Помогите с тождеством
tg(π-α)÷cos(π+α)×sin(3π/2+α)/tg(3π/2+α)=tg²α

Решениеtg(π-α)/cos(π+α)×sin(3π/2+α)/tg(3π/2+α) = tg²αtg(π-α)/cos(π+α)×sin(3π/2+α)/tg(3π/2+α) =[ - tgα / ( - cosα)*( - cosα)] / (- ctgα) = (sinx*sinx*cosx) / (cos^2x*cosx) = tg^2x