Question

помогите сделать пожалуйста 89 номер (1,2)

Answer 1

(1+cos2α)/2cosα =2cos²α/2cosα = cosα.
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(tqα-sinα)/(tqα +sinα) = (tqα-tqα*cosα)/(tqα+tqα*cosα)=
tqα*(1 -cosα)/tqα*(1 -cosα) =(1 -cosα)/(1 -cosα) =(2sin²(α/2))/(2cos²(α/2)) =tq²(α/2).