Paano mo pinasimple (sec ^ 2x-1) / sin ^ 2x?

Paano mo pinasimple (sec ^ 2x-1) / sin ^ 2x?
Anonim

Sagot:

# (sec ^ 2 (x) -1) / sin ^ 2 (x) = sec ^ 2 (x) #

Paliwanag:

Una, i-convert ang lahat ng mga trigonometriko function sa #sin (x) # at #cos (x) #:

# (sec ^ 2 (x) -1) / sin ^ 2 (x) #

# = (1 / cos ^ 2 (x) -1) / sin ^ 2 (x) #

# = ((1-cos ^ 2 (x)) / cos ^ 2 (x)) / sin ^ 2 (x) #

Gamitin ang pagkakakilanlan # sin ^ 2 (x) + cos ^ 2 (x) = 1 #:

# = (sin ^ 2 (x) / cos ^ 2 (x)) / sin ^ 2 (x) #

Kinakansela ang # sin ^ 2 (x) # naroroon sa parehong numerator at denamineytor:

# = 1 / cos ^ 2 (x) #

# = sec ^ 2 (x) #

Sagot:

Ang sagot ay # sec ^ 2x #.

Paliwanag:

Alam namin na, # sec ^ 2x-1 = tan ^ 2x #

Samakatuwid,# (sec ^ 2x-1) / sin ^ 2x #

=# tan ^ 2x / sin ^ 2x #

=# sin ^ 2x / cos ^ 2x * 1 / sin ^ 2x #

=# 1 / cos ^ 2x #

=# sec ^ 2x #

Sagot:

# sec ^ 2x #

Paliwanag:

# "gamit ang" kulay (bughaw) "trigonometriko identities" #

# • kulay (puti) (x) secx = 1 / cosx #

# • kulay (puti) (x) sin ^ 2x + cos ^ 2x = 1 #

#rArr (1 / cos ^ 2x-cos ^ 2x / cos ^ 2x) / sin ^ 2x #

# = ((1-cos ^ 2x) / cos ^ 2x) / sin ^ 2x #

# = (sin ^ 2x / cos ^ 2x) / sin ^ 2x #

# = kanselahin (sin ^ 2x) / cos ^ 2x xx1 / kanselahin (sin ^ 2x) #

# = 1 / cos ^ 2x = sec ^ 2x #