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verify (sec x - tan x)^2 = 1-cos x/1+cos x
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which side do you want to start playing with?
\(\large\color{black}{ \displaystyle (\sec x - \tan x)^2=\frac{1-\cos x}{1+\cos x} }\) playing with the left side 1. Multiply top and bottom times \(\large\color{black}{ \displaystyle 1-\cos x }\). 2. re-write the bottom in terms of sines, using \(\large\color{black}{ \sin^2x+\cos^2x=1 }\). 3. re-write each term in the fraction separately, simplifying as much as you can. 4. use the \(\large\color{black}{ a^2-2ab+b^2=(a+b)^2 }\).
Oh, this is playing with the right side, excuse me.
we can do a step at a time. what do you get after doing step 1?
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