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Hannah_Ahn
prove: (sin2x over 2-2cos^2x) = cotx
i think you mistyped something...
it should be sin2x over (2-2cos^2x) = cotx
\[\left( \sin2x \over 2-2\cos^2x \right) = cotx\]
first look at the numerator sin2x=2sinx cosx then the deno 2-2cos^2 x = 2(1-cos^2 x)=2 sin^2 x= 2 sinx sinx so LHS=nume/demo=2sinxcosx / 2sinxsinx = cosx/sinx=cotx