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 preverite tan ^ 2θ- sin ^ 2θ = tan ^ 2θsin ^ 2θ?

Preverite razlago Žal mi je za moje pisanje;)

Kako preverite identiteto sec ^ 2 (x / 2) = (2secx + 2) / (secx + 2 + cosx)?

Potrebno je dokazati: sec ^ 2 (x / 2) = (2secx + 2) / (secx + 2 + cosx) "Desna stran" = (2secx + 2) / (secx + 2 + cosx) Ne pozabite, da je sekx = 1 / cosx => (2 * 1 / cosx + 2) / (1 / cosx + 2 + cosx) Zdaj, pomnožimo vrh in dno s cosx => (cosx xx (2 * 1 / cosx + 2)) / (cosx xx) (1 / cosx + 2 + cosx)) => (2 + 2cosx) / (1 + 2cosx + cos ^ 2x) Faktoriziraj dno, => (2 (1 + cosx)) / (1 + cosx) ^ 2 = > 2 / (1 + cosx) Recall identiteta: cos2x = 2cos ^ 2x-1 => 1 + cos2x = 2cos ^ 2x Podobno: 1 + cosx = 2cos ^ 2 (x / 2) => "Desna stran" = 2 / (2cos ^ 2 (x / 2)) = 1 / cos ^ 2 (x / 2) = barva (m

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

Uporabite naslednja pravila: tanx = sinx / cosx 1 / sinx = cscx 1 / cosx = secx 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