(1-ctg^2(-x))/(tg(x-pi)-1)/(ctg(3pi-x)/ctg(pi+x)) - вопрос №12134302332 от олеся14111 16.01.2020 20:56

Question

Алгебра: (1-ctg^2(-x))/(tg(x-pi)-1)/(ctg(3pi-x)/ctg(pi+x))

олеся14111 · Answer

( 1-ctq²(-x) ) /(tq(x -π) -1) : (ctq(3π -x)/ctq(π +x) = (1-ctq²x ) /(tqx -1) : (-ctqx)/ctqx) =(1-ctq²x ) /(tqx -1) : (-1) =- (1-ctq²x)/(1/ctqx -1) = -(1-ctqx)(1+ctqx)ctqx /(1- ctqx) =
-ctqx(1+ctqx).
ответ : -ctqx(ctqx+1).

a если ( 1-ctq²(-x) ) /(ctq(x -π) -1) : (ctq(3π -x)/ctq(π +x) = (1-ctq²x ) /(ctqx -1) : (-ctqx)/ctqx) = - (ctqx +1) : (-1) =ctqx +1.