守夜者皮甲:tan2(B+C)/2怎麽降次？

tan2(B+C)/2怎麽降次？ 化简:(2a-b-c)/(a-b)(a-c)+(2b-c-a)/(b-c)(b-a)+(2c-a-b)/(c-a)(c-b) 2a-b-c/(a-b)(a-c)+ab -c-a/(b-c)(b-a)+2c-a-b/(c-a)(c-b) b/(a+b)=(a+c+-b)/(b+c-a)=(a+b+c)/(2a+b+2c)求a:b:c= (a+b)/c=(a+c)/b=(b+c)/a =2(a+b+c)/ tan(π/4+α)-tan(π/4-α)=2tan2α 求证:tan(π/4+α)-tan(π/4-α)=2tan2α cos4α×tan2α-sin4α=2tanα/(tan^2-1) 求证a^2/(b+c-a)+b^2/(a+c-b)+c^2/(a+b-c)=>a+b+c 三角形ABC三边abc，求证：a^2/（b+c-a)+b^2/(c+a-b)+c^2/(a+b-c)>=a+b+c