(3cos(π-β) + sin(π/2+β)) ÷ сos(β+3π)

Question

(3cos(π-β) + sin(π/2+β)) ÷ сos(β+3π)

Answer 1

(3cos(π-β)+sin(π/2+β)):cos(β+3π)=
=(-3cosβ+cosβ):(-cosβ)=(-3cosβ:(-cosβ))+(cosβ:(-cosβ))=3-1=2