^2(x)-1 = (\tan(x)-1)(\tan(x)+1) \]
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\[ 1-\tan(x) = -(\tan(x)-1) \]
idk if this will help but 1+tan x=cosx
....
\[ \frac{\tan^2x-1}{1-\tan x} = \frac{(\tan x-1)(\tan x+1)}{-(\tan x-1)} \]
I've tried to get help with this problem before but I put part of the question wrong. I think the other guy was on the right track but was wrong cause I squared something that shouldn't have been squared
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ooooo that should work
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