Patunayan ito: sqrt ((1-cosx) / (1 + cosx)) + sqrt ((1 + cosx) / (1-cosx)) = 2 / abs (sinx)?

Patunayan ito: sqrt ((1-cosx) / (1 + cosx)) + sqrt ((1 + cosx) / (1-cosx)) = 2 / abs (sinx)?
Anonim

Sagot:

Katunayan sa ibaba

gamit ang mga conjugate at trigonometriko na bersyon ng Pythagorean Theorem.

Paliwanag:

Bahagi 1

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

#color (white) ("XXX") = sqrt (1-cosx) / sqrt (1 + cosx) #

#color (white) ("XXX") = sqrt ((1-cosx)) / sqrt (1 + cosx) * sqrt (1-cosx) / sqrt (1-cosx)

#color (white) ("XXX") = (1-cosx) / sqrt (1-cos ^ 2x) #

Bahagi 2

Katulad nito

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

#color (white) ("XXX") = (1 + cosx) / sqrt (1-cos ^ 2x) #

Bahagi 3: Pinagsasama ang mga tuntunin

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

#color (white) ("XXX") = (1-cosx) / sqrt (1-cos ^ 2x) + (1 + cosx) / sqrt (1-cos ^ 2x) #

#color (white) ("XXX") = 2 / sqrt (1-cos ^ 2x) #

#color (white) ("XXXXXX") #at mula noon # sin ^ 2x + cos ^ 2x = 1 # (batay sa Pythagorean Teorama)

#color (white) ("XXXXXXXXX") sin ^ 2x = 1-cos ^ 2x #

#color (white) ("XXXXXXXXX") sqrt (1-cos ^ 2x) = abs (sinx) #

#sqrt ((1-cosx) / (1 + cosx)) + sqrt ((1 + cosx) / (1-cosx)) = 2 / sqrt (1-cos ^ 2x)