Vprašanje # c3e29

Vprašanje # c3e29
Anonim

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

Zdaj

cscA + posteljica A = (csc ^ 2A-cot ^ 2A) / (cscA + cotA) cscA+posteljicaA=csc2Acot2AcscA+cotA

=> cscA + posteljica A = x …… (2) cscA+posteljicaA=x(2)

Dodamo (1) in (2)

2cscx = x + 1 / x 2cscx=x+1x

=> cscx = 1/2 (x + 1 / x) = 1/2 (x ^ 2 + 1) / x cscx=12(x+1x)=12x2+1x

Odštejemo (1) od (2)

2cotA = x-1 / x 2cotA=x1x

cotA = 1/2 (x-1 / x) = 1/2 (x ^ 2-1) / x cotA=12(x1x)=12x21x

Zdaj

sec A = cscA / cotA = (x ^ 2 + 1) / (x ^ 2 - 1) secA=cscAcotA=x2+1x21

Odgovor:

Glej spodaj.

Pojasnilo:

Let cscA-cotA = 1 / x cscAcotA=1x…….1

To vemo, rarrcsc ^ 2A-cot ^ 2A = 1 csc2Acot2A=1

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

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

rarrcscA + cotA = x cscA+cotA=x….2

Dodajanje enačb 1 in 2,

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

rarr2cscA = (x ^ 2 + 1) / x 2cscA=x2+1x…..3

Odvzemanje enačbe 1 iz 2, rarrcscA + cotA- (cscA-cotA) = x-1 / x cscA+cotA(cscAcotA)=x1x

rarrcscA + cotA-cscA + cotA = (x ^ 2-1) / x cscA+cotAcscA+cotA=x21x

rarr2cotA = (x ^ 2-1) / x 2cotA=x21x…….4

Delitev enačbe 3 z 4, rarr (2cscA) / (2cotA) = ((x ^ 2 + 1) / x) / ((x ^ 2-1) / x) 2cscA2cotA=x2+1xx21x

rarr (1 / sinA) / (cosA / sinA) = (x ^ 2 + 1) / (x ^ 2-1) 1sinAcosAsinA=x2+1x21

rarrsecA = (x ^ 2 + 1) / (x ^ 2-1) secA=x2+1x21 Dokazano …

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