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I'm working on manipulating trig identities and using Wolfram Alpha to check the identity still holds.

I'm going from this:

$\frac{1-\cos x}{1+\cos x} = \frac{1}{tan^2x}-\frac{2}{\tan x \sin x} + \frac{1}{\sin^2 x}$

which WA verifies is an identity to this:

$\frac{1+\cos x}{1-\cos x} = tan^2x-\frac{\tan x \sin x}{2} + \sin^2 x$

which WA seems to think is only true for certain values of x.

Based on my workings out on paper, I think I'm safe in flipping both sides. Wonder if someone might weigh in on this to help me out please?

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    @AméricoTavares: Mike's answer has cleared it up for me now, thanks.2012-11-13

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The reciprocal of a sum is not equal to the sum of the reciprocals. If $\frac1a=\frac1b+\frac1c, a=\frac{bc}{b+c}$, not $b+c$

Your right side is equal to $(\csc x-\cot x)^2$. If you really want to flip both sides, you'll get

$\frac{1+\cos x}{1-\cos x}=\frac1{(\csc x-\cot x)^2}=\frac{(\csc x+\cot x)^2}{(\csc^2x-\cot^2x)^2}=$ $(\csc x+\cot x)^2=\frac1{\tan^2x}+\frac2{\tan x\sin x}+\frac1{\sin^2x}$

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    Got it. The "workings out on paper" I mentioned started with $\frac{a}{b}=\frac{c}{d} - \frac{e}{f}$ which I erroneously manipulated from $\frac{b}{a}=\frac{df}{cf-de}$ to $\frac{b}{a}=\frac{df}{cf} - \frac{df}{de}$2012-11-13