Question

1-cos2x-sinx/cosx-sin2x

Answer 1

$boldsymbol{dfrac{1-cos(2x)-sin x}{cos x-sin(2x)}=}\\\=dfrac{1-(1-2sin^2x)-sin x}{cos x-2sin x cos x}=dfrac{2sin^2 x-sin x}{cos x(1-2sin x)}=\\\=dfrac{-sin x(1-2sin x)}{cos x(1-2sin x)}=dfrac{-sin x}{cos x}boldsymbol{=-tg~x}$

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Использованы формулы

cos (2α) = 1 - 2 sin²α

sin (2α) = 2 sin α · cos α