Kako preverite (1 + tanx) / (sinx) = cscx + secx?

Odgovor:

Uporabite naslednja pravila:

# tanx = sinx / cosx #

# 1 / sinx = cscx #

# 1 / cosx = secx #

Pojasnilo:

Začnite z leve strani# ("LHS") #:

# => "LHS" = (1 + tanx) / sinx = 1 / sinx + tanx / sinx #

# = cscx + tanx xx1 / sinx = cscx + preklic (sinx) / cosx xx1 / preklic (sinx) = cscx + 1 / cosx = barva (modra) (cscx + secx) #

# QED #

Preverite secx • cscx + cotx = tanx + 2cosx • cscx?

RHS = tanx + 2cosx * cscx = sinx / cosx + (2cosx) / sinx = (sin ^ 2x + 2cos ^ 2x) / (sinx * cosx) = (sin ^ 2x + cos ^ 2x + cos ^ 2x) / (sinx * cosx) = (1 + cos ^ 2x) / (sinx * cosx) = 1 / (sinx * cosx) + (cos ^ 2x) / (sinx * cosx) = cscx * secx + cotx = LHS

Kako preverite (tan ^ 2x) / (secx-1) -1 = secx?

"Leva stran" = tan ^ 2x / (secx-1) -1 Uporabite identiteto: cos ^ 2x + sin ^ 2x = 1 => 1 + tan ^ 2x = sek ^ 2x => tan ^ 2x = sek ^ 2x -1 => "Leva stran" = (sec ^ 2x-1) / (secx-1) -1 = (prekliči ((secx-1)) (secx + 1)) / prekliči (secx-1) -1 => secx + 1-1 = barva (modra) secx = "desna stran"

Kako dokazati to identiteto? (cosxcotx-tanx) / cscx = cosx / secx-sinx / cotx

Identiteta mora biti resnična za katero koli število x, ki preprečuje delitev na nič. (cosxcotx-tanx) / cscx = {cos x (cos x / sin x) - sin x / cos x} / (1 / sin x) = cos ^ 2x - sin ^ 2 x / cos x = cos x / (1 / cos x) - sin x / (cos x / sin x) = cosx / secx-sinx / cotx