Paano mo patunayan (cotx + cscx / sinx + tanx) = (cotx) (cscx)?

Sagot:

Na-verify sa ibaba

Paliwanag:

# (cotx + cscx) / (sinx + tanx) = (cotx) (cscx) #

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

# ((cosx + 1) / sinx) / ((sinxcosx) / cosx + sinx / cosx) = (cotx) (cscx) #

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

# (kanselahin (cosx + 1) / sinx) * (cosx / (sinxcancel ((cosx + 1)))) = (cotx) (cscx) #

# (cosx / sinx * 1 / sinx) = (cotx) (cscx) #

# (cotx) (cscx) = (cotx) (cscx) #

Sinusubukan naming patunayan iyon # (cotx + cscx) / (sinx + tanx) = cotxcscx #. Narito ang mga pagkakakilanlan na kakailanganin mo:

# tanx = sinx / cosx #

# cotx = cosx / sinx #

# cscx = 1 / sinx #

Magsisimula ako sa kaliwang bahagi at manipulahin ito hanggang sa ito ay katumbas ng kanang bahagi:

#color (white) = (cotx + cscx) / (sinx + tanx) #

# = (qquadcosx / sinx + 1 / sinxqquad) / (qquadsinx / 1 + sinx / cosxqquad) #

# = (qquad (cosx + 1) / sinxqquad) / (qquad (sinxcosx) / cosx + sinx / cosxqquad) #

# = (qquad (cosx + 1) / sinxqquad) / (qquad (sinxcosx + sinx) / cosxqquad) #

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

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

# = (cosx (cosx + 1)) / (sin ^ 2x (cosx + 1)) #

# = (cosxcolor (red) cancelcolor (black) ((cosx + 1))) / (sin ^ 2xcolor (red) cancelcolor (black) ((cosx + 1)

# = cosx / sin ^ 2x #

# = cosx / sinx * 1 / sinx #

# = cotx * cscx #

# = RHS #

Iyan ang patunay. Sana nakakatulong ito!

# LHS = (cotx + cscx) / (sinx + tanx) #

# = (cotx + cscx) / (sinx + tanx) * ((cotx * cscx) / (cotx * cscx)) #

# = cotx * cscx (cotx + cscx) / ((sinx + tanx) * cotx * cscx) #

# = cotx * cscx (cotx + cscx) / ((sinx * cscx * cotx + tanx * cotx * cscx)) #

# = cotx * cscxcancel ((cotx + cscx) / (cotx + cscx)) = cotx * cscx = RHS #