Com es verifica la identitat 3sec ^ 2thetatan ^ 2theta + 1 = sec ^ 6theta-tan ^ 6theta?

Resposta:

Mirar abaix

Explicació:

# 3sec ^ 2thetatan ^ 2theta + 1 = sec ^ 6theta-tan ^ 6theta #

Costat dret# = sec ^ 6theta-tan ^ 6theta #

# = (seg ^ 2theta) ^ 3- (tan ^ 2theta) ^ 3 #-> utilitzeu la fórmula de diferència de dos cubs

# = (sec ^ 2theta-tan ^ 2theta) (sec ^ 4theta + sec ^ 2thetatan ^ 2theta + tan ^ 4theta) #

# = 1 * (seg. ^ 4a + sec ^ 2thetatan ^ 2theta + tan ^ 4tha) #

# = sec ^ 4theta + sec ^ 2thetatan ^ 2theta + tan ^ 4theta #

# = sec ^ 2theta sec ^ 2 theta + sec ^ 2thetatan ^ 2theta + tan ^ 2theta tan ^ 2 theta #

# = sec ^ 2theta (tan ^ 2theta + 1) + sec ^ 2thetatan ^ 2theta + tan ^ 2theta (seg ^ 2theta-1) #

# = sec ^ 2thetatan ^ 2theta + sec ^ 2theta + sec ^ 2thetatan ^ 2theta + sec ^ 2thetatan ^ 2theta-tan ^ 2theta #

# = sec ^ 2thetatan ^ 2theta + sec ^ 2thetatan ^ 2theta + sec ^ 2thetatan ^ 2theta + sec ^ 2theta-tan ^ 2theta #

# = 3sec ^ 2thetatan ^ 2theta + 1 #

#=# Costat esquerre