Kako kažeš tanx / tanx + sinx = 1/1 + cosx?

# LHS = tanx / (tanx + sinx) #

# = prekliči (tanx) / (prekliči (tanx) (1 + sinx / tanx)) #

# = 1 / (1 + sinx * cosx / sinx) = 1 / (1 + cosx) = RHS #

Odgovor:

Oglejte si razlago.

Pojasnilo:

Tukaj, # tanx / (tanx + sinx) = 1 / (1 + cosx) #

# LHS = tanx / (tanx + sinx) = (sinx / cosx) / (sinx / cosx + sinx #)

# LHS = sinx / (sinx + sinxcosx) #

# = sinx / (sinx (1 + cosx)) #

# = 1 / (1 + cosx) #

# = RHS #

Kako dokazati (1 + sinx-cosx) / (1 + cosx + sinx) = tan (x / 2)?

Glej spodaj. LHS = (1-cosx + sinx) / (1 + cosx + sinx) = (2sin ^ 2 (x / 2) + 2sin (x / 2) * cos (x / 2)) / (2cos ^ 2 (x / 2) + 2sin (x / 2) * cos (x / 2) = (2sin (x / 2) [sin (x / 2) + cos (x / 2)]) / (2cos (x / 2) * [ sin (x / 2) + cos (x / 2)]) = tan (x / 2) = RHS

Dokaži: sqrt ((1-cosx) / (1 + cosx)) + sqrt ((1 + cosx) / (1-cosx)) = 2 / abs (sinx)?

Dokaz spodaj z uporabo konjugatov in trigonometrične različice Pitagorejeve teoreme. Barva dela 1 ((1-cosx) / (1 + cosx)) (bela) ("XXX") = sqrt (1-cosx) / sqrt (1 + cosx) barva (bela) ("XXX") = sqrt ((1-cosx)) / sqrt (1 + cosx) * sqrt (1-cosx) / sqrt (1-cosx) barva (bela) ("XXX") = (1-cosx) / sqrt (1-cos ^ 2x) 2. del Podobno sqrt ((1 + cosx) / (1-cosx) barva (bela) ("XXX") = (1 + cosx) / sqrt (1-cos ^ 2x) 3. del: Združevanje izrazov sqrt ( (1-cosx) / (1 + cosx)) + sqrt ((1 + cosx) / (1-cosx) barva (bela) ("XXX") = (1-cosx) / sqrt (1-cos ^ 2x) + (1 + cosx) / sqrt (1-cos ^ 2x

Kako dokazujete (cosx / (1 + sinx)) + ((1 + sinx) / cosx) = 2secx?

Pretvorite levo stran v izraze s skupnim imenovalcem in dodajte (pretvorite cos ^ 2 + sin ^ 2 v 1 na poti); poenostavimo in se sklicujemo na definicijo sek = 1 / cos (cos (x) / (1 + sin (x))) + ((1 + sin (x)) / cos (x)) = (cos ^ 2 (x) + 1 + 2sin (x) + sin ^ 2 (x)) / (cos (x) (1 + sin (x) = (2 + 2sin (x)) / (cos (x) (1 + sin (x)) ) = 2 / cos (x) = 2 * 1 / cos (x) = 2sek (x)