trig
posted by samuel r. edward on .
tan x = n tan y, sin x = m sin y then prove that m^2/n^21 = cos^2x

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 = 1sin^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  (1cos^2x)
n^2 cos^2 x = m^2  1 + cos^2 x
(n^21)cos^2 x = m^21
so
cos^2 x = (m^21)/(n^21)