Matematika

Pertanyaan

buktikan bahwa
1) 2-sec²B/sec²B = 1-2 sin²B
2) sin²A - sin²B/cos²A cos²B = tan²A - tan²B

2 Jawaban

  • 2-sec2 B/sec2 B = 1-2sin2 B
    2/sec2 B - 1 = 2cos2 B - 1 = 2(1-sin2 B) - 1 = 2 - 2sin2 B - 1 = 1 - 2 sin2 B (Terbukti)

    sin2 A - sin2 B/cos2 A cos2 B = tan2 A - tan2 B
    sin2 A - cos2 A . sin2 B/cos2 B = sin2 A - cos2A . tan2 B = 1/cos2 A (sin2 A - cos2 A tan2 B = sin2 A/cos2 A - tan2 B = tan2 A - tan2 B (Terbukti)
  • (2 - sec^2 B)/(sec^2 B)
    = 2/(sec^2 B) - (sec^2 B)/(sec^2 B)
    = 2 cos^2 B - 1
    = 2(1 - sin^2 B) - 1
    = 2 - 2 sin^2 B - 1
    = 1 - 2 sin^2 B

    (sin^2 A - sin^2 B) / (cos^2 A cos^2 B)
    = (sin^2 A)/(cos^2 A . cos^2 B) - (sin^2 B)/(cos^2 B cos^2 A)
    = tan^2 A . sec^2 B - tan^2 B sec^2 B
    = tan^2 A (1 + tan^2 B) - tan^2 B(1 + tan^2 A)
    = tan^2 A + tan^2 A tan^2 B - tan^2 B - tan^2 A tan^2 B
    = tan^2 A - tan^2 B

    ^ = pangkat

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