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hines57 · Dec 13, 2011, 04:55 PM

(cot^2x-1)/(1 cot^2x)=1-2sin^2

hines57 · Dec 13, 2011, 05:12 PM

OK I found it by turning the denominator into sec2 then moving it to the top using recipricol identity followed by distribution then used a quotient identity and came to the answer

ebaines · Dec 14, 2011, 06:56 AM

Good work. In general for these types of problems I find it easiest to convert the tangenet, cotangent, secant, and cosecant functions into their sine and cosine equivalents. I do it this way because I can remember the basic sine and cosine identies. So for this problem I would have done this:

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

Multiply through by sin^2x:

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

Then apply identities:
$ \cos^2x - \sin^2x = (1-\sin^2x) - \sin^2x = 1 - 2\sin^2x\\ \sin^2x + \cos^2x = 1 $