Sederhanakan (1- cos theta + sin theta) / (1+ cos theta + sin theta)?

Sederhanakan (1- cos theta + sin theta) / (1+ cos theta + sin theta)?
Anonim

Menjawab:

# = sin (theta) / (1 + cos (theta)) #

Penjelasan:

# (1-cos (theta) + sin (theta)) / (1 + cos (theta) + sin (theta)) #

# = (1-cos (theta) + dosa (theta)) * (1 + cos (theta) + dosa (theta)) / (1 + cos (theta) + sin (theta)) ^ 2 #

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

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

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

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

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

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

# = (1/2) (1 + sin (theta)) / (1 + cos (theta)) - (1/2) ((1-sin (theta)) * (1 + sin (theta))) / ((1 + cos (theta)) (1 + sin (theta))) #

# = (1/2) (1 + sin (theta)) / (1 + cos (theta)) - (1/2) (1-sin (theta)) / (1 + cos (theta)) #

# = sin (theta) / (1 + cos (theta)) #