Question

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

Answer 1

$1)\\\psi=\alpha\\----------\\\\\mathtt{=\dfrac{Sin2\psi*Cos2\psi}{1-Sin2\psi}}\\\\\\\mathtt{=\dfrac{1}{2}*\dfrac{2(Sin2\psi*Cos2\psi)}{(1-Sin2\psi)}}\\\\\\\mathtt{=\dfrac{Sin4\psi}{2(1-Sin2\psi)}}\\\\\\\mathtt{=\dfrac{Sin4\psi}{2-2Sin2\psi)}}\\\\----------\\\\2)\\=\dfrac{1-Cos2\psi}{Cos2\psi*Sin2\psi}\\\\\\=\dfrac{Sin^2\psi+Cos^2\psi-Cos^2\psi+Sin^2\psi}{Cos2\psi*Sin2\psi}\\\\\\=\dfrac{2Sin^2\psi}{Cos2\psi*2*Sin\psi*Cos\psi}\\\\\\=\dfrac{Sin\psi}{Cos2\psi*Cos\psi}\\---------$
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