Vprašanje # c3e29

Glede na #csc A - posteljica A = 1 / x.. (1) #

Zdaj

# cscA + posteljica A = (csc ^ 2A-cot ^ 2A) / (cscA + cotA) #

# => cscA + posteljica A = x …… (2) #

Dodamo (1) in (2)

# 2cscx = x + 1 / x #

# => cscx = 1/2 (x + 1 / x) = 1/2 (x ^ 2 + 1) / x #

Odštejemo (1) od (2)

# 2cotA = x-1 / x #

# cotA = 1/2 (x-1 / x) = 1/2 (x ^ 2-1) / x #

Zdaj

#sec A = cscA / cotA = (x ^ 2 + 1) / (x ^ 2 - 1) #

Odgovor:

Glej spodaj.

Pojasnilo:

Let # cscA-cotA = 1 / x #…….1

To vemo, # rarrcsc ^ 2A-cot ^ 2A = 1 #

#rarr (cscA-cotA) * (cscA + cotA) = 1 #

# rarr1 / x (cscA + cotA) = 1 #

# rarrcscA + cotA = x #….2

Dodajanje enačb 1 in 2,

# rarrcscA-cotA + cscA + cotA = 1 / x + x #

# rarr2cscA = (x ^ 2 + 1) / x #…..3

Odvzemanje enačbe 1 iz 2, # rarrcscA + cotA- (cscA-cotA) = x-1 / x #

# rarrcscA + cotA-cscA + cotA = (x ^ 2-1) / x #

# rarr2cotA = (x ^ 2-1) / x #…….4

Delitev enačbe 3 z 4, #rarr (2cscA) / (2cotA) = ((x ^ 2 + 1) / x) / ((x ^ 2-1) / x) #

#rarr (1 / sinA) / (cosA / sinA) = (x ^ 2 + 1) / (x ^ 2-1) #

# rarrsecA = (x ^ 2 + 1) / (x ^ 2-1) # Dokazano …

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