Question

Помогите решить:
а) (sin2альфа*cos2альфа)/1-sin2альфа
б) (1-cos2альфа)/sin2альфа*cos2альфа

Answer 1

$1)\psi=alpha\----------\\mathtt{=dfrac{Sin2psi*Cos2psi}{1-Sin2psi}}\\\mathtt{=dfrac{1}{2}*dfrac{2(Sin2psi*Cos2psi)}{(1-Sin2psi)}}\\\mathtt{=dfrac{Sin4psi}{2(1-Sin2psi)}}\\\mathtt{=dfrac{Sin4psi}{2-2Sin2psi)}}\\----------\\2)\=dfrac{1-Cos2psi}{Cos2psi*Sin2psi}\\\=dfrac{Sin^2psi+Cos^2psi-Cos^2psi+Sin^2psi}{Cos2psi*Sin2psi}\\\=dfrac{2Sin^2psi}{Cos2psi*2*Sinpsi*Cospsi}\\\=dfrac{Sinpsi}{Cos2psi*Cospsi}\---------$
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