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

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

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