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✨ Best Answer ✨

(1)
A+B+C=180°
B+C=180°-A
cos{(B+C)/2}=cos{(180°-A)/2}
      =cos(90°-A/2)
=sin(A/2)

(2)
tan(A/2)=sin(A/2)/cos(A/2)
tan{(B+C)/2}=sin{(B+C)/2}/cos{(B+C)/2}
      =cos(A/2)/sin(A/2)
よってtan(A/2)tan{(B+C)/2}
  =sin(A/2)/cos(A/2)・cos(A/2)/sin(A/2)
  =1

sin(90°-θ)=cosθとcos(90°-θ)=sinθ を使います。

まつ。

ありがとうございます!!解き方が分かりました‪⸜‪‪☺︎‬⸝‬‪‪

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