trig

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tan x = n tan y, sin x = m sin y then prove that m^2-/n^2-1 = cos^2x

  • trig -

    You have a typo which makes it kind of confusing.
    tan x = n tan y
    so
    sin x/cos x = n sin y/cos y
    n = (cos y/cos x)(sin x/sin y)
    but
    m = (sin x/sin y)
    sin y = (1/m) sin x
    cos^2 y = 1-sin^2y = 1-(1/m^2) sin^2x
    so
    n = (cos y / cos x)m
    n^2 = (m^2)(cos^2 y/cos^2 x)
    n^2 = (m^2/cos^2x) [ 1-(1/m^2)sin^2 x]
    n^2 = m^2/cos^2x - sin^2x/cos^2x
    n^2 cos^2x = m^2 - (1-cos^2x)
    n^2 cos^2 x = m^2 - 1 + cos^2 x
    (n^2-1)cos^2 x = m^2-1
    so
    cos^2 x = (m^2-1)/(n^2-1)

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