Question

ctg x - ((sin x) / (1 - cos x)) = -((1)/ (sin x))
тотожность

Answer 1

$ctgx-\frac{sinx}{1-cosx} =-\frac{1}{sinx} \\ \frac{cosx}{sinx} -\frac{sinx}{1-cosx} +\frac{1}{sinx}=0\\ \frac{(1-cosx)*cosx-sin^{2}+1-cosx }{(1-cosx)sinx} =0\\\frac{cosx-cos^{2}x+cos^{2} x-cosx }{(1-cosx)sinx} =0\\\frac{0}{(1-cosx)sinx} =0\\0=0$

х ∈ R