Sec thita -1 ÷ sec thita +1 = (sin thita ÷ 1+ costhita) ^ 2?

Odgovor:

Oglejte si spodnji dokaz

Potrebujemo

# sectheta = 1 / costheta #

# sin ^ 2theta + cos ^ 2theta = 1 #

Zato

# LHS = (sectheta-1) / (sectheta + 1) #

# = (1 / costheta-1) / (1 / costheta + 1) #

# = (1-costheta) / (1 + costheta) #

# = ((1-costheta) (1 + costheta)) / ((1 + costheta) (1 + costheta)) #

# = (1-cos ^ 2theta) / (1 + costheta) ^ 2 #

# sin ^ 2theta / (1 + costheta) ^ 2 #

# = (sintheta / (1 + costheta)) ^ 2 #

# = RHS #

# QED #

# LHS = (secx-1) / (secx + 1) #

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

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

# = (1-cos ^ 2x) / (1 + cosx) ^ 2 = sin ^ 2x / (1 + cosx) ^ 2 = (sinx / (1 + cosx)) ^ 2 = RHS #

Razlaga spodaj

# (secx-1) / (secx + 1) #

=# ((secx-1) * (secx + 1)) / (secx + 1) ^ 2 #

=# ((secx) ^ 2-1) / (secx + 1) ^ 2 #

=# (tanx) ^ 2 / (secx + 1) ^ 2 #

=# (sinx / cosx) ^ 2 / (1 / cosx + 1) ^ 2 #

=# ((sinx) ^ 2 / (cosx) ^ 2) / ((1 + cosx) ^ 2 / (cosx) ^ 2) #

=# (sinx) ^ 2 // (1 + cosx) ^ 2 #

=# (sinx / (1 + cosx)) ^ 2 #