Question

Помогите решить, пожалуйста

Answer 1

$2\, ctg270^\circ -2\, cos60^\circ +sin^260^\circ =0-2\cdot \dfrac{1}{2}+\Big(\dfrac{\sqrt3}{2}\Big)^2=-1+\dfrac{3}{4}=-\dfrac{1}{4}\\\\\\\Big(2\, ctg270^\circ -2\, cos60^\circ +sin^260^\circ \Big)^{-1}=\Big(-\dfrac{1}{4}\Big)^{-1}=-4$

Answer 2

(2ctg270°-2cos60°+sin²60°)⁻¹=(0-1+3/4)⁻¹=(-1/4)⁻¹=(-4)¹=-4